Introduction
The theory of the Diophantine tuple started with the observation of Diophantus, who noted that the product of any two distinct elements of the set { 1 16 , 33 16 , 17 4 , 105 16 } \left\{\frac{1}{16},\frac{33}{16},\frac{17}{4},\frac{105}{16}\right\} { 16 1 , 16 33 , 4 17 , 16 105 } { 1 , 3 , 8 , 120 } \{1,3,8,120\} { 1 , 3 , 8 , 120 }
Definition 1.1. Let k k k m m m D ( k ) D(k) D ( k ) m m m k k k
When m = 2 , 3 , 4 m = 2, 3, 4 m = 2 , 3 , 4 D ( k ) D(k) D ( k ) [8 ] proved that if an integer k ≢ 2 ( m o d 4 ) k \not\equiv2 \pmod{4} k ≡ 2 ( mod 4 ) k ∉ { − 4 , − 3 , − 1 , 3 , 5 , 8 , 12 , 20 } k \notin\{-4,-3,-1,3,5,8,12,20\} k ∈ / { − 4 , − 3 , − 1 , 3 , 5 , 8 , 12 , 20 } D ( k ) D(k) D ( k )
Conjecture 1.2 (Dujella). If k ∈ { − 4 , − 3 , − 1 , 3 , 5 , 8 , 12 , 20 } k \in\{-4,-3,-1,3,5,8,12,20\} k ∈ { − 4 , − 3 , − 1 , 3 , 5 , 8 , 12 , 20 } D ( k ) D(k) D ( k )
He, Togbé and Ziegler [17 ] proved the folklore conjecture of non-existence of quintuple when k = 1 k = 1 k = 1 [1 ] proved the conjecture of non-existence quadruple when k = − 1 , − 4 k = -1,-4 k = − 1 , − 4 [2 ] proved that there do not exist D ( 4 ) D(4) D ( 4 ) { F n : n ≥ 0 } \{F_n : n \ge0\} { F n : n ≥ 0 }
MSC2020: 11B37, 11D09, 11J86 defined by F 0 = 0 F_0 = 0 F 0 = 0 F 1 = 1 F_1 = 1 F 1 = 1 F n + 2 = F n + 1 + F n F_{n+2} = F_{n+1} + F_n F n + 2 = F n + 1 + F n n ≥ 0 n \ge0 n ≥ 0 { P n : n ≥ 0 } \{P_n : n \ge0\} { P n : n ≥ 0 }
P n = { 0 if n = 0 , 1 if n = 1 , 2 P n − 1 + P n − 2 if n ≥ 2. P_n =
\begin{cases}
0 & \text{if } n = 0,\\
1 & \text{if } n = 1,\\
2P_{n-1} + P_{n-2} & \text{if } n \ge2.
\end{cases} P n = ⎩ ⎨ ⎧ 0 1 2 P n − 1 + P n − 2 if n = 0 , if n = 1 , if n ≥ 2. Diophantine triples and quadruples involving Fibonacci and Pell numbers have been studied in [15 , 18 , 22 , 23 ] . In particular, the non-existence of Diophantine quadruples consisting of Fibonacci or Pell numbers was established in [15 , 23 ] . A Triangular number is a number of the form T n = n ( n + 1 ) 2 T_n = \frac{n(n+1)}{2} T n = 2 n ( n + 1 ) n ≥ 1 n \ge1 n ≥ 1 m m m [6 ] . Subsequent work by Bliznac Trebješanin [3 , 4 ] addressed the properties D ( u 2 ) D(u^2) D ( u 2 ) D ( a ) D(a) D ( a ) [5 , 24 , 16 ] . More than 600 papers have been published on this topic. The latest progress and all references can be found at Dujella’s webpage [9 ] .
Motivated by these developments involving Fibonacci, Pell, triangular numbers, in this paper, we consider the Diophantine tuple problem in the setting of star numbers. The n n n S n S_n S n S n = 6 n ( n − 1 ) + 1 S_n = 6n(n - 1) + 1 S n = 6 n ( n − 1 ) + 1 n ≥ 1 n \ge1 n ≥ 1 S n = 1 + 12 T n − 1 S_n = 1 + 12T_{n-1} S n = 1 + 12 T n − 1
The following proposition establishes the form of admissible integers k ≠ 0 k \ne0 k = 0 { S a , S b } \{S_a, S_b\} { S a , S b } D ( k ) D(k) D ( k )
Proposition 1.3. If { S a , S b } \{S_a, S_b\} { S a , S b } D ( k ) D(k) D ( k ) k ≡ 0 , 3 , 8 , 11 ( m o d 12 ) k \equiv0, 3, 8, 11 \pmod{12} k ≡ 0 , 3 , 8 , 11 ( mod 12 )
Proof. Since { S a , S b } \{S_a, S_b\} { S a , S b } D ( k ) D(k) D ( k ) m m m S a S b + k = m 2 S_aS_b + k = m^2 S a S b + k = m 2 m 2 ≡ 0 , 1 , 4 , 9 ( m o d 12 ) m^2 \equiv0, 1, 4, 9 \pmod{12} m 2 ≡ 0 , 1 , 4 , 9 ( mod 12 ) S a , S b ≡ 1 ( m o d 12 ) S_a, S_b \equiv1 \pmod{12} S a , S b ≡ 1 ( mod 12 ) k ≡ 0 , 3 , 8 , 11 ( m o d 12 ) k \equiv0, 3, 8, 11 \pmod{12} k ≡ 0 , 3 , 8 , 11 ( mod 12 )
In this paper, we mainly consider certain admissible values of k k k k ∈ { − 12 , − 9 , − 4 , − 1 , 3 , 8 , 11 , 12 } k \in\{-12, -9, -4, -1, 3, 8, 11, 12\} k ∈ { − 12 , − 9 , − 4 , − 1 , 3 , 8 , 11 , 12 } C ± = S a + S b ± 2 S a S b + k C_\pm= S_a + S_b \pm2\sqrt{S_aS_b + k} C ± = S a + S b ± 2 S a S b + k { S a , S b , C ± } \{S_a, S_b, C_\pm\} { S a , S b , C ± } D ( k ) D(k) D ( k ) { S a , S b } \{S_a, S_b\} { S a , S b } S c ± = C ± S_{c_\pm} = C_\pm S c ± = C ±
S a S b + k ≡ k + 1 ( m o d 6 ) , S_aS_b + k \equiv k + 1 \pmod{6}, S a S b + k ≡ k + 1 ( mod 6 ) , which gives
S a S b + k ( m o d 6 ) = { 1 , 5 k = − 12 , 12 2 , 4 k = − 9 , 3 , 3 k = − 4 , 8 , 0 k = − 1 , 11. \sqrt{S_aS_b + k} \pmod{6} =
\begin{cases}
1, 5 & k = -12, 12\\
2, 4 & k = -9, 3,\\
3 & k = -4, 8,\\
0 & k = -1, 11.
\end{cases} S a S b + k ( mod 6 ) = ⎩ ⎨ ⎧ 1 , 5 2 , 4 3 0 k = − 12 , 12 k = − 9 , 3 , k = − 4 , 8 , k = − 1 , 11. Therefore,
1 = S c ± ( m o d 6 ) = { 4 , 0 k = − 12 , 12 0 , 4 k = − 9 , 3 , 2 k = − 4 , 8 , 2 k = − 1 , 11 , 1 = S_{c_\pm} \pmod{6} =
\begin{cases}
4, 0 & k = -12, 12\\
0, 4 & k = -9, 3,\\
2 & k = -4, 8,\\
2 & k = -1, 11,
\end{cases} 1 = S c ± ( mod 6 ) = ⎩ ⎨ ⎧ 4 , 0 0 , 4 2 2 k = − 12 , 12 k = − 9 , 3 , k = − 4 , 8 , k = − 1 , 11 , and
1 = S c − ( m o d 6 ) = { 0 , 4 k = − 12 , 12 , 4 , 0 k = − 9 , 3 , 2 k = − 4 , 8 , 2 k = − 1 , 11 , 1 = S_{c-} \pmod{6} =
\begin{cases}
0, 4 & k = -12, 12,\\
4, 0 & k = -9, 3,\\
2 & k = -4, 8,\\
2 & k = -1, 11,
\end{cases} 1 = S c − ( mod 6 ) = ⎩ ⎨ ⎧ 0 , 4 4 , 0 2 2 k = − 12 , 12 , k = − 9 , 3 , k = − 4 , 8 , k = − 1 , 11 , which is a contradiction.
Our first theorem in this direction establishes the existence and nonexistence of pairs { S a , S b } \{S_a, S_b\} { S a , S b }
Theorem 1.4. For k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 k = 3, 8, 12, -1, -4, -9, -12 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 D ( k ) D(k) D ( k ) D ( 11 ) D(11) D ( 11 )
The next theorem produces necessary lower bounds for the index b b b a a a D ( k ) D(k) D ( k )
Theorem 1.5. Let 2 ≤ a < b 2 \le a < b 2 ≤ a < b { S a , S b } \{S_a, S_b\} { S a , S b } D ( k ) D(k) D ( k )
(1) b = a + 1 b = a + 1 b = a + 1 b ≥ 4 a + 2 b \ge4a + 2 b ≥ 4 a + 2 k = 3 , k = 3, k = 3 ,
(2) b ≥ 4 a + 2 b \ge4a + 2 b ≥ 4 a + 2 k = 8 , − 1 , − 9 , − 12 , k = 8, -1, -9, -12, k = 8 , − 1 , − 9 , − 12 ,
(3) b = a + 2 b = a + 2 b = a + 2 b ≥ 4 a + 2 b \ge4a + 2 b ≥ 4 a + 2 k = 12 , k = 12, k = 12 ,
(4) a = 2 , b = 9 a = 2, b = 9 a = 2 , b = 9 b ≥ 4 a + 2 b \ge4a + 2 b ≥ 4 a + 2 k = − 4 k = -4 k = − 4
In the next theorem, we use the above theorem to get infinitely many D ( k ) D(k) D ( k ) S 1 S_1 S 1 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 k = 3, 8, 12, -1, -4, -9, -12 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12
Theorem 1.6. For k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 k = 3, 8, 12, -1, -4, -9, -12 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 D ( k ) D(k) D ( k ) S 1 S_1 S 1
Our next result establishes the non-existence of Diophantine D ( k ) D(k) D ( k ) k ≠ 0 k \ne0 k = 0
Theorem 1.7. If k ≡ 12 , 20 ( m o d 24 ) k \equiv12, 20 \pmod{24} k ≡ 12 , 20 ( mod 24 ) k ≡ 23 , 32 ( m o d 36 ) k \equiv23, 32 \pmod{36} k ≡ 23 , 32 ( mod 36 ) { S a , S b , S c } \{S_a, S_b, S_c\} { S a , S b , S c } D ( k ) D(k) D ( k )
As an immediate consequence, we have the following corollary.
Corollary 1.8. For k = − 12 , − 4 , 12 k = -12, -4, 12 k = − 12 , − 4 , 12 a , b , c a, b, c a , b , c { S a , S b , S c } \{S_a, S_b, S_c\} { S a , S b , S c } D ( k ) D(k) D ( k )
In view of the computation evidence, we strongly believe that the following conjecture is true:
Conjecture 1.10. For k = 3 , 8 , − 9 , − 1 k = 3, 8, -9, -1 k = 3 , 8 , − 9 , − 1 D ( k ) D(k) D ( k ) { S a , S b , S c } \{S_a, S_b, S_c\} { S a , S b , S c }
Proof. If possible, assume that Conjecture 1.2 holds, but Conjecture 1.10 is false for k = 3 k = 3 k = 3 a , b , c ≥ 1 a, b, c \ge1 a , b , c ≥ 1 { S a , S b , S c } \{S_a, S_b, S_c\} { S a , S b , S c } D ( 3 ) D(3) D ( 3 ) 6 6 6 { 6 , S a , S b , S c } \{6, S_a, S_b, S_c\} { 6 , S a , S b , S c } D ( 3 ) D(3) D ( 3 ) Conjecture 1.2 . This completes the proof. □
:::
However, there are several values of k k k { S a , S b , S c } \{S_a,S_b,S_c\} { S a , S b , S c } D ( k ) D(k) D ( k ) ( a , b , c , k ) ∈ { ( 1 , 2 , 57 , 1011 ) , ( 1 , 3 , 126 , 363 ) , ( 1 , 6 , 65 , 1935 ) , ( 1 , 10 , 362 , 903 ) } (a,b,c,k) \in\{(1,2,57,1011),(1,3,126,363),(1,6,65,1935),(1,10,362,903)\} ( a , b , c , k ) ∈ {( 1 , 2 , 57 , 1011 ) , ( 1 , 3 , 126 , 363 ) , ( 1 , 6 , 65 , 1935 ) , ( 1 , 10 , 362 , 903 )}
From Theorem 1.5 , it is easy to prove that for a ≥ 2 a \ge2 a ≥ 2 { S a , S a + 1 } \{S_a,S_{a+1}\} { S a , S a + 1 } { S a , S a + 2 } \{S_a,S_{a+2}\} { S a , S a + 2 } D ( 3 ) D(3) D ( 3 ) D ( 12 ) D(12) D ( 12 ) Theorem 1.6 ). Corollary 1.8 proves that { S a , S a + 2 } \{S_a,S_{a+2}\} { S a , S a + 2 } D ( 12 ) D(12) D ( 12 ) Conjecture 1.10 .
Theorem 1.12. Let a > b ≥ 1 a>b\ge1 a > b ≥ 1 { S b , S a , S a + 1 } \{S_b,S_a,S_{a+1}\} { S b , S a , S a + 1 } D ( 3 ) D(3) D ( 3 )
In particular, { S a − 1 , S a , S a + 1 } \{S_{a-1},S_a,S_{a+1}\} { S a − 1 , S a , S a + 1 } D ( 3 ) D(3) D ( 3 )
On the other hand, an analogous problem involving classical Diophantine tuples has also been studied extensively in the setting of triangular numbers. The definition of a triangular Diophantine D ( k ) D(k) D ( k )
Definition 1.13. Let k k k m m m D ( k ) D(k) D ( k ) m m m k k k
In this set up, Filipin and Szalay [13 ] studied triangular Diophantine tuples containing { 1 , 2 } \{1,2\} { 1 , 2 } D ( 1 ) D(1) D ( 1 ) { 1 , 2 , a , b } \{1,2,a,b\} { 1 , 2 , a , b } [14 ] studied the extension of the triangular D ( 3 ) D(3) D ( 3 ) { 3 , 6 } \{3,6\} { 3 , 6 } { 3 , 6 , a , b } \{3,6,a,b\} { 3 , 6 , a , b } [12 ] studied the triangular D ( k ) D(k) D ( k ) { k , 2 k } \{k,2k\} { k , 2 k } k = 10 k=10 k = 10 k = 28 k=28 k = 28 { k , 2 k , a , b } \{k,2k,a,b\} { k , 2 k , a , b } k k k [19 , 11 ] .
Motivated by these developments in the study of triangular Diophantine D ( k ) D(k) D ( k )
Definition 1.14. Let k k k m m m D ( k ) D(k) D ( k ) m m m k k k
In this direction, our main theorems are stated below.
Theorem 1.15. For all
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 1 2 : k ≥ 1 , ξ 1 , 0 = 17 , ξ 2 , 0 = 7 } , a \in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-1}{2}: k\ge1,\ \xi_{1,0}=17,\xi_{2,0}=7\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 1 : k ≥ 1 , ξ 1 , 0 = 17 , ξ 2 , 0 = 7 } , where
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}, ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , the set { a , a + 1 , 12 ( 2 a + 1 ) } \{a,a+1,12(2a+1)\} { a , a + 1 , 12 ( 2 a + 1 )} D ( 1 ) D(1) D ( 1 ) 12 ∤ b 12\nmid b 12 ∤ b { a , a + 1 , 12 ( 2 a + 1 ) , b } \{a,a+1,12(2a+1),b\} { a , a + 1 , 12 ( 2 a + 1 ) , b } D ( 1 ) D(1) D ( 1 )
Theorem 1.16. For all
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 2 2 : k ≥ 1 , ξ 1 , 0 = 22 , ξ 2 , 0 = 9 } , a \in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-2}{2}: k\ge1,\ \xi_{1,0}=22,\xi_{2,0}=9\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 2 : k ≥ 1 , ξ 1 , 0 = 22 , ξ 2 , 0 = 9 } , where
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}, ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , the set { a , a + 2 , 6 ( a + 1 ) } \{a,a+2,6(a+1)\} { a , a + 2 , 6 ( a + 1 )} D ( 1 ) D(1) D ( 1 ) 6 ∤ b 6\nmid b 6 ∤ b { a , a + 2 , 6 ( a + 1 ) , b } \{a,a+2,6(a+1),b\} { a , a + 2 , 6 ( a + 1 ) , b } D ( 1 ) D(1) D ( 1 )
Theorem 1.17. For all
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 3 2 : k ≥ 1 , ξ 1 , 0 = 27 , ξ 2 , 0 = 11 } , a \in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-3}{2}: k \ge1,\ \xi_{1,0}=27,\ \xi_{2,0}=11\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 3 : k ≥ 1 , ξ 1 , 0 = 27 , ξ 2 , 0 = 11 } , where
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}, ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , the set { a , a + 3 , 8 a + 12 3 } \{a,a+3,\frac{8a+12}{3}\} { a , a + 3 , 3 8 a + 12 } D ( 1 ) D(1) D ( 1 ) 4 ∤ b 4 \nmid b 4 ∤ b { a , a + 3 , 8 a + 12 3 , b } \{a,a+3,\frac{8a+12}{3},b\} { a , a + 3 , 3 8 a + 12 , b } D ( 1 ) D(1) D ( 1 )
Theorem 1.18. For all
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 4 2 : k ≥ 1 , ξ 1 , 0 = 8 , ξ 2 , 0 = 3 } , a \in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-4}{2}: k \ge1,\ \xi_{1,0}=8,\ \xi_{2,0}=3\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 4 : k ≥ 1 , ξ 1 , 0 = 8 , ξ 2 , 0 = 3 } , where
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}, ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , the set { a , a + 4 , 3 a + 6 2 } \{a,a+4,\frac{3a+6}{2}\} { a , a + 4 , 2 3 a + 6 } D ( 1 ) D(1) D ( 1 ) 3 ∤ b 3 \nmid b 3 ∤ b { a , a + 4 , 3 a + 6 2 , b } \{a,a+4,\frac{3a+6}{2},b\} { a , a + 4 , 2 3 a + 6 , b } D ( 1 ) D(1) D ( 1 )
It is easy to check that the set { 4 , 108 } \{4,108\} { 4 , 108 } D ( 1 ) D(1) D ( 1 ) { 4 , 108 } \{4,108\} { 4 , 108 } D ( 1 ) D(1) D ( 1 )
Theorem 1.19. For all
a ∈ { 3 ( ξ 2 , k 2 − 1 ) 8 : k ≥ 1 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } ∪ { 3 ( ξ 2 , k 2 − 1 ) 8 : k ≥ 1 , ξ 1 , 0 = 1 , ξ 2 , 0 = 1 } , a \in\left\{\frac{3\left(\xi_{2,k}^{2}-1\right)}{8}: k \ge1,\xi_{1,0}=-1,\xi_{2,0}=1\right\}\cup\left\{\frac{3\left(\xi_{2,k}^{2}-1\right)}{8}: k \ge1,\xi_{1,0}=1,\xi_{2,0}=1\right\}, a ∈ ⎩ ⎨ ⎧ 8 3 ( ξ 2 , k 2 − 1 ) : k ≥ 1 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 ⎭ ⎬ ⎫ ∪ ⎩ ⎨ ⎧ 8 3 ( ξ 2 , k 2 − 1 ) : k ≥ 1 , ξ 1 , 0 = 1 , ξ 2 , 0 = 1 ⎭ ⎬ ⎫ , where
ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 , \xi_{1,k}=26\xi_{1,k-1}+135\xi_{2,k-1},\qquad\xi_{2,k}=5\xi_{1,k-1}+26\xi_{2,k-1}, ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 , the set { 4 , 108 , a } \{4,108,a\} { 4 , 108 , a } D ( 1 ) D(1) D ( 1 )
From the above theorem, it is clear that { 4 , 108 , 165 } \{4,108,165\} { 4 , 108 , 165 } D ( 1 ) D(1) D ( 1 ) D ( 1 ) D(1) D ( 1 )
Theorem 1.20. Let { 4 , 108 , 165 , a } \{4,108,165,a\} { 4 , 108 , 165 , a } D ( 1 ) D(1) D ( 1 ) a = 360 a=360 a = 360
As an immediate consequence, we obtain the following corollary.
Corollary 1.21. There does not exist a star Diophantine D ( 1 ) D(1) D ( 1 ) { 4 , 108 , 165 , a , b } \{4,108,165,a,b\} { 4 , 108 , 165 , a , b }
In view of the computation evidence and results obtained in this article, we strongly believe that the following conjecture is true:
Conjecture 1.22. There exists no star Diophantine D ( 1 ) D(1) D ( 1 )
We also pose the following open question.
Question 1.23. Does there exist a star Diophantine D ( 1 ) D(1) D ( 1 ) { 4 , 108 , 165 , 360 } \{4,108,165,360\} { 4 , 108 , 165 , 360 }
Preliminaries
In this section, we prove several auxiliary lemmas and recall a few useful results, which will play a pivotal role in the proof of our main theorems.
Lemma 2.1. Let a , b , c , n a, b, c, n a , b , c , n b > a ≥ 1 b>a\ge1 b > a ≥ 1 S a S b + n = c 2 S_aS_b+n=c^2 S a S b + n = c 2
X 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) Y 2 = 4 n − 3 ( 2 a − 1 ) 2 + 1 , (1) X^2-\left(9(2a-1)^2-3\right)Y^2=4n-3(2a-1)^2+1, \tag*{(1)} X 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) Y 2 = 4 n − 3 ( 2 a − 1 ) 2 + 1 , (1) where X = 2 c X=2c X = 2 c Y = 2 b − 1 Y=2b-1 Y = 2 b − 1
Proof. Observe that
S a = 6 a ( a − 1 ) + 1 = 3 ( 2 a − 1 ) 2 − 1 2 , S b = 6 b ( b − 1 ) + 1 = 3 ( 2 b − 1 ) 2 − 1 2 . S_a=6a(a-1)+1=\frac{3(2a-1)^2-1}{2},\qquad S_b=6b(b-1)+1=\frac{3(2b-1)^2-1}{2}. S a = 6 a ( a − 1 ) + 1 = 2 3 ( 2 a − 1 ) 2 − 1 , S b = 6 b ( b − 1 ) + 1 = 2 3 ( 2 b − 1 ) 2 − 1 . Since S a S b + n = c 2 S_aS_b+n=c^2 S a S b + n = c 2
( 3 ( 2 a − 1 ) 2 − 1 ) ( 3 ( 2 b − 1 ) 2 − 1 ) 4 + n = c 2 , \frac{\left(3(2a-1)^2-1\right)\left(3(2b-1)^2-1\right)}{4}+n=c^2, 4 ( 3 ( 2 a − 1 ) 2 − 1 ) ( 3 ( 2 b − 1 ) 2 − 1 ) + n = c 2 , which gives
( 2 c ) 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) ( 2 b − 1 ) 2 = 4 n − 3 ( 2 a − 1 ) 2 + 1. (2c)^2-\left(9(2a-1)^2-3\right)(2b-1)^2=4n-3(2a-1)^2+1. ( 2 c ) 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) ( 2 b − 1 ) 2 = 4 n − 3 ( 2 a − 1 ) 2 + 1. Lemma 2.2. Under the assumptions in Lemma 2.1, for fixed a ≥ 1 a\ge1 a ≥ 1 n n n ( x , y ) (x,y) ( x , y ) (1) , then S a S b + n = c 2 S_aS_b+n=c^2 S a S b + n = c 2
b ∈ { ξ 2 , k + 1 2 : k ≥ 0 } , b\in\left\{\frac{\xi_{2,k}+1}{2}:k\ge0\right\}, b ∈ { 2 ξ 2 , k + 1 : k ≥ 0 } , where ξ 1 , 0 = x , ξ 2 , 0 = y \xi_{1,0}=x,\xi_{2,0}=y ξ 1 , 0 = x , ξ 2 , 0 = y k ≥ 1 k\ge1 k ≥ 1
ξ 1 , k = ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 1 , k − 1 + 2 ( 2 a − 1 ) ( 9 ( 2 a − 1 ) 2 − 3 ) ξ 2 , k − 1 , \xi_{1,k}=\left(6(2a-1)^2-1\right)\xi_{1,k-1}+2(2a-1)\left(9(2a-1)^2-3\right)\xi_{2,k-1}, ξ 1 , k = ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 1 , k − 1 + 2 ( 2 a − 1 ) ( 9 ( 2 a − 1 ) 2 − 3 ) ξ 2 , k − 1 , ξ 2 , k = 2 ( 2 a − 1 ) ξ 1 , k − 1 + ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 2 , k − 1 . \xi_{2,k}=2(2a-1)\xi_{1,k-1}+\left(6(2a-1)^2-1\right)\xi_{2,k-1}. ξ 2 , k = 2 ( 2 a − 1 ) ξ 1 , k − 1 + ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 2 , k − 1 . Proof. We first prove that ξ = 6 ( 2 a − 1 ) 2 − 1 + 2 ( 2 a − 1 ) 9 ( 2 a − 1 ) 2 − 3 \xi=6(2a-1)^2-1+2(2a-1)\sqrt{9(2a-1)^2-3} ξ = 6 ( 2 a − 1 ) 2 − 1 + 2 ( 2 a − 1 ) 9 ( 2 a − 1 ) 2 − 3 1 1 1 Z [ 9 ( 2 a − 1 ) 2 − 3 ] \mathbb{Z}[\sqrt{9(2a-1)^2-3}] Z [ 9 ( 2 a − 1 ) 2 − 3 ]
N ( ξ ) = ( 6 ( 2 a − 1 ) 2 − 1 ) 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) ( 2 ( 2 a − 1 ) ) 2 = 1. N(\xi)=\left(6(2a-1)^2-1\right)^2-\left(9(2a-1)^2-3\right)\left(2(2a-1)\right)^2=1. N ( ξ ) = ( 6 ( 2 a − 1 ) 2 − 1 ) 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) ( 2 ( 2 a − 1 ) ) 2 = 1. Since x 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) y 2 = 4 n − 3 ( 2 a − 1 ) 2 + 1 x^2-\left(9(2a-1)^2-3\right)y^2=4n-3(2a-1)^2+1 x 2 − ( 9 ( 2 a − 1 ) 2 − 3 ) y 2 = 4 n − 3 ( 2 a − 1 ) 2 + 1
N ( ( x + y 9 ( 2 a − 1 ) 2 − 3 ) ξ k ) = N ( x + y 9 ( 2 a − 1 ) 2 − 3 ) N ( ξ k ) N\left(\left(x+y\sqrt{9(2a-1)^2-3}\right)\xi^k\right)=N\left(x+y\sqrt{9(2a-1)^2-3}\right)N(\xi^k) N ( ( x + y 9 ( 2 a − 1 ) 2 − 3 ) ξ k ) = N ( x + y 9 ( 2 a − 1 ) 2 − 3 ) N ( ξ k ) = 4 n − 3 ( 2 a − 1 ) 2 + 1 , =4n-3(2a-1)^2+1, = 4 n − 3 ( 2 a − 1 ) 2 + 1 , for each integer k ≥ 0 k\ge0 k ≥ 0
ξ 1 , k + ξ 2 , k 9 ( 2 a − 1 ) 2 − 3 = ( x + y 9 ( 2 a − 1 ) 2 − 3 ) ξ k , (2) \xi_{1,k}+\xi_{2,k}\sqrt{9(2a-1)^2-3}=\left(x+y\sqrt{9(2a-1)^2-3}\right)\xi^k, \tag*{(2)} ξ 1 , k + ξ 2 , k 9 ( 2 a − 1 ) 2 − 3 = ( x + y 9 ( 2 a − 1 ) 2 − 3 ) ξ k , (2) we can observe that ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) (1) . Now, iterating (2) , for k ≥ 1 k\ge1 k ≥ 1
ξ 1 , k = ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 1 , k − 1 + 2 ( 2 a − 1 ) ( 9 ( 2 a − 1 ) 2 − 3 ) ξ 2 , k − 1 , \xi_{1,k}=\left(6(2a-1)^2-1\right)\xi_{1,k-1}+2(2a-1)\left(9(2a-1)^2-3\right)\xi_{2,k-1}, ξ 1 , k = ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 1 , k − 1 + 2 ( 2 a − 1 ) ( 9 ( 2 a − 1 ) 2 − 3 ) ξ 2 , k − 1 , ξ 2 , k = 2 ( 2 a − 1 ) ξ 1 , k − 1 + ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 2 , k − 1 , \xi_{2,k}=2(2a-1)\xi_{1,k-1}+\left(6(2a-1)^2-1\right)\xi_{2,k-1}, ξ 2 , k = 2 ( 2 a − 1 ) ξ 1 , k − 1 + ( 6 ( 2 a − 1 ) 2 − 1 ) ξ 2 , k − 1 , with ξ 1 , 0 = x \xi_{1,0}=x ξ 1 , 0 = x ξ 2 , 0 = y \xi_{2,0}=y ξ 2 , 0 = y ξ 2 , k \xi_{2,k} ξ 2 , k b b b 2 b − 1 = ξ 2 , k 2b-1=\xi_{2,k} 2 b − 1 = ξ 2 , k a ≥ 1 a\geq1 a ≥ 1
b ∈ { ξ 2 , k + 1 2 : k ≥ 0 } , b\in\left\{\frac{\xi_{2,k}+1}{2}: k\geq0\right\}, b ∈ { 2 ξ 2 , k + 1 : k ≥ 0 } , we have S a S b + n = c 2 S_aS_b+n=c^2 S a S b + n = c 2 c c c b b b
Lemma 2.3. Let a , b , n ≥ 1 a,b,n\geq1 a , b , n ≥ 1 b > a b>a b > a a b + 1 = S n ab+1=S_n ab + 1 = S n
X 2 − 6 Y 2 = ( b − a ) 2 − 6 , (3) X^2-6Y^2=(b-a)^2-6, \tag*{(3)} X 2 − 6 Y 2 = ( b − a ) 2 − 6 , (3) where X = a + b X=a+b X = a + b Y = 2 n − 1 Y=2n-1 Y = 2 n − 1
Proof. Since
S n = 6 n ( n − 1 ) + 1 = 3 ( 2 n − 1 ) 2 − 1 2 , S_n=6n(n-1)+1=\frac{3(2n-1)^2-1}{2}, S n = 6 n ( n − 1 ) + 1 = 2 3 ( 2 n − 1 ) 2 − 1 , and a b + 1 = S n ab+1=S_n ab + 1 = S n
a b + 1 = 3 ( 2 n − 1 ) 2 − 1 2 , ab+1=\frac{3(2n-1)^2-1}{2}, ab + 1 = 2 3 ( 2 n − 1 ) 2 − 1 , which gives 2 a b + 3 = 3 ( 2 n − 1 ) 2 2ab+3=3(2n-1)^2 2 ab + 3 = 3 ( 2 n − 1 ) 2
( a + b ) 2 − 6 ( 2 n − 1 ) 2 = ( b − a ) 2 − 6. (a+b)^2-6(2n-1)^2=(b-a)^2-6. ( a + b ) 2 − 6 ( 2 n − 1 ) 2 = ( b − a ) 2 − 6. Lemma 2.4. Under the assumptions in Lemma 2.3 , if ( x , y ) (x,y) ( x , y ) (3) , then for each integer k ≥ 0 k\geq0 k ≥ 0 ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) (3) , where ξ 1 , 0 = x \xi_{1,0}=x ξ 1 , 0 = x ξ 2 , 0 = y \xi_{2,0}=y ξ 2 , 0 = y k ≥ 1 k\geq1 k ≥ 1
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}. ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . Proof. We first prove that ξ = 5 + 2 6 \xi=5+2\sqrt{6} ξ = 5 + 2 6 Z [ 6 ] \mathbb{Z}[\sqrt{6}] Z [ 6 ]
N ( ξ ) = 5 2 − 6 ⋅ 2 2 = 1. N(\xi)=5^2-6\cdot2^2=1. N ( ξ ) = 5 2 − 6 ⋅ 2 2 = 1. Since x 2 − 6 y 2 = ( b − a ) 2 − 6 x^2-6y^2=(b-a)^2-6 x 2 − 6 y 2 = ( b − a ) 2 − 6
N ( ( x + y 6 ) ξ k ) = N ( x + y 6 ) N ( ξ k ) = ( b − a ) 2 − 6 , N\left((x+y\sqrt{6})\xi^k\right)=N(x+y\sqrt{6})N(\xi^k)=(b-a)^2-6, N ( ( x + y 6 ) ξ k ) = N ( x + y 6 ) N ( ξ k ) = ( b − a ) 2 − 6 , for each integer k ≥ 0 k\geq0 k ≥ 0
ξ 1 , k + ξ 2 , k 6 = ( x + y 6 ) ξ k , (4) \xi_{1,k}+\xi_{2,k}\sqrt{6}=(x+y\sqrt{6})\xi^k, \tag*{(4)} ξ 1 , k + ξ 2 , k 6 = ( x + y 6 ) ξ k , (4) we can observe that ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) (3) . Now, iterating (4) , for k ≥ 1 k\geq1 k ≥ 1
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}, ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 , with ξ 1 , 0 = x \xi_{1,0}=x ξ 1 , 0 = x ξ 2 , 0 = y \xi_{2,0}=y ξ 2 , 0 = y
To prove Theorem 1.20 , we need the following results due to Nagell, Matveev, and an immediate variation of a result due to Dujella and Pethö.
Theorem 2.5 (Nagell, 1951). Let u + v d u+v\sqrt{d} u + v d x 2 − d y 2 = 1 x^2-dy^2=1 x 2 − d y 2 = 1 x ∗ + y ∗ d x^*+y^*\sqrt{d} x ∗ + y ∗ d x 2 − d y 2 = N x^2-dy^2=N x 2 − d y 2 = N
0 ≤ y ∗ ≤ v 2 ( u + ϵ ) ∣ N ∣ , 0 \leq y^* \leq\frac{v}{\sqrt{2(u+\epsilon)}}\sqrt{\lvert N\rvert}, 0 ≤ y ∗ ≤ 2 ( u + ϵ ) v ∣ N ∣ , ∣ x ∗ ∣ ≤ 1 2 ( u + ϵ ) ∣ N ∣ , \lvert x^*\rvert\leq\sqrt{\frac{1}{2}(u+\epsilon)\lvert N\rvert}, ∣ x ∗ ∣ ≤ 2 1 ( u + ϵ ) ∣ N ∣ , where ϵ = 1 \epsilon=1 ϵ = 1 N > 0 N>0 N > 0 ϵ = − 1 \epsilon=-1 ϵ = − 1 N < 0 N<0 N < 0
Before stating other two results, let us first recall some basic notions from algebraic number theory.
Let η \eta η d d d
a 0 x d + a 1 x d − 1 + … + a d = a 0 ∏ i = 1 d ( x − η ( i ) ) . a_0x^d+a_1x^{d-1}+\ldots+a_d=a_0\prod_{i=1}^{d}\left(x-\eta^{(i)}\right). a 0 x d + a 1 x d − 1 + … + a d = a 0 i = 1 ∏ d ( x − η ( i ) ) . where the a i a_i a i a 0 > 0 a_0>0 a 0 > 0 η ( i ) \eta^{(i)} η ( i ) η \eta η
h ( η ) = 1 d ( log a 0 + ∑ i = 1 d log ( max { ∣ η ( i ) ∣ , 1 } ) ) h(\eta)=\frac{1}{d}\left(\log a_0+\sum_{i=1}^{d}\log\left(\max\{\lvert\eta^{(i)}\rvert,1\}\right)\right) h ( η ) = d 1 ( log a 0 + i = 1 ∑ d log ( max {∣ η ( i ) ∣ , 1 } ) ) is called the logarithmic height of η \eta η η = a / b \eta=a/b η = a / b gcd ( a , b ) = 1 \gcd(a,b)=1 g cd( a , b ) = 1 b > 0 b>0 b > 0 h ( η ) = log ( max { ∣ a ∣ , b } ) h(\eta)=\log(\max\{\lvert a\rvert,b\}) h ( η ) = log ( max {∣ a ∣ , b })
(1) h ( η ± γ ) ≤ h ( η ) + h ( γ ) + log 2 , h(\eta\pm\gamma)\leq h(\eta)+h(\gamma)+\log2, h ( η ± γ ) ≤ h ( η ) + h ( γ ) + log 2 ,
(2) h ( η γ ± 1 ) ≤ h ( η ) + h ( γ ) , h(\eta\gamma^{\pm1})\leq h(\eta)+h(\gamma), h ( η γ ± 1 ) ≤ h ( η ) + h ( γ ) ,
(3) h ( η s ) = ∣ s ∣ h ( η ) h(\eta^s)=\lvert s\rvert h(\eta) h ( η s ) = ∣ s ∣ h ( η ) s ∈ Z s\in\mathbb{Z} s ∈ Z
Theorem 2.6 ([20 ] ). Assume that γ 1 , γ 2 , … , γ r \gamma_1,\gamma_2,\ldots,\gamma_r γ 1 , γ 2 , … , γ r K \mathbb{K} K d d d b 1 , b 2 , … , b r ∈ Z b_1,b_2,\ldots,b_r\in\mathbb{Z} b 1 , b 2 , … , b r ∈ Z
Λ : = γ 1 b 1 ⋯ γ r b r − 1 \Lambda:=\gamma_1^{b_1}\cdots\gamma_r^{b_r}-1 Λ := γ 1 b 1 ⋯ γ r b r − 1 is not zero. Then
∣ Λ ∣ > exp ( − 1.4 ⋅ 30 r + 3 ⋅ r 4.5 ⋅ d 2 ⋅ ( 1 + log d ) ⋅ ( 1 + log B ) A 1 A 2 … A r ) , \lvert\Lambda\rvert>\exp\left(-1.4\cdot30^{r+3}\cdot r^{4.5}\cdot d^2\cdot(1+\log d)\cdot(1+\log B)A_1A_2\ldots A_r\right), ∣ Λ ∣ > exp ( − 1.4 ⋅ 3 0 r + 3 ⋅ r 4.5 ⋅ d 2 ⋅ ( 1 + log d ) ⋅ ( 1 + log B ) A 1 A 2 … A r ) , where
B ≥ max { ∣ b 1 ∣ , … , ∣ b r ∣ } , B\geq\max\{\lvert b_1\rvert,\ldots,\lvert b_r\rvert\}, B ≥ max {∣ b 1 ∣ , … , ∣ b r ∣} , and
A i ≥ max { d h ( γ i ) , ∣ log γ i ∣ , 0.16 } for all i = 1 , … , r . A_i\geq\max\{dh(\gamma_i),\lvert\log\gamma_i\rvert,0.16\}\quad\text{for all }i=1,\ldots,r. A i ≥ max { d h ( γ i ) , ∣ log γ i ∣ , 0.16 } for all i = 1 , … , r . Lemma 2.7 . [10 ] Let M M M p / q p/q p / q γ \gamma γ q > 6 M q>6M q > 6 M A A A B B B μ \mu μ A > 0 A>0 A > 0 B > 1 B>1 B > 1 ϵ : = ∥ μ q ∥ − M ∥ γ q ∥ \epsilon:=\lVert\mu q\rVert-M\lVert\gamma q\rVert ϵ := ∥ μ q ∥ − M ∥ γ q ∥ ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ x x x ∥ x ∥ = min { ∣ x − n ∣ : n ∈ Z } \lVert x\rVert=\min\{\lvert x-n\rvert:n\in\mathbb{Z}\} ∥ x ∥ = min {∣ x − n ∣ : n ∈ Z } x x x ϵ > 0 \epsilon>0 ϵ > 0
0 < ∣ u γ − v + μ ∣ < A B − w , 0<\lvert u\gamma-v+\mu\rvert<AB^{-w}, 0 < ∣ u γ − v + μ ∣ < A B − w , in positive integers u u u v v v w w w
u ≤ M and w ≥ log ( A q / ϵ ) log B . u \le M \text{ and } w \ge\frac{\log(Aq/\epsilon)}{\log B}. u ≤ M and w ≥ log B log ( A q / ϵ ) . Proof of the main theorems
Proof of Theorem 1.4 . The proof is divided into the following eight cases depending on the value k k k
Case-1: Let k = 3 k=3 k = 3 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( 3 ) D(3) D ( 3 ) S b + 3 = m 2 S_b+3=m^2 S b + 3 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = 10 X^2-6Y^2=10 X 2 − 6 Y 2 = 10 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 4 , 1 ) (X,Y)=(4,1) ( X , Y ) = ( 4 , 1 ) X 2 − 6 Y 2 = 10 X^2-6Y^2=10 X 2 − 6 Y 2 = 10 Lemma 2.2 , we have infinitely many b b b
Case-2: Let k = 8 k=8 k = 8 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( 8 ) D(8) D ( 8 ) S b + 8 = m 2 S_b+8=m^2 S b + 8 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = 30 X^2-6Y^2=30 X 2 − 6 Y 2 = 30 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 6 , 1 ) (X,Y)=(6,1) ( X , Y ) = ( 6 , 1 ) X 2 − 6 Y 2 = 30 X^2-6Y^2=30 X 2 − 6 Y 2 = 30 Lemma 2.2 , we have infinitely many b b b
Case-3: Let k = 12 k=12 k = 12 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( 12 ) D(12) D ( 12 ) S b + 12 = m 2 S_b+12=m^2 S b + 12 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = 46 X^2-6Y^2=46 X 2 − 6 Y 2 = 46 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 10 , 3 ) (X,Y)=(10,3) ( X , Y ) = ( 10 , 3 ) X 2 − 6 Y 2 = 46 X^2-6Y^2=46 X 2 − 6 Y 2 = 46 Lemma 2.2 , we have infinitely many b b b
Case-4: Let k = 11 k=11 k = 11 { S a , S b } \{S_a,S_b\} { S a , S b } D ( 11 ) D(11) D ( 11 ) S a S b + 11 = m 2 S_aS_b+11=m^2 S a S b + 11 = m 2 m m m S n ≡ 1 , 10 , 13 , 19 ( m o d 27 ) S_n\equiv1,10,13,19\pmod{27} S n ≡ 1 , 10 , 13 , 19 ( mod 27 ) S a S b + 11 ≡ 3 , 6 , 12 , 15 , 18 , 21 , 24 ( m o d 27 ) S_aS_b+11\equiv3,6,12,15,18,21,24\pmod{27} S a S b + 11 ≡ 3 , 6 , 12 , 15 , 18 , 21 , 24 ( mod 27 ) m 2 ≡ 0 , 1 , 4 , 7 , 9 , 10 , 13 , 16 , 19 , 22 , 25 ( m o d 27 ) m^2\equiv0,1,4,7,9,10,13,16,19,22,25\pmod{27} m 2 ≡ 0 , 1 , 4 , 7 , 9 , 10 , 13 , 16 , 19 , 22 , 25 ( mod 27 ) S a S b + 11 S_aS_b+11 S a S b + 11 a , b a,b a , b
Case-5: Let k = − 1 k=-1 k = − 1 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( − 1 ) D(-1) D ( − 1 ) S b − 1 = m 2 S_b-1=m^2 S b − 1 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = − 6 X^2-6Y^2=-6 X 2 − 6 Y 2 = − 6 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 12 , 5 ) (X,Y)=(12,5) ( X , Y ) = ( 12 , 5 ) X 2 − 6 Y 2 = − 6 X^2-6Y^2=-6 X 2 − 6 Y 2 = − 6 Lemma 2.2 , we have infinitely many b b b
Case-6: Let k = − 4 k=-4 k = − 4 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( − 4 ) D(-4) D ( − 4 ) S b − 4 = m 2 S_b-4=m^2 S b − 4 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = − 18 X^2-6Y^2=-18 X 2 − 6 Y 2 = − 18 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 6 , 3 ) (X,Y)=(6,3) ( X , Y ) = ( 6 , 3 ) X 2 − 6 Y 2 = − 18 X^2-6Y^2=-18 X 2 − 6 Y 2 = − 18 Lemma 2.2 , we have infinitely many b b b
Case-7: Let k = − 9 k=-9 k = − 9 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( − 9 ) D(-9) D ( − 9 ) S b − 9 = m 2 S_b-9=m^2 S b − 9 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = − 38 X^2-6Y^2=-38 X 2 − 6 Y 2 = − 38 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 16 , 7 ) (X,Y)=(16,7) ( X , Y ) = ( 16 , 7 ) X 2 − 6 Y 2 = − 38 X^2-6Y^2=-38 X 2 − 6 Y 2 = − 38 Lemma 2.2 , we have infinitely many b b b
Case-8: Let k = − 12 k=-12 k = − 12 { S 1 , S b } = { 1 , S b } \{S_1,S_b\}=\{1,S_b\} { S 1 , S b } = { 1 , S b } D ( − 12 ) D(-12) D ( − 12 ) S b − 12 = m 2 S_b-12=m^2 S b − 12 = m 2 m m m Lemma 2.1 , we get X 2 − 6 Y 2 = − 50 X^2-6Y^2=-50 X 2 − 6 Y 2 = − 50 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 10 , 5 ) (X,Y)=(10,5) ( X , Y ) = ( 10 , 5 ) X 2 − 6 Y 2 = − 50 X^2-6Y^2=-50 X 2 − 6 Y 2 = − 50 Lemma 2.2 , we have infinitely many b b b □ \square □
Proof of Theorem 1.5 . Observe that
S a S b + k = ( 6 a b − 3 ( b + a ) + 1 ) 2 − 3 ( b − a ) 2 + k . (5) S_aS_b+k=(6ab-3(b+a)+1)^2-3(b-a)^2+k. \tag*{(5)} S a S b + k = ( 6 ab − 3 ( b + a ) + 1 ) 2 − 3 ( b − a ) 2 + k . (5) If b = a + 1 b=a+1 b = a + 1 3 ( b − a ) 2 − k = 0 3(b-a)^2-k=0 3 ( b − a ) 2 − k = 0 k = 3 k=3 k = 3 S a S a + 1 + k S_aS_{a+1}+k S a S a + 1 + k k = 8 , 12 k=8,12 k = 8 , 12 (5) , we have
( 6 a 2 − 2 ) 2 < S a S a + 1 + k < ( 6 a 2 − 1 ) 2 , (6a^2-2)^2<S_aS_{a+1}+k<(6a^2-1)^2, ( 6 a 2 − 2 ) 2 < S a S a + 1 + k < ( 6 a 2 − 1 ) 2 , and S a S a + 1 + k S_aS_{a+1}+k S a S a + 1 + k k = − 1 , − 4 , − 9 , − 12 k=-1,-4,-9,-12 k = − 1 , − 4 , − 9 , − 12 (5) , we have
( 6 a 2 − 3 ) 2 < S a S a + 1 + k < ( 6 a 2 − 2 ) 2 . (6a^2-3)^2<S_aS_{a+1}+k<(6a^2-2)^2. ( 6 a 2 − 3 ) 2 < S a S a + 1 + k < ( 6 a 2 − 2 ) 2 . For b = a + 2 b=a+2 b = a + 2 3 ( b − a ) 2 − k = 0 3(b-a)^2-k=0 3 ( b − a ) 2 − k = 0 k = 12 k=12 k = 12 S a S a + 2 + k S_aS_{a+2}+k S a S a + 2 + k k = 3 , 8 , − 1 , − 4 , − 9 , − 12 k=3,8,-1,-4,-9,-12 k = 3 , 8 , − 1 , − 4 , − 9 , − 12 (5) , we have
( 6 a 2 + 6 a − 6 ) 2 < S a S a + 2 + k < ( 6 a 2 + 6 a − 5 ) 2 . (6a^2+6a-6)^2<S_aS_{a+2}+k<(6a^2+6a-5)^2. ( 6 a 2 + 6 a − 6 ) 2 < S a S a + 2 + k < ( 6 a 2 + 6 a − 5 ) 2 . Now, for b ≥ a + 3 b\ge a+3 b ≥ a + 3 a = 2 a=2 a = 2 4 a + 2 = 10 4a+2=10 4 a + 2 = 10 5 ≤ b ≤ 9 5\le b\le9 5 ≤ b ≤ 9 { S 2 , S 9 } \{S_2,S_9\} { S 2 , S 9 } D ( − 4 ) D(-4) D ( − 4 ) k = 3 , 8 , 12 , − 1 , − 9 , − 12 k=3,8,12,-1,-9,-12 k = 3 , 8 , 12 , − 1 , − 9 , − 12 5 ≤ b ≤ 9 5\le b\le9 5 ≤ b ≤ 9 { S 2 , S b } \{S_2,S_b\} { S 2 , S b } D ( k ) D(k) D ( k )
Now we claim that if a + 3 ≤ b ≤ 4 a + 1 a+3\le b\le4a+1 a + 3 ≤ b ≤ 4 a + 1 a ≥ 3 a\ge3 a ≥ 3 { S a , S b } \{S_a,S_b\} { S a , S b } D ( k ) D(k) D ( k ) k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 k=3,8,12,-1,-4,-9,-12 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 S a S b + k S_aS_b+k S a S b + k a + 3 ≤ b ≤ 4 a + 1 a+3\le b\le4a+1 a + 3 ≤ b ≤ 4 a + 1 a ≥ 3 a\ge3 a ≥ 3 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12 k=3,8,12,-1,-4,-9,-12 k = 3 , 8 , 12 , − 1 , − 4 , − 9 , − 12
Set d = b − a d=b-a d = b − a 3 ≤ d ≤ 3 a + 1 3\le d\le3a+1 3 ≤ d ≤ 3 a + 1 f a : [ 3 , 3 a + 1 ] → R f_a:[3,3a+1]\to\mathbb{R} f a : [ 3 , 3 a + 1 ] → R f a ( d ) = 12 a d + 12 a 2 − 6 d − 12 a − 3 d 2 + k + 1 f_a(d)=12ad+12a^2-6d-12a-3d^2+k+1 f a ( d ) = 12 a d + 12 a 2 − 6 d − 12 a − 3 d 2 + k + 1 f a ′ ′ ( d ) = − 6 < 0 f_a''(d)=-6<0 f a ′′ ( d ) = − 6 < 0 a ≥ 3 a\ge3 a ≥ 3
f a ( 3 ) = 12 a 2 + 24 a + k − 44 > 0 , f a ( 3 a + 1 ) = 21 a 2 − 36 a + k − 8 > 0 , f_a(3)=12a^2+24a+k-44>0,\qquad f_a(3a+1)=21a^2-36a+k-8>0, f a ( 3 ) = 12 a 2 + 24 a + k − 44 > 0 , f a ( 3 a + 1 ) = 21 a 2 − 36 a + k − 8 > 0 , we have f a ( d ) > 0 f_a(d)>0 f a ( d ) > 0 2 ( 6 a b − 3 ( a + b ) ) + 1 > 3 d 2 − k 2(6ab-3(a+b))+1>3d^2-k 2 ( 6 ab − 3 ( a + b )) + 1 > 3 d 2 − k 3 ≤ d ≤ 3 a + 1 3\le d\le3a+1 3 ≤ d ≤ 3 a + 1 (5) that
( 6 a b − 3 ( b + a ) ) 2 < S a S b + k < ( 6 a b − 3 ( b + a ) + 1 ) 2 . (6ab-3(b+a))^2<S_aS_b+k<(6ab-3(b+a)+1)^2. ( 6 ab − 3 ( b + a ) ) 2 < S a S b + k < ( 6 ab − 3 ( b + a ) + 1 ) 2 . This completes the proof. □
Proof of Theorem 1.6. The proof is divided into the following seven cases depending on the value k k k
Case-1: Let k = 3 k=3 k = 3 a ≥ 2 a\ge2 a ≥ 2
S a S a + 1 + 3 = ( 6 a 2 − 2 ) 2 . S_aS_{a+1}+3=(6a^2-2)^2. S a S a + 1 + 3 = ( 6 a 2 − 2 ) 2 . It follows that { S a , S a + 1 } \{S_a,S_{a+1}\} { S a , S a + 1 } D ( 3 ) D(3) D ( 3 ) a ≥ 2 a\ge2 a ≥ 2
Case-2: Let k = 8 k=8 k = 8 { S 4 , S b } = { 73 , S b } \{S_4,S_b\}=\{73,S_b\} { S 4 , S b } = { 73 , S b } D ( 8 ) D(8) D ( 8 ) 73 S b + 8 = m 2 73S_b+8=m^2 73 S b + 8 = m 2 m m m Lemma 2.1 , we get X 2 − 438 Y 2 = − 114 X^2-438Y^2=-114 X 2 − 438 Y 2 = − 114 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 18 , 1 ) (X,Y)=(18,1) ( X , Y ) = ( 18 , 1 ) X 2 − 438 Y 2 = − 114 X^2-438Y^2=-114 X 2 − 438 Y 2 = − 114 Lemma 2.2 , we have infinitely many b b b
Case-3: Let k = 12 k=12 k = 12 a ≥ 2 a\ge2 a ≥ 2
S a S a + 2 + 12 = ( 6 a 2 + 6 a − 5 ) 2 . S_aS_{a+2}+12=(6a^2+6a-5)^2. S a S a + 2 + 12 = ( 6 a 2 + 6 a − 5 ) 2 . It follows that { S a , S a + 2 } \{S_a,S_{a+2}\} { S a , S a + 2 } D ( 12 ) D(12) D ( 12 ) a ≥ 2 a\ge2 a ≥ 2
Case-4: Let k = − 1 k=-1 k = − 1 { S 3 , S b } = { 37 , S b } \{S_3,S_b\}=\{37,S_b\} { S 3 , S b } = { 37 , S b } D ( − 1 ) D(-1) D ( − 1 ) 37 S b − 1 = m 2 37S_b-1=m^2 37 S b − 1 = m 2 m m m Lemma 2.1 , we get X 2 − 222 Y 2 = − 78 X^2-222Y^2=-78 X 2 − 222 Y 2 = − 78 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 12 , 1 ) (X,Y)=(12,1) ( X , Y ) = ( 12 , 1 ) X 2 − 222 Y 2 = − 78 X^2-222Y^2=-78 X 2 − 222 Y 2 = − 78 Lemma 2.2 , we have infinitely many b b b
Case-5: Let k = − 4 k=-4 k = − 4 { S 2 , S b } = { 13 , S b } \{S_2,S_b\}=\{13,S_b\} { S 2 , S b } = { 13 , S b } D ( − 4 ) D(-4) D ( − 4 ) 13 S b − 4 = m 2 13S_b-4=m^2 13 S b − 4 = m 2 m m m Lemma 2.1 , we get X 2 − 78 Y 2 = − 42 X^2-78Y^2=-42 X 2 − 78 Y 2 = − 42 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 6 , 1 ) (X,Y)=(6,1) ( X , Y ) = ( 6 , 1 ) X 2 − 78 Y 2 = − 42 X^2-78Y^2=-42 X 2 − 78 Y 2 = − 42 Lemma 2.2 , we have infinitely many b b b
Case-6: Let k = − 9 k=-9 k = − 9 { S 2 , S b } = { 13 , S b } \{S_2,S_b\}=\{13,S_b\} { S 2 , S b } = { 13 , S b } D ( − 9 ) D(-9) D ( − 9 ) 13 S b − 9 = m 2 13S_b-9=m^2 13 S b − 9 = m 2 m m m Lemma 2.1 , we get X 2 − 78 Y 2 = − 62 X^2-78Y^2=-62 X 2 − 78 Y 2 = − 62 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 4 , 1 ) (X,Y)=(4,1) ( X , Y ) = ( 4 , 1 ) X 2 − 78 Y 2 = − 62 X^2-78Y^2=-62 X 2 − 78 Y 2 = − 62 Lemma 2.2 , we have infinitely many b b b
Case-7: Let k = − 12 k=-12 k = − 12 { S 2 , S b } = { 13 , S b } \{S_2,S_b\}=\{13,S_b\} { S 2 , S b } = { 13 , S b } D ( − 12 ) D(-12) D ( − 12 ) 13 S b − 12 = m 2 13S_b-12=m^2 13 S b − 12 = m 2 m m m Lemma 2.1 , we get X 2 − 78 Y 2 = − 74 X^2-78Y^2=-74 X 2 − 78 Y 2 = − 74 X = 2 m X=2m X = 2 m Y = 2 b − 1 Y=2b-1 Y = 2 b − 1 ( X , Y ) = ( 2 , 1 ) (X,Y)=(2,1) ( X , Y ) = ( 2 , 1 ) X 2 − 78 Y 2 = − 74 X^2-78Y^2=-74 X 2 − 78 Y 2 = − 74 Lemma 2.2 , we have infinitely many b b b □ \square □
Proof of Theorem 1.7. Note that
S n ( m o d 8 ) = { 1 if n ≡ 0 , 1 ( m o d 4 ) , 5 if n ≡ 2 , 3 ( m o d 4 ) . S_n\pmod{8}=
\begin{cases}
1 & \text{if } n\equiv0,1\pmod{4},\\
5 & \text{if } n\equiv2,3\pmod{4}.
\end{cases} S n ( mod 8 ) = { 1 5 if n ≡ 0 , 1 ( mod 4 ) , if n ≡ 2 , 3 ( mod 4 ) . Therefore,
S n S m ( m o d 8 ) = { 1 if S n ≡ S m ( m o d 8 ) , 5 if S n ≢ S m ( m o d 8 ) . S_nS_m\pmod{8}=
\begin{cases}
1 & \text{if } S_n\equiv S_m\pmod{8},\\
5 & \text{if } S_n\not\equiv S_m\pmod{8}.
\end{cases} S n S m ( mod 8 ) = { 1 5 if S n ≡ S m ( mod 8 ) , if S n ≡ S m ( mod 8 ) . Since k ≡ 12 , 20 ( m o d 24 ) k\equiv12,20\pmod{24} k ≡ 12 , 20 ( mod 24 ) k ≡ 4 ( m o d 8 ) k\equiv4\pmod{8} k ≡ 4 ( mod 8 )
S n S m + k ( m o d 8 ) = { 5 if S n ≡ S m ( m o d 8 ) , 1 if S n ≢ S m ( m o d 8 ) . S_nS_m+k\pmod{8}=
\begin{cases}
5 & \text{if } S_n\equiv S_m\pmod{8},\\
1 & \text{if } S_n\not\equiv S_m\pmod{8}.
\end{cases} S n S m + k ( mod 8 ) = { 5 1 if S n ≡ S m ( mod 8 ) , if S n ≡ S m ( mod 8 ) . Since m 2 ≡ 0 , 1 , 4 ( m o d 8 ) m^2\equiv0,1,4\pmod{8} m 2 ≡ 0 , 1 , 4 ( mod 8 ) { S a , S b , S c } \{S_a,S_b,S_c\} { S a , S b , S c } D ( k ) D(k) D ( k ) S i ≢ S j ( m o d 8 ) S_i\not\equiv S_j\pmod{8} S i ≡ S j ( mod 8 ) i ≠ j i\ne j i = j i , j ∈ { a , b , c } i,j\in\{a,b,c\} i , j ∈ { a , b , c } S a ≡ S b ( m o d 8 ) S_a\equiv S_b\pmod{8} S a ≡ S b ( mod 8 ) S a ≡ S c ( m o d 8 ) S_a\equiv S_c\pmod{8} S a ≡ S c ( mod 8 )
Again, observe that
S n ( m o d 9 ) = { 1 if n ≡ 0 , 1 ( m o d 3 ) , 4 if n ≡ 2 ( m o d 3 ) . S_n\pmod{9}=
\begin{cases}
1 & \text{if } n\equiv0,1\pmod{3},\\
4 & \text{if } n\equiv2\pmod{3}.
\end{cases} S n ( mod 9 ) = { 1 4 if n ≡ 0 , 1 ( mod 3 ) , if n ≡ 2 ( mod 3 ) . Therefore,
S n S m ( m o d 9 ) = { 7 if S n ≡ S m ≡ 4 ( m o d 9 ) , 1 if S n ≡ S m ≡ 1 ( m o d 9 ) , 4 if S n ≢ S m ( m o d 9 ) . S_nS_m\pmod{9}=
\begin{cases}
7 & \text{if } S_n\equiv S_m\equiv4\pmod{9},\\
1 & \text{if } S_n\equiv S_m\equiv1\pmod{9},\\
4 & \text{if } S_n\not\equiv S_m\pmod{9}.
\end{cases} S n S m ( mod 9 ) = ⎩ ⎨ ⎧ 7 1 4 if S n ≡ S m ≡ 4 ( mod 9 ) , if S n ≡ S m ≡ 1 ( mod 9 ) , if S n ≡ S m ( mod 9 ) . Since k ≡ 23 , 32 ( m o d 36 ) k\equiv23,32\pmod{36} k ≡ 23 , 32 ( mod 36 ) k ≡ 5 ( m o d 9 ) k\equiv5\pmod{9} k ≡ 5 ( mod 9 )
S n S m + k ( m o d 9 ) = { 3 if S n ≡ S m ≡ 4 ( m o d 9 ) , 6 if S n ≡ S m ≡ 1 ( m o d 9 ) , 0 if S n ≢ S m ( m o d 9 ) . S_nS_m+k\pmod{9}=
\begin{cases}
3 & \text{if } S_n\equiv S_m\equiv4\pmod{9},\\
6 & \text{if } S_n\equiv S_m\equiv1\pmod{9},\\
0 & \text{if } S_n\not\equiv S_m\pmod{9}.
\end{cases} S n S m + k ( mod 9 ) = ⎩ ⎨ ⎧ 3 6 0 if S n ≡ S m ≡ 4 ( mod 9 ) , if S n ≡ S m ≡ 1 ( mod 9 ) , if S n ≡ S m ( mod 9 ) . Since m 2 ≡ 0 , 1 , 4 , 7 ( m o d 9 ) m^2\equiv0,1,4,7\pmod{9} m 2 ≡ 0 , 1 , 4 , 7 ( mod 9 ) { S a , S b , S c } \{S_a,S_b,S_c\} { S a , S b , S c } D ( k ) D(k) D ( k ) S i ≢ S j ( m o d 9 ) S_i\not\equiv S_j\pmod{9} S i ≡ S j ( mod 9 ) i ≠ j i\ne j i = j i , j ∈ { a , b , c } i,j\in\{a,b,c\} i , j ∈ { a , b , c } S a ≡ S b ( m o d 9 ) S_a\equiv S_b\pmod{9} S a ≡ S b ( mod 9 ) S a ≡ S c ( m o d 9 ) S_a\equiv S_c\pmod{9} S a ≡ S c ( mod 9 ) □ \square □
Proof of Corollary 1.8. The proof follows from Theorem 1.7. □ \square □
Proof of Theorem 1.12. If possible, assume that { S b , S a , S a + 1 } \{S_b,S_a,S_{a+1}\} { S b , S a , S a + 1 } D ( 3 ) D(3) D ( 3 ) { S b , S a } \{S_b,S_a\} { S b , S a } { S b , S a + 1 } \{S_b,S_{a+1}\} { S b , S a + 1 } D ( 3 ) D(3) D ( 3 )
S b S a + 3 = c 2 , S b S a + 1 + 3 = d 2 , (6) S_bS_a+3=c^2,\qquad S_bS_{a+1}+3=d^2, \tag*{(6)} S b S a + 3 = c 2 , S b S a + 1 + 3 = d 2 , (6) where c c c d d d
Case-1: Let b = 1 b=1 b = 1 (6) and Lemma 2.1 , we have two corresponding generalized Pell equations
X 1 2 − 6 Y 1 2 = 10 , X 2 2 − 6 Y 2 2 = 10 , (7) X_1^2-6Y_1^2=10,\qquad X_2^2-6Y_2^2=10, \tag*{(7)} X 1 2 − 6 Y 1 2 = 10 , X 2 2 − 6 Y 2 2 = 10 , (7) where ( X 1 , Y 1 ) = ( 2 c , 2 a − 1 ) (X_1,Y_1)=(2c,2a-1) ( X 1 , Y 1 ) = ( 2 c , 2 a − 1 ) ( X 2 , Y 2 ) = ( 2 d , 2 a + 1 ) (X_2,Y_2)=(2d,2a+1) ( X 2 , Y 2 ) = ( 2 d , 2 a + 1 ) Y 2 = Y 1 + 2 Y_2=Y_1+2 Y 2 = Y 1 + 2 Y 1 ≥ 3 Y_1\ge3 Y 1 ≥ 3
10 + 6 Y 1 2 < ( 6 Y 1 + 1 ) 2 , 10+6Y_1^2<(\sqrt{6}Y_1+1)^2, 10 + 6 Y 1 2 < ( 6 Y 1 + 1 ) 2 , it follows that
6 Y 1 < X 1 < 6 Y 1 + 1 , 6 ( Y 1 + 2 ) < X 2 < 6 ( Y 1 + 2 ) + 1. \sqrt{6}Y_1<X_1<\sqrt{6}Y_1+1,\qquad\sqrt{6}(Y_1+2)<X_2<\sqrt{6}(Y_1+2)+1. 6 Y 1 < X 1 < 6 Y 1 + 1 , 6 ( Y 1 + 2 ) < X 2 < 6 ( Y 1 + 2 ) + 1. Therefore,
2 6 − 1 < X 2 − X 1 < 2 6 + 1 , 2\sqrt{6}-1<X_2-X_1<2\sqrt{6}+1, 2 6 − 1 < X 2 − X 1 < 2 6 + 1 , that is X 2 − X 1 = 4 X_2-X_1=4 X 2 − X 1 = 4 X 2 − X 1 = 5 X_2-X_1=5 X 2 − X 1 = 5 X 2 − X 1 = 5 X_2-X_1=5 X 2 − X 1 = 5 X i X_i X i X 2 − X 1 = 4 X_2-X_1=4 X 2 − X 1 = 4 (7) , we get
X 1 2 − 6 Y 1 2 = ( X 1 + 4 ) 2 − 6 ( Y 1 + 2 ) 2 , X_1^2-6Y_1^2=(X_1+4)^2-6(Y_1+2)^2, X 1 2 − 6 Y 1 2 = ( X 1 + 4 ) 2 − 6 ( Y 1 + 2 ) 2 , which gives X 1 = 3 Y 1 + 1 X_1=3Y_1+1 X 1 = 3 Y 1 + 1 c = 3 a − 1 c=3a-1 c = 3 a − 1
S a + 3 = 6 a ( a − 1 ) + 4 = ( 3 a − 1 ) 2 , S_a+3=6a(a-1)+4=(3a-1)^2, S a + 3 = 6 a ( a − 1 ) + 4 = ( 3 a − 1 ) 2 , we get a = 1 a=1 a = 1 a ≥ 2 a\ge2 a ≥ 2
Case-2: Let b ≥ 2 b\ge2 b ≥ 2 D = 9 ( 2 b − 1 ) 2 − 3 D=9(2b-1)^2-3 D = 9 ( 2 b − 1 ) 2 − 3 R = 3 ( 2 b − 1 ) 2 − 13 R=3(2b-1)^2-13 R = 3 ( 2 b − 1 ) 2 − 13 a > b a>b a > b a ≥ b + 2 a\ge b+2 a ≥ b + 2 a = b + 1 a=b+1 a = b + 1 { S b , S b + 1 , S b + 2 } \{S_b,S_{b+1},S_{b+2}\} { S b , S b + 1 , S b + 2 } D ( 3 ) D(3) D ( 3 ) (6) , we have
X 1 2 − D Y 1 2 = − R , X 2 2 − D Y 2 2 = − R , (8) X_1^2-DY_1^2=-R,\qquad X_2^2-DY_2^2=-R, \tag*{(8)} X 1 2 − D Y 1 2 = − R , X 2 2 − D Y 2 2 = − R , (8) where ( X 1 , Y 1 ) = ( 2 c , 2 a − 1 ) (X_1,Y_1)=(2c,2a-1) ( X 1 , Y 1 ) = ( 2 c , 2 a − 1 ) ( X 2 , Y 2 ) = ( 2 d , 2 a + 1 ) (X_2,Y_2)=(2d,2a+1) ( X 2 , Y 2 ) = ( 2 d , 2 a + 1 ) Y 2 = Y 1 + 2 Y_2=Y_1+2 Y 2 = Y 1 + 2 b ≥ 2 b\ge2 b ≥ 2 R > 0 R>0 R > 0 i = 1 , 2 i=1,2 i = 1 , 2
X i 2 − ( D Y i ) 2 = − R , X_i^2-(\sqrt{D}Y_i)^2=-R, X i 2 − ( D Y i ) 2 = − R , which gives
X i = D Y i − R D Y i + X i . (9) X_i=\sqrt{D}Y_i-\frac{R}{\sqrt{D}Y_i+X_i}. \tag*{(9)} X i = D Y i − D Y i + X i R . (9) Since Y 2 > Y 1 Y_2>Y_1 Y 2 > Y 1 X 2 > X 1 X_2>X_1 X 2 > X 1
D Y 2 + X 2 > D Y 1 + X 1 . (10) \sqrt{D}Y_2+X_2>\sqrt{D}Y_1+X_1. \tag*{(10)} D Y 2 + X 2 > D Y 1 + X 1 . (10) Using (9) , and the fact that Y 2 − Y 1 = 2 Y_2-Y_1=2 Y 2 − Y 1 = 2
X 2 − X 1 = 2 D + R ( 1 D Y 1 + X 1 − 1 D Y 2 + X 2 ) , (11) X_2-X_1=2\sqrt{D}+R\left(\frac{1}{\sqrt{D}Y_1+X_1}-\frac{1}{\sqrt{D}Y_2+X_2}\right), \tag*{(11)} X 2 − X 1 = 2 D + R ( D Y 1 + X 1 1 − D Y 2 + X 2 1 ) , (11) using (10) , and the above equation, we get
X 2 − X 1 > 2 D . (12) X_2-X_1>2\sqrt{D}. \tag*{(12)} X 2 − X 1 > 2 D . (12) For b ≥ 2 b\ge2 b ≥ 2
( 3 ( 2 b − 1 ) − 1 ) 2 = 9 ( 2 b − 1 ) 2 − 6 ( 2 b − 1 ) + 1 < 9 ( 2 b − 1 ) 2 − 3 = D < ( 3 ( 2 b − 1 ) ) 2 , (3(2b-1)-1)^2=9(2b-1)^2-6(2b-1)+1<9(2b-1)^2-3=D<(3(2b-1))^2, ( 3 ( 2 b − 1 ) − 1 ) 2 = 9 ( 2 b − 1 ) 2 − 6 ( 2 b − 1 ) + 1 < 9 ( 2 b − 1 ) 2 − 3 = D < ( 3 ( 2 b − 1 ) ) 2 , from which we obtain
6 ( 2 b − 1 ) − 2 < 2 D < 6 ( 2 b − 1 ) . (13) 6(2b-1)-2 < 2\sqrt{D} < 6(2b-1). \tag*{(13)} 6 ( 2 b − 1 ) − 2 < 2 D < 6 ( 2 b − 1 ) . (13) Now, using (12) and (13) , we get
6 ( 2 b − 1 ) − 2 < X 2 − X 1 . (14) 6(2b-1)-2 < X_2-X_1. \tag*{(14)} 6 ( 2 b − 1 ) − 2 < X 2 − X 1 . (14) Since a ≥ b + 2 a \ge b+2 a ≥ b + 2 Y 1 = 2 a − 1 Y_1=2a-1 Y 1 = 2 a − 1 Y 1 = 2 a − 1 ≥ 2 b + 2 Y_1=2a-1 \ge2b+2 Y 1 = 2 a − 1 ≥ 2 b + 2 (11) , we have
R ( 1 D Y 1 + X 1 − 1 D Y 2 + X 2 ) < R D Y 1 + X 1 < 3 ( 2 b − 1 ) 2 ( 3 ( 2 b − 1 ) − 1 ) ( 2 b + 2 ) ≤ 1 , R\left(\frac{1}{\sqrt{D}Y_1+X_1}-\frac{1}{\sqrt{D}Y_2+X_2}\right)<\frac{R}{\sqrt{D}Y_1+X_1}<\frac{3(2b-1)^2}{(3(2b-1)-1)(2b+2)}\le1, R ( D Y 1 + X 1 1 − D Y 2 + X 2 1 ) < D Y 1 + X 1 R < ( 3 ( 2 b − 1 ) − 1 ) ( 2 b + 2 ) 3 ( 2 b − 1 ) 2 ≤ 1 , which together with (11) imply that
X 2 − X 1 < 2 D + 1 < 6 ( 2 b − 1 ) + 1. X_2-X_1<2\sqrt{D}+1<6(2b-1)+1. X 2 − X 1 < 2 D + 1 < 6 ( 2 b − 1 ) + 1. Consequently, from (14) , we obtain
6 ( 2 b − 1 ) − 2 < X 2 − X 1 < 2 D + 1 < 6 ( 2 b − 1 ) + 1 , 6(2b-1)-2<X_2-X_1<2\sqrt{D}+1<6(2b-1)+1, 6 ( 2 b − 1 ) − 2 < X 2 − X 1 < 2 D + 1 < 6 ( 2 b − 1 ) + 1 , it follows that X 2 − X 1 = 6 ( 2 b − 1 ) X_2-X_1=6(2b-1) X 2 − X 1 = 6 ( 2 b − 1 ) X 2 − X 1 X_2-X_1 X 2 − X 1 (8) , we get
X 1 2 − D Y 1 2 = ( 6 ( 2 b − 1 ) + X 1 ) 2 − D ( Y 1 + 2 ) 2 , X_1^2-DY_1^2=(6(2b-1)+X_1)^2-D(Y_1+2)^2, X 1 2 − D Y 1 2 = ( 6 ( 2 b − 1 ) + X 1 ) 2 − D ( Y 1 + 2 ) 2 , which gives
3 ( 2 b − 1 ) X 1 + 9 ( 2 b − 1 ) 2 = D ( Y 1 + 1 ) = 2 a D . 3(2b-1)X_1+9(2b-1)^2=D(Y_1+1)=2aD. 3 ( 2 b − 1 ) X 1 + 9 ( 2 b − 1 ) 2 = D ( Y 1 + 1 ) = 2 a D . Therefore
X 1 = 2 D a − 9 ( 2 b − 1 ) 2 3 ( 2 b − 1 ) = 2 a ( 3 ( 2 b − 1 ) 2 − 1 ) − 3 ( 2 b − 1 ) 2 2 b − 1 , X_1=\frac{2Da-9(2b-1)^2}{3(2b-1)}=\frac{2a(3(2b-1)^2-1)-3(2b-1)^2}{2b-1}, X 1 = 3 ( 2 b − 1 ) 2 D a − 9 ( 2 b − 1 ) 2 = 2 b − 1 2 a ( 3 ( 2 b − 1 ) 2 − 1 ) − 3 ( 2 b − 1 ) 2 , since X 1 = 2 c X_1=2c X 1 = 2 c X 1 ≡ 0 ( m o d 2 ) X_1\equiv0\pmod{2} X 1 ≡ 0 ( mod 2 )
2 a ( 3 ( 2 b − 1 ) 2 − 1 ) − 3 ( 2 b − 1 ) 2 ≢ 0 ( m o d 2 ) , 2a(3(2b-1)^2-1)-3(2b-1)^2\not\equiv0\pmod{2}, 2 a ( 3 ( 2 b − 1 ) 2 − 1 ) − 3 ( 2 b − 1 ) 2 ≡ 0 ( mod 2 ) , which is absurd. Hence { S b , S a , S a + 1 } \{S_b,S_a,S_{a+1}\} { S b , S a , S a + 1 } D ( 3 ) D(3) D ( 3 )
Proof of Theorem 1.15. If { a , a + 1 , 12 ( 2 a + 1 ) } \{a,a+1,12(2a+1)\} { a , a + 1 , 12 ( 2 a + 1 )} D ( 1 ) D(1) D ( 1 )
a ( a + 1 ) + 1 = S n , 12 a ( 2 a + 1 ) + 1 = S m , 12 ( a + 1 ) ( 2 a + 1 ) + 1 = S r , a(a+1)+1=S_n,\qquad12a(2a+1)+1=S_m,\qquad12(a+1)(2a+1)+1=S_r, a ( a + 1 ) + 1 = S n , 12 a ( 2 a + 1 ) + 1 = S m , 12 ( a + 1 ) ( 2 a + 1 ) + 1 = S r , where n , m , r n,m,r n , m , r 12 a ( 2 a + 1 ) + 1 = S m 12a(2a+1)+1=S_m 12 a ( 2 a + 1 ) + 1 = S m m = 2 a + 1 m=2a+1 m = 2 a + 1 12 ( a + 1 ) ( 2 a + 1 ) + 1 = S r 12(a+1)(2a+1)+1=S_r 12 ( a + 1 ) ( 2 a + 1 ) + 1 = S r r = 2 a + 2 r=2a+2 r = 2 a + 2 a , n a,n a , n a ( a + 1 ) + 1 = S n a(a+1)+1=S_n a ( a + 1 ) + 1 = S n a ( a + 1 ) + 1 = S n a(a+1)+1=S_n a ( a + 1 ) + 1 = S n
X 2 − 6 Y 2 = − 5 , X^2-6Y^2=-5, X 2 − 6 Y 2 = − 5 , where X = 2 a + 1 X=2a+1 X = 2 a + 1 Y = 2 n − 1 Y=2n-1 Y = 2 n − 1 ( x , y ) = ( 17 , 7 ) (x,y)=(17,7) ( x , y ) = ( 17 , 7 ) X 2 − 6 Y 2 = − 5 X^2-6Y^2=-5 X 2 − 6 Y 2 = − 5 ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) X 2 − 6 Y 2 = − 5 X^2-6Y^2=-5 X 2 − 6 Y 2 = − 5 ξ 1 , 0 = 17 \xi_{1,0}=17 ξ 1 , 0 = 17 ξ 2 , 0 = 7 \xi_{2,0}=7 ξ 2 , 0 = 7 k ≥ 1 k\ge1 k ≥ 1
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}. ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . Since X = 2 a + 1 X=2a+1 X = 2 a + 1
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 1 2 : k ≥ 1 , ξ 1 , 0 = 17 , ξ 2 , 0 = 7 } , a\in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-1}{2}: k\ge1,\ \xi_{1,0}=17,\ \xi_{2,0}=7\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 1 : k ≥ 1 , ξ 1 , 0 = 17 , ξ 2 , 0 = 7 } , { a , a + 1 , 12 ( 2 a + 1 ) } \{a, a + 1, 12(2a + 1)\} { a , a + 1 , 12 ( 2 a + 1 )} D ( 1 ) D(1) D ( 1 ) { a , a + 1 , 12 ( 2 a + 1 ) , b } \{a, a + 1, 12(2a + 1), b\} { a , a + 1 , 12 ( 2 a + 1 ) , b } D ( 1 ) D(1) D ( 1 ) 12 ∤ b 12 \nmid b 12 ∤ b n , m n, m n , m
a b + 1 = S n = 6 n ( n − 1 ) + 1 , ( a + 1 ) b + 1 = S m = 6 m ( m − 1 ) + 1 , ab + 1 = S_n = 6n(n - 1) + 1,\qquad(a + 1)b + 1 = S_m = 6m(m - 1) + 1, ab + 1 = S n = 6 n ( n − 1 ) + 1 , ( a + 1 ) b + 1 = S m = 6 m ( m − 1 ) + 1 , since 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( m o d 12 ) 6n(n - 1), 6m(m - 1) \equiv0 \pmod{12} 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( mod 12 ) 12 ∣ b 12 \mid b 12 ∣ b 12 ∤ b 12 \nmid b 12 ∤ b { a , a + 1 , 12 ( 2 a + 1 ) , b } \{a, a + 1, 12(2a + 1), b\} { a , a + 1 , 12 ( 2 a + 1 ) , b } D ( 1 ) D(1) D ( 1 ) 12 ∤ b 12 \nmid b 12 ∤ b
Proof of Theorem 1.16 . If { a , a + 2 , 6 ( a + 1 ) } \{a, a + 2, 6(a + 1)\} { a , a + 2 , 6 ( a + 1 )} D ( 1 ) D(1) D ( 1 )
a ( a + 2 ) + 1 = S n , 6 a ( a + 1 ) + 1 = S m , 6 ( a + 2 ) ( a + 1 ) + 1 = S r , a(a + 2) + 1 = S_n,\qquad6a(a + 1) + 1 = S_m,\qquad6(a + 2)(a + 1) + 1 = S_r, a ( a + 2 ) + 1 = S n , 6 a ( a + 1 ) + 1 = S m , 6 ( a + 2 ) ( a + 1 ) + 1 = S r , where n , m , r n, m, r n , m , r 6 a ( a + 1 ) + 1 = S m 6a(a + 1) + 1 = S_m 6 a ( a + 1 ) + 1 = S m m = a + 1 m = a + 1 m = a + 1 6 ( a + 2 ) ( a + 1 ) + 1 = S r 6(a + 2)(a + 1) + 1 = S_r 6 ( a + 2 ) ( a + 1 ) + 1 = S r r = a + 2 r = a + 2 r = a + 2 a , n a, n a , n a ( a + 2 ) + 1 = S n a(a + 2) + 1 = S_n a ( a + 2 ) + 1 = S n Lemma 2.3 , a ( a + 2 ) + 1 = S n a(a + 2) + 1 = S_n a ( a + 2 ) + 1 = S n
X 2 − 6 Y 2 = − 2 , X^2 - 6Y^2 = -2, X 2 − 6 Y 2 = − 2 , where X = 2 a + 2 X = 2a + 2 X = 2 a + 2 Y = 2 n − 1 Y = 2n - 1 Y = 2 n − 1 ( x , y ) = ( 22 , 9 ) (x, y) = (22, 9) ( x , y ) = ( 22 , 9 ) X 2 − 6 Y 2 = − 2 X^2 - 6Y^2 = -2 X 2 − 6 Y 2 = − 2 Lemma 2.4 , ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k}, \xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) X 2 − 6 Y 2 = − 2 X^2 - 6Y^2 = -2 X 2 − 6 Y 2 = − 2 ξ 1 , 0 = 22 \xi_{1,0} = 22 ξ 1 , 0 = 22 ξ 2 , 0 = 9 \xi_{2,0} = 9 ξ 2 , 0 = 9 k ≥ 1 k \ge1 k ≥ 1
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . \xi_{1,k} = 5\xi_{1,k-1} + 12\xi_{2,k-1},\qquad\xi_{2,k} = 2\xi_{1,k-1} + 5\xi_{2,k-1}. ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . Since X = 2 a + 2 X = 2a + 2 X = 2 a + 2
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 2 2 : k ≥ 1 , ξ 1 , 0 = 22 , ξ 2 , 0 = 9 } , a \in\left\{\frac{5\xi_{1,k-1} + 12\xi_{2,k-1} - 2}{2}: k \ge1,\ \xi_{1,0} = 22,\ \xi_{2,0} = 9\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 2 : k ≥ 1 , ξ 1 , 0 = 22 , ξ 2 , 0 = 9 } , { a , a + 2 , 6 ( a + 1 ) } \{a, a + 2, 6(a + 1)\} { a , a + 2 , 6 ( a + 1 )} D ( 1 ) D(1) D ( 1 ) { a , a + 2 , 6 ( a + 1 ) , b } \{a, a + 2, 6(a + 1), b\} { a , a + 2 , 6 ( a + 1 ) , b } D ( 1 ) D(1) D ( 1 ) 6 ∤ b 6 \nmid b 6 ∤ b n , m n, m n , m
a b + 1 = S n , ( a + 2 ) b + 1 = S m , ab + 1 = S_n,\qquad(a + 2)b + 1 = S_m, ab + 1 = S n , ( a + 2 ) b + 1 = S m , since 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( m o d 12 ) 6n(n - 1), 6m(m - 1) \equiv0 \pmod{12} 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( mod 12 ) 6 ∣ b 6 \mid b 6 ∣ b 6 ∤ b 6 \nmid b 6 ∤ b { a , a + 2 , 6 ( a + 1 ) , b } \{a, a + 2, 6(a + 1), b\} { a , a + 2 , 6 ( a + 1 ) , b } D ( 1 ) D(1) D ( 1 ) 6 ∤ b 6 \nmid b 6 ∤ b
Proof of Theorem 1.17 . If { a , a + 3 , 8 a + 12 3 } \{a, a + 3, \frac{8a + 12}{3}\} { a , a + 3 , 3 8 a + 12 } D ( 1 ) D(1) D ( 1 )
a ( a + 3 ) + 1 = S n , a ( 8 a + 12 ) 3 + 1 = S m , ( a + 3 ) ( 8 a + 12 ) 3 + 1 = S r , a(a + 3) + 1 = S_n,\qquad\frac{a(8a + 12)}{3} + 1 = S_m,\qquad\frac{(a + 3)(8a + 12)}{3} + 1 = S_r, a ( a + 3 ) + 1 = S n , 3 a ( 8 a + 12 ) + 1 = S m , 3 ( a + 3 ) ( 8 a + 12 ) + 1 = S r , where n , m , r n, m, r n , m , r a ( 8 a + 12 ) 3 + 1 = S m \frac{a(8a + 12)}{3} + 1 = S_m 3 a ( 8 a + 12 ) + 1 = S m m = 2 a + 3 3 m = \frac{2a + 3}{3} m = 3 2 a + 3 ( a + 3 ) ( 8 a + 12 ) 3 + 1 = S r \frac{(a + 3)(8a + 12)}{3} + 1 = S_r 3 ( a + 3 ) ( 8 a + 12 ) + 1 = S r r = 2 a + 6 3 r = \frac{2a + 6}{3} r = 3 2 a + 6 a , n a, n a , n a ( a + 3 ) + 1 = S n a(a + 3) + 1 = S_n a ( a + 3 ) + 1 = S n 3 ∣ a 3 \mid a 3 ∣ a Lemma 2.3 , a ( a + 3 ) + 1 = S n a(a + 3) + 1 = S_n a ( a + 3 ) + 1 = S n
X 2 − 6 Y 2 = 3 , X^2 - 6Y^2 = 3, X 2 − 6 Y 2 = 3 , where X = 2 a + 3 X=2a+3 X = 2 a + 3 Y = 2 n − 1 Y=2n-1 Y = 2 n − 1 ( x , y ) = ( 27 , 11 ) (x,y)=(27,11) ( x , y ) = ( 27 , 11 ) X 2 − 6 Y 2 = 3 X^2-6Y^2=3 X 2 − 6 Y 2 = 3 Lemma 2.4 , ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) X 2 − 6 Y 2 = 3 X^2-6Y^2=3 X 2 − 6 Y 2 = 3 ξ 1 , 0 = 27 \xi_{1,0}=27 ξ 1 , 0 = 27 ξ 2 , 0 = 11 \xi_{2,0}=11 ξ 2 , 0 = 11 k ≥ 1 k\geq1 k ≥ 1
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}. ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . Since X = 2 a + 3 X=2a+3 X = 2 a + 3 3 ∣ X 2 − 6 Y 2 3\mid X^2-6Y^2 3 ∣ X 2 − 6 Y 2 3 ∣ a 3\mid a 3 ∣ a
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 3 2 : k ≥ 1 , ξ 1 , 0 = 27 , ξ 2 , 0 = 11 } , a\in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-3}{2}:k\geq1,\ \xi_{1,0}=27,\ \xi_{2,0}=11\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 3 : k ≥ 1 , ξ 1 , 0 = 27 , ξ 2 , 0 = 11 } , { a , a + 3 , 8 a + 12 3 } \left\{a,a+3,\frac{8a+12}{3}\right\} { a , a + 3 , 3 8 a + 12 } D ( 1 ) D(1) D ( 1 )
If possible, assume that { a , a + 3 , 8 a + 12 3 , b } \left\{a,a+3,\frac{8a+12}{3},b\right\} { a , a + 3 , 3 8 a + 12 , b } D ( 1 ) D(1) D ( 1 ) 4 ∤ b 4\nmid b 4 ∤ b n , m n,m n , m
a b + 1 = S n , ( a + 3 ) b + 1 = S m , ab+1=S_n,\qquad(a+3)b+1=S_m, ab + 1 = S n , ( a + 3 ) b + 1 = S m , since 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( m o d 12 ) 6n(n-1),6m(m-1)\equiv0\pmod{12} 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( mod 12 ) 4 ∣ b 4\mid b 4 ∣ b 4 ∤ b 4\nmid b 4 ∤ b { a , a + 3 , 8 a + 12 3 , b } \left\{a,a+3,\frac{8a+12}{3},b\right\} { a , a + 3 , 3 8 a + 12 , b } D ( 1 ) D(1) D ( 1 ) 4 ∤ b 4\nmid b 4 ∤ b
Proof of Theorem 1.18 . If { a , a + 4 , 3 a + 6 2 } \left\{a,a+4,\frac{3a+6}{2}\right\} { a , a + 4 , 2 3 a + 6 } D ( 1 ) D(1) D ( 1 )
a ( a + 4 ) + 1 = S n , 3 a ( a + 2 ) 2 + 1 = S m , 3 ( a + 4 ) ( a + 2 ) 2 + 1 = S r , a(a+4)+1=S_n,\qquad\frac{3a(a+2)}{2}+1=S_m,\qquad\frac{3(a+4)(a+2)}{2}+1=S_r, a ( a + 4 ) + 1 = S n , 2 3 a ( a + 2 ) + 1 = S m , 2 3 ( a + 4 ) ( a + 2 ) + 1 = S r , where n , m , r n,m,r n , m , r 3 a ( a + 2 ) 2 + 1 = S m \frac{3a(a+2)}{2}+1=S_m 2 3 a ( a + 2 ) + 1 = S m m = a + 2 2 m=\frac{a+2}{2} m = 2 a + 2 3 ( a + 4 ) ( a + 2 ) 2 + 1 = S r \frac{3(a+4)(a+2)}{2}+1=S_r 2 3 ( a + 4 ) ( a + 2 ) + 1 = S r r = a + 4 2 r=\frac{a+4}{2} r = 2 a + 4 a , n a,n a , n a ( a + 4 ) + 1 = S n a(a+4)+1=S_n a ( a + 4 ) + 1 = S n 2 ∣ a 2\mid a 2 ∣ a Lemma 2.3 , a ( a + 4 ) + 1 = S n a(a+4)+1=S_n a ( a + 4 ) + 1 = S n
X 2 − 6 Y 2 = 10. X^2-6Y^2=10. X 2 − 6 Y 2 = 10. where X = 2 a + 4 X=2a+4 X = 2 a + 4 Y = 2 n − 1 Y=2n-1 Y = 2 n − 1 ( x , y ) = ( 8 , 3 ) (x,y)=(8,3) ( x , y ) = ( 8 , 3 ) X 2 − 6 Y 2 = 10 X^2-6Y^2=10 X 2 − 6 Y 2 = 10 Lemma 2.4 , ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) X 1 2 − 6 Y 1 2 = 10 X_1^2-6Y_1^2=10 X 1 2 − 6 Y 1 2 = 10 ξ 1 , 0 = 8 \xi_{1,0}=8 ξ 1 , 0 = 8 ξ 2 , 0 = 3 \xi_{2,0}=3 ξ 2 , 0 = 3 k ≥ 1 k\geq1 k ≥ 1
ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . \xi_{1,k}=5\xi_{1,k-1}+12\xi_{2,k-1},\qquad\xi_{2,k}=2\xi_{1,k-1}+5\xi_{2,k-1}. ξ 1 , k = 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 , ξ 2 , k = 2 ξ 1 , k − 1 + 5 ξ 2 , k − 1 . Since X = 2 a + 4 X=2a+4 X = 2 a + 4 4 ∣ 10 + 6 Y 2 = X 2 4\mid10+6Y^2=X^2 4 ∣ 10 + 6 Y 2 = X 2 2 ∣ a 2\mid a 2 ∣ a
a ∈ { 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 4 2 : k ≥ 1 , ξ 1 , 0 = 8 , ξ 2 , 0 = 3 } , a\in\left\{\frac{5\xi_{1,k-1}+12\xi_{2,k-1}-4}{2}:k\geq1,\ \xi_{1,0}=8,\ \xi_{2,0}=3\right\}, a ∈ { 2 5 ξ 1 , k − 1 + 12 ξ 2 , k − 1 − 4 : k ≥ 1 , ξ 1 , 0 = 8 , ξ 2 , 0 = 3 } , { a , a + 4 , 3 a + 6 2 } \left\{a,a+4,\frac{3a+6}{2}\right\} { a , a + 4 , 2 3 a + 6 } D ( 1 ) D(1) D ( 1 )
If possible, assume that { a , a + 4 , 3 a + 6 2 , b } \left\{a,a+4,\frac{3a+6}{2},b\right\} { a , a + 4 , 2 3 a + 6 , b } D ( 1 ) D(1) D ( 1 ) 3 ∤ b 3\nmid b 3 ∤ b n , m n,m n , m
a b + 1 = S n , ( a + 4 ) b + 1 = S m , (15) ab+1=S_n,\qquad(a+4)b+1=S_m, \tag*{(15)} ab + 1 = S n , ( a + 4 ) b + 1 = S m , (15) Since 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( m o d 12 ) 6n(n-1), 6m(m-1) \equiv0 \pmod{12} 6 n ( n − 1 ) , 6 m ( m − 1 ) ≡ 0 ( mod 12 ) 3 ∣ b 3 \mid b 3 ∣ b 3 ∤ b 3 \nmid b 3 ∤ b { a , a + 4 , 3 a + 6 2 , b } \{a, a+4, \frac{3a+6}{2}, b\} { a , a + 4 , 2 3 a + 6 , b } D ( 1 ) D(1) D ( 1 ) 3 ∤ b 3 \nmid b 3 ∤ b
Proof of Theorem 1.19 . If { 4 , 108 , a } \{4,108,a\} { 4 , 108 , a } D ( 1 ) D(1) D ( 1 ) n , m n,m n , m
4 a + 1 = S n , 108 a + 1 = S m . 4a+1=S_n,\qquad108a+1=S_m. 4 a + 1 = S n , 108 a + 1 = S m . Using Lemma 2.3 , the above equations can be written as
X 1 2 − 6 Y 1 2 = ( a − 4 ) 2 − 6 , X 2 2 − 6 Y 2 2 = ( a − 108 ) 2 − 6 , X_1^2-6Y_1^2=(a-4)^2-6,\qquad X_2^2-6Y_2^2=(a-108)^2-6, X 1 2 − 6 Y 1 2 = ( a − 4 ) 2 − 6 , X 2 2 − 6 Y 2 2 = ( a − 108 ) 2 − 6 , where X 1 = a + 4 X_1=a+4 X 1 = a + 4 Y 1 = 2 n − 1 Y_1=2n-1 Y 1 = 2 n − 1 X 2 = a + 108 X_2=a+108 X 2 = a + 108 Y 2 = 2 m − 1 Y_2=2m-1 Y 2 = 2 m − 1
a = 3 Y 1 2 − 3 8 , a = Y 2 2 − 1 72 . (16) a=\frac{3Y_1^2-3}{8},\qquad a=\frac{Y_2^2-1}{72}. \tag*{(16)} a = 8 3 Y 1 2 − 3 , a = 72 Y 2 2 − 1 . (16) From the above equations, we get
24 a + 9 = 9 Y 1 2 , 72 a + 1 = Y 2 2 , 24a+9=9Y_1^2,\qquad72a+1=Y_2^2, 24 a + 9 = 9 Y 1 2 , 72 a + 1 = Y 2 2 , it follows that
Y 2 2 − 27 Y 1 2 = − 26. (17) Y_2^2-27Y_1^2=-26. \tag*{(17)} Y 2 2 − 27 Y 1 2 = − 26. (17) Since α = 26 + 5 27 \alpha=26+5\sqrt{27} α = 26 + 5 27 x 2 − 27 y 2 = 1 x^2-27y^2=1 x 2 − 27 y 2 = 1 Theorem 2.5 , for each fundamental solution x ∗ + y ∗ 27 x^*+y^*\sqrt{27} x ∗ + y ∗ 27 (17) satisfies
0 ≤ y ∗ ≤ 5 2 ( 26 − 1 ) 26 < 4 , 0\leq y^*\leq\frac{5}{\sqrt{2(26-1)}}\sqrt{26}<4, 0 ≤ y ∗ ≤ 2 ( 26 − 1 ) 5 26 < 4 , ∣ x ∗ ∣ ≤ 1 2 ( 26 − 1 ) 26 < 19. |x^*|\leq\sqrt{\frac{1}{2}(26-1)26}<19. ∣ x ∗ ∣ ≤ 2 1 ( 26 − 1 ) 26 < 19. It is easy to check that (17) is satisfied only for ( x ∗ , y ∗ ) = ( ± 1 , 1 ) (x^*,y^*)=(\pm1,1) ( x ∗ , y ∗ ) = ( ± 1 , 1 ) (17) are ( 1 , 1 ) (1,1) ( 1 , 1 ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 ) 26 + 5 27 26+5\sqrt{27} 26 + 5 27 Z [ 27 ] \mathbb{Z}[\sqrt{27}] Z [ 27 ] k ≥ 0 k\geq0 k ≥ 0 ( ξ 1 , k , ξ 2 , k ) (\xi_{1,k},\xi_{2,k}) ( ξ 1 , k , ξ 2 , k ) ( ξ 1 , k ′ , ξ 2 , k ′ ) (\xi'_{1,k},\xi'_{2,k}) ( ξ 1 , k ′ , ξ 2 , k ′ ) (17) , where
ξ 1 , k + ξ 2 , k 27 = ( − 1 + 27 ) ( 26 + 5 27 ) k . (18) \xi_{1,k}+\xi_{2,k}\sqrt{27}=(-1+\sqrt{27})(26+5\sqrt{27})^k. \tag*{(18)} ξ 1 , k + ξ 2 , k 27 = ( − 1 + 27 ) ( 26 + 5 27 ) k . (18) and
ξ 1 , k ′ + ξ 2 , k ′ 27 = ( 1 + 27 ) ( 26 + 5 27 ) k . (19) \xi'_{1,k}+\xi'_{2,k}\sqrt{27}=(1+\sqrt{27})(26+5\sqrt{27})^k. \tag*{(19)} ξ 1 , k ′ + ξ 2 , k ′ 27 = ( 1 + 27 ) ( 26 + 5 27 ) k . (19) Now, iterating (18) and (19) , for k ≥ 1 k\geq1 k ≥ 1
ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 , \xi_{1,k}=26\xi_{1,k-1}+135\xi_{2,k-1},\qquad\xi_{2,k}=5\xi_{1,k-1}+26\xi_{2,k-1}, ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 , with ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 \xi_{1,0}=-1,\xi_{2,0}=1 ξ 1 , 0 = − 1 , ξ 2 , 0 = 1
ξ 1 , k ′ = 26 ξ 1 , k − 1 ′ + 135 ξ 2 , k − 1 ′ , ξ 2 , k ′ = 5 ξ 1 , k − 1 ′ + 26 ξ 2 , k − 1 ′ , \xi'_{1,k}=26\xi'_{1,k-1}+135\xi'_{2,k-1},\qquad\xi'_{2,k}=5\xi'_{1,k-1}+26\xi'_{2,k-1}, ξ 1 , k ′ = 26 ξ 1 , k − 1 ′ + 135 ξ 2 , k − 1 ′ , ξ 2 , k ′ = 5 ξ 1 , k − 1 ′ + 26 ξ 2 , k − 1 ′ , with ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 \xi'_{1,0}=1,\xi'_{2,0}=1 ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1
Y 1 ∈ A 1 = { ξ 2 , k : k ≥ 0 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } Y_1\in A_1=\{\xi_{2,k}:k\geq0,\ \xi_{1,0}=-1,\ \xi_{2,0}=1\} Y 1 ∈ A 1 = { ξ 2 , k : k ≥ 0 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } where ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 \xi_{1,k}=26\xi_{1,k-1}+135\xi_{2,k-1},\xi_{2,k}=5\xi_{1,k-1}+26\xi_{2,k-1} ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1
Y 1 ∈ A 2 = { ξ 2 , k ′ : k ≥ 0 , ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 } , Y_1\in A_2=\{\xi'_{2,k}:k\geq0,\ \xi'_{1,0}=1,\ \xi'_{2,0}=1\}, Y 1 ∈ A 2 = { ξ 2 , k ′ : k ≥ 0 , ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 } , where ξ 1 , k ′ = 26 ξ 1 , k − 1 ′ + 135 ξ 2 , k − 1 ′ \xi'_{1,k}=26\xi'_{1,k-1}+135\xi'_{2,k-1} ξ 1 , k ′ = 26 ξ 1 , k − 1 ′ + 135 ξ 2 , k − 1 ′ ξ 2 , k ′ = 5 ξ 1 , k − 1 ′ + 26 ξ 2 , k − 1 ′ \xi'_{2,k}=5\xi'_{1,k-1}+26\xi'_{2,k-1} ξ 2 , k ′ = 5 ξ 1 , k − 1 ′ + 26 ξ 2 , k − 1 ′ (16) , we have a = 3 Y 1 2 − 3 8 a=\frac{3Y_1^2-3}{8} a = 8 3 Y 1 2 − 3 ξ 2 , 0 , ξ 2 , 0 ′ = 1 \xi_{2,0},\xi'_{2,0}=1 ξ 2 , 0 , ξ 2 , 0 ′ = 1 a = 0 a=0 a = 0
a ∈ { 3 ( ξ 2 , k 2 − 1 ) 8 : k ≥ 1 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } ∪ { 3 ( ξ 2 , k ′ 2 − 1 ) 8 : k ≥ 1 , ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 } . a \in\left\{\frac{3\left(\xi_{2,k}^2-1\right)}{8}: k\geq1,\xi_{1,0}=-1,\xi_{2,0}=1\right\}\cup\left\{\frac{3\left(\xi'_{2,k}{}^2-1\right)}{8}: k\geq1,\xi'_{1,0}=1,\xi'_{2,0}=1\right\}. a ∈ ⎩ ⎨ ⎧ 8 3 ( ξ 2 , k 2 − 1 ) : k ≥ 1 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 ⎭ ⎬ ⎫ ∪ ⎩ ⎨ ⎧ 8 3 ( ξ 2 , k ′ 2 − 1 ) : k ≥ 1 , ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 ⎭ ⎬ ⎫ . where
ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 , \xi_{1,k}=26\xi_{1,k-1}+135\xi_{2,k-1},\qquad\xi_{2,k}=5\xi_{1,k-1}+26\xi_{2,k-1}, ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 , { 4 , 108 , a } \{4,108,a\} { 4 , 108 , a } D ( 1 ) D(1) D ( 1 )
Proof of Theorem 1.20 . If { 4 , 108 , 165 , a } \{4,108,165,a\} { 4 , 108 , 165 , a } D ( 1 ) D(1) D ( 1 ) n , m , r n,m,r n , m , r
4 a + 1 = S n , 108 a + 1 = S m , 165 a + 1 = S r . 4a+1=S_n,\qquad108a+1=S_m,\qquad165a+1=S_r. 4 a + 1 = S n , 108 a + 1 = S m , 165 a + 1 = S r . Then, using Lemma 2.3 , the above three equations can be written as
X 1 2 − 6 Y 1 2 = ( a − 4 ) 2 − 6 , X 2 2 − 6 Y 2 2 = ( a − 108 ) 2 − 6 , X 3 2 − 6 Y 3 2 = ( a − 165 ) 2 − 6. X_1^2-6Y_1^2=(a-4)^2-6,\qquad X_2^2-6Y_2^2=(a-108)^2-6,\qquad X_3^2-6Y_3^2=(a-165)^2-6. X 1 2 − 6 Y 1 2 = ( a − 4 ) 2 − 6 , X 2 2 − 6 Y 2 2 = ( a − 108 ) 2 − 6 , X 3 2 − 6 Y 3 2 = ( a − 165 ) 2 − 6. where X 1 = a + 4 X_1=a+4 X 1 = a + 4 Y 1 = 2 n − 1 Y_1=2n-1 Y 1 = 2 n − 1 X 2 = a + 108 X_2=a+108 X 2 = a + 108 Y 2 = 2 m − 1 Y_2=2m-1 Y 2 = 2 m − 1 X 3 = a + 165 X_3=a+165 X 3 = a + 165 Y 3 = 2 r − 1 Y_3=2r-1 Y 3 = 2 r − 1
a = 3 Y 1 2 − 3 8 , a = Y 2 2 − 1 72 , a = Y 3 2 − 1 110 . a=\frac{3Y_1^2-3}{8},\qquad a=\frac{Y_2^2-1}{72},\qquad a=\frac{Y_3^2-1}{110}. a = 8 3 Y 1 2 − 3 , a = 72 Y 2 2 − 1 , a = 110 Y 3 2 − 1 . Therefore, from the above three equations, we obtain
Y 2 2 − 27 Y 1 2 = − 26 , 55 Y 2 2 − 36 Y 3 2 = 19. Y_2^2-27Y_1^2=-26,\qquad55Y_2^2-36Y_3^2=19. Y 2 2 − 27 Y 1 2 = − 26 , 55 Y 2 2 − 36 Y 3 2 = 19. Let us consider the following two cases.
Case-1: For the generalized Pell equation Y 2 2 − 27 Y 1 2 = − 26 Y_2^2-27Y_1^2=-26 Y 2 2 − 27 Y 1 2 = − 26 Theorem 1.19 , we get
Y 2 ∈ A 1 = { ξ 1 , k : k ≥ 0 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } Y_2\in A_1=\{\xi_{1,k}:k\geq0,\ \xi_{1,0}=-1,\xi_{2,0}=1\} Y 2 ∈ A 1 = { ξ 1 , k : k ≥ 0 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } where ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1 \xi_{1,k}=26\xi_{1,k-1}+135\xi_{2,k-1},\xi_{2,k}=5\xi_{1,k-1}+26\xi_{2,k-1} ξ 1 , k = 26 ξ 1 , k − 1 + 135 ξ 2 , k − 1 , ξ 2 , k = 5 ξ 1 , k − 1 + 26 ξ 2 , k − 1
Y 2 ∈ A 2 = { ξ 1 , k ′ : k ≥ 0 , ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 } , Y_2\in A_2=\{\xi'_{1,k}:k\geq0,\ \xi'_{1,0}=1,\xi'_{2,0}=1\}, Y 2 ∈ A 2 = { ξ 1 , k ′ : k ≥ 0 , ξ 1 , 0 ′ = 1 , ξ 2 , 0 ′ = 1 } , where ξ 1 , k ′ = 26 ξ 1 , k − 1 ′ + 135 ξ 2 , k − 1 ′ , ξ 2 , k ′ = 5 ξ 1 , k − 1 ′ + 26 ξ 2 , k − 1 ′ \xi'_{1,k}=26\xi'_{1,k-1}+135\xi'_{2,k-1},\quad\xi'_{2,k}=5\xi'_{1,k-1}+26\xi'_{2,k-1} ξ 1 , k ′ = 26 ξ 1 , k − 1 ′ + 135 ξ 2 , k − 1 ′ , ξ 2 , k ′ = 5 ξ 1 , k − 1 ′ + 26 ξ 2 , k − 1 ′ Y 2 = 2 m − 1 ≥ 1 Y_2=2m-1\geq1 Y 2 = 2 m − 1 ≥ 1 ξ 1 , 0 ′ \xi'_{1,0} ξ 1 , 0 ′
Y 2 ∈ A 1 = { ξ 1 , k : k ≥ 1 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } . Y_2\in A_1=\{\xi_{1,k}:k\geq1,\ \xi_{1,0}=-1,\xi_{2,0}=1\}. Y 2 ∈ A 1 = { ξ 1 , k : k ≥ 1 , ξ 1 , 0 = − 1 , ξ 2 , 0 = 1 } . Observe that for k ≥ 0 k\geq0 k ≥ 0
ξ 1 , k + 2 = 52 ξ 1 , k + 1 − ξ 1 , k , ξ 1 , k + 2 ′ = 52 ξ 1 , k + 1 ′ − ξ 1 , k ′ . \xi_{1,k+2}=52\xi_{1,k+1}-\xi_{1,k},\qquad\xi'_{1,k+2}=52\xi'_{1,k+1}-\xi'_{1,k}. ξ 1 , k + 2 = 52 ξ 1 , k + 1 − ξ 1 , k , ξ 1 , k + 2 ′ = 52 ξ 1 , k + 1 ′ − ξ 1 , k ′ . Therefore,
ξ 1 , k ( m o d 8 ) = { 5 if k ≡ 1 , 2 ( m o d 4 ) , 7 if k ≡ 0 , 3 ( m o d 4 ) . \xi_{1,k}\pmod8=
\begin{cases}
5 & \text{if } k\equiv1,2\pmod4,\\
7 & \text{if } k\equiv0,3\pmod4.
\end{cases} ξ 1 , k ( mod 8 ) = { 5 7 if k ≡ 1 , 2 ( mod 4 ) , if k ≡ 0 , 3 ( mod 4 ) . Also, note that for k ≥ 0 k\geq0 k ≥ 0
ξ 1 , k ′ = ( 1 + 27 ) α k + ( 1 − 27 ) α ˉ k 2 , where α = 26 + 5 27 . \xi'_{1,k}=\frac{(1+\sqrt{27})\alpha^k+(1-\sqrt{27})\bar{\alpha}^k}{2},\qquad\text{where }\alpha=26+5\sqrt{27}. ξ 1 , k ′ = 2 ( 1 + 27 ) α k + ( 1 − 27 ) α ˉ k , where α = 26 + 5 27 . Case-2: The generalized Pell equation 55 Y 2 2 − 36 Y 3 2 = 19 55Y_2^2-36Y_3^2=19 55 Y 2 2 − 36 Y 3 2 = 19 Y 4 2 − 1980 Y 2 2 = − 684 Y_4^2-1980Y_2^2=-684 Y 4 2 − 1980 Y 2 2 = − 684 Y 4 = 36 Y 3 Y_4=36Y_3 Y 4 = 36 Y 3 β = 89 + 2 1980 \beta=89+2\sqrt{1980} β = 89 + 2 1980 x 2 − 1980 y 2 = 1 x^2-1980y^2=1 x 2 − 1980 y 2 = 1 Theorem 2.5 , for each fundamental solution x ∗ + y ∗ 1980 x^*+y^*\sqrt{1980} x ∗ + y ∗ 1980 x 2 − 1980 y 2 = − 684 x^2-1980y^2=-684 x 2 − 1980 y 2 = − 684
0 ≤ y ∗ ≤ 2 2 ( 89 − 1 ) 684 < 4 , 0\le y^*\le\frac{2}{\sqrt{2(89-1)}}\sqrt{684}<4, 0 ≤ y ∗ ≤ 2 ( 89 − 1 ) 2 684 < 4 , ∣ x ∗ ∣ ≤ 1 2 ( 89 − 1 ) 684 < 174. |x^*|\le\sqrt{\frac{1}{2}(89-1)684}<174. ∣ x ∗ ∣ ≤ 2 1 ( 89 − 1 ) 684 < 174. It is easy to check that x 2 − 1980 y 2 = − 684 x^2-1980y^2=-684 x 2 − 1980 y 2 = − 684 ( x ∗ , y ∗ ) = ( ± 36 , 1 ) (x^*,y^*)=(\pm36,1) ( x ∗ , y ∗ ) = ( ± 36 , 1 ) 89 + 2 1980 89+2\sqrt{1980} 89 + 2 1980 Z [ 1980 ] \mathbb{Z}[\sqrt{1980}] Z [ 1980 ] k ≥ 0 k\ge0 k ≥ 0 ( η 1 , k , η 2 , k ) (\eta_{1,k},\eta_{2,k}) ( η 1 , k , η 2 , k ) ( η 1 , k ′ , η 2 , k ′ ) (\eta'_{1,k},\eta'_{2,k}) ( η 1 , k ′ , η 2 , k ′ ) Y 4 2 − 1980 Y 2 2 = − 684 Y_4^2-1980Y_2^2=-684 Y 4 2 − 1980 Y 2 2 = − 684
η 1 , k + η 2 , k 1980 = ( − 36 + 1980 ) ( 89 + 2 1980 ) k (20) \eta_{1,k}+\eta_{2,k}\sqrt{1980}=(-36+\sqrt{1980})(89+2\sqrt{1980})^k \tag*{(20)} η 1 , k + η 2 , k 1980 = ( − 36 + 1980 ) ( 89 + 2 1980 ) k (20) and
η 1 , k ′ + η 2 , k ′ 1980 = ( 36 + 1980 ) ( 89 + 2 1980 ) k . (21) \eta'_{1,k}+\eta'_{2,k}\sqrt{1980}=(36+\sqrt{1980})(89+2\sqrt{1980})^k. \tag*{(21)} η 1 , k ′ + η 2 , k ′ 1980 = ( 36 + 1980 ) ( 89 + 2 1980 ) k . (21) Now, iterating (20) and (21) , for k ≥ 1 k\ge1 k ≥ 1
η 1 , k = 89 η 1 , k − 1 + 3960 η 2 , k − 1 , η 2 , k = 2 η 1 , k − 1 + 89 η 2 , k − 1 , \eta_{1,k}=89\eta_{1,k-1}+3960\eta_{2,k-1},\qquad\eta_{2,k}=2\eta_{1,k-1}+89\eta_{2,k-1}, η 1 , k = 89 η 1 , k − 1 + 3960 η 2 , k − 1 , η 2 , k = 2 η 1 , k − 1 + 89 η 2 , k − 1 , with η 1 , 0 = − 36 , η 2 , 0 = 1 \eta_{1,0}=-36,\eta_{2,0}=1 η 1 , 0 = − 36 , η 2 , 0 = 1
η 1 , k ′ = 89 η 1 , k − 1 ′ + 3960 η 2 , k − 1 ′ , η 2 , k ′ = 2 η 1 , k − 1 ′ + 89 η 2 , k − 1 ′ , \eta'_{1,k}=89\eta'_{1,k-1}+3960\eta'_{2,k-1},\qquad\eta'_{2,k}=2\eta'_{1,k-1}+89\eta'_{2,k-1}, η 1 , k ′ = 89 η 1 , k − 1 ′ + 3960 η 2 , k − 1 ′ , η 2 , k ′ = 2 η 1 , k − 1 ′ + 89 η 2 , k − 1 ′ , with η 1 , 0 ′ = 36 , η 2 , 0 ′ = 1 \eta'_{1,0}=36,\eta'_{2,0}=1 η 1 , 0 ′ = 36 , η 2 , 0 ′ = 1
Y 2 ∈ B 1 = { η 2 , k : k ≥ 0 , η 1 , 0 = − 36 , η 2 , 0 = 1 } , Y_2\in B_1=\{\eta_{2,k}:k\ge0,\ \eta_{1,0}=-36,\eta_{2,0}=1\}, Y 2 ∈ B 1 = { η 2 , k : k ≥ 0 , η 1 , 0 = − 36 , η 2 , 0 = 1 } , where
η 1 , k = 89 η 1 , k − 1 + 3960 η 2 , k − 1 , η 2 , k = 2 η 1 , k − 1 + 89 η 2 , k − 1 . \eta_{1,k}=89\eta_{1,k-1}+3960\eta_{2,k-1},\qquad\eta_{2,k}=2\eta_{1,k-1}+89\eta_{2,k-1}. η 1 , k = 89 η 1 , k − 1 + 3960 η 2 , k − 1 , η 2 , k = 2 η 1 , k − 1 + 89 η 2 , k − 1 . Now, using the above two relations, we get
η 2 , k + 2 = 178 η 2 , k + 1 − η 2 , k , \eta_{2,k+2}=178\eta_{2,k+1}-\eta_{2,k}, η 2 , k + 2 = 178 η 2 , k + 1 − η 2 , k , thus, for k ≥ 1 k\ge1 k ≥ 1 η 2 , k ≡ 1 ( m o d 8 ) \eta_{2,k}\equiv1\pmod{8} η 2 , k ≡ 1 ( mod 8 )
Y 2 ∈ B 2 = { η 2 , k ′ : k ≥ 0 , η 1 , 0 ′ = 36 , η 2 , 0 ′ = 1 } , Y_2\in B_2=\{\eta'_{2,k}:k\ge0,\ \eta'_{1,0}=36,\eta'_{2,0}=1\}, Y 2 ∈ B 2 = { η 2 , k ′ : k ≥ 0 , η 1 , 0 ′ = 36 , η 2 , 0 ′ = 1 } , where
η 1 , k ′ = 89 η 1 , k − 1 ′ + 3960 η 2 , k − 1 ′ , η 2 , k ′ = 2 η 1 , k − 1 ′ + 89 η 2 , k − 1 ′ . \eta'_{1,k}=89\eta'_{1,k-1}+3960\eta'_{2,k-1},\qquad\eta'_{2,k}=2\eta'_{1,k-1}+89\eta'_{2,k-1}. η 1 , k ′ = 89 η 1 , k − 1 ′ + 3960 η 2 , k − 1 ′ , η 2 , k ′ = 2 η 1 , k − 1 ′ + 89 η 2 , k − 1 ′ . Now, using the above two relations, we get
η 2 , k + 2 ′ = 178 η 2 , k + 1 ′ − η 2 , k ′ , \eta'_{2,k+2}=178\eta'_{2,k+1}-\eta'_{2,k}, η 2 , k + 2 ′ = 178 η 2 , k + 1 ′ − η 2 , k ′ , thus, for k ≥ 1 k\ge1 k ≥ 1 η 2 , k ′ ≡ 1 ( m o d 8 ) \eta'_{2,k}\equiv1\pmod{8} η 2 , k ′ ≡ 1 ( mod 8 ) A 1 ∩ ( B 1 ∪ B 2 ) = ∅ A_1\cap(B_1\cup B_2)=\varnothing A 1 ∩ ( B 1 ∪ B 2 ) = ∅ k ≥ 0 k\ge0 k ≥ 0
ξ 2 , k = ( − 36 + 1980 ) β k + ( 36 + 1980 ) β ‾ k 2 1980 , ξ 2 , k ′ = ( 36 + 1980 ) β k − ( 36 − 1980 ) β ‾ k 2 1980 , \xi_{2,k}=\frac{(-36+\sqrt{1980})\beta^k+(36+\sqrt{1980})\overline{\beta}^k}{2\sqrt{1980}},\qquad\xi'_{2,k}=\frac{(36+\sqrt{1980})\beta^k-(36-\sqrt{1980})\overline{\beta}^k}{2\sqrt{1980}}, ξ 2 , k = 2 1980 ( − 36 + 1980 ) β k + ( 36 + 1980 ) β k , ξ 2 , k ′ = 2 1980 ( 36 + 1980 ) β k − ( 36 − 1980 ) β k , where β = 89 + 2 1980 \beta=89+2\sqrt{1980} β = 89 + 2 1980
First few elements of A 2 A_2 A 2 B 1 B_1 B 1 B 2 B_2 B 2
A 2 = { 1 , 161 , 8371 , 435131 , 22618441 , 1175723801 , … } , A_2=\{1,161,8371,435131,22618441,1175723801,\ldots\}, A 2 = { 1 , 161 , 8371 , 435131 , 22618441 , 1175723801 , … } , B 1 = { 1 , 17 , 3025 , 538433 , 95838049 , 17058634289 , … } , B_1=\{1,17,3025,538433,95838049,17058634289,\ldots\}, B 1 = { 1 , 17 , 3025 , 538433 , 95838049 , 17058634289 , … } , B 2 = { 1 , 161 , 28657 , 5100785 , 907911073 , 161603070209 , … } . B_2=\{1,161,28657,5100785,907911073,161603070209,\ldots\}. B 2 = { 1 , 161 , 28657 , 5100785 , 907911073 , 161603070209 , … } . Now, we prove that A 2 ∩ B 2 = { 1 , 161 } A_2 \cap B_2 = \{1,161\} A 2 ∩ B 2 = { 1 , 161 } A 2 ∩ B 1 = { 1 } A_2 \cap B_1 = \{1\} A 2 ∩ B 1 = { 1 } Y 2 = 1 Y_2 = 1 Y 2 = 1 Y 2 = 1 Y_2 = 1 Y 2 = 1 a = 0 a = 0 a = 0 a = 360 a = 360 a = 360 Y 2 = 161 Y_2 = 161 Y 2 = 161
C = ( 1 + 27 ) 1980 , C ′ = ( 1 − 27 ) 1980 , D = ( 36 + 1980 ) , D ′ = ( 36 − 1980 ) . C = (1+\sqrt{27})\sqrt{1980}, \qquad C' = (1-\sqrt{27})\sqrt{1980}, \qquad D = (36+\sqrt{1980}), \qquad D' = (36-\sqrt{1980}). C = ( 1 + 27 ) 1980 , C ′ = ( 1 − 27 ) 1980 , D = ( 36 + 1980 ) , D ′ = ( 36 − 1980 ) . Consider the following two cases.
Case-3: Since Y 2 ≠ 1 Y_2 \ne1 Y 2 = 1 ξ 1 , n ′ = η 2 , m ′ \xi'_{1,n} = \eta'_{2,m} ξ 1 , n ′ = η 2 , m ′ n , m ≥ 2 n,m \ge2 n , m ≥ 2 β n / 2 < ξ 1 , n ′ = ξ 2 , m ′ < β m \beta^{n/2} < \xi'_{1,n} = \xi'_{2,m} < \beta^m β n /2 < ξ 1 , n ′ = ξ 2 , m ′ < β m n < 2 m n < 2m n < 2 m
( 1 + 27 ) α n + ( 1 − 27 ) α ˉ n 2 = D β m − D ′ β ˉ m 2 1980 , \frac{(1+\sqrt{27})\alpha^n + (1-\sqrt{27})\bar{\alpha}^n}{2} = \frac{D\beta^m-D'\bar{\beta}^m}{2\sqrt{1980}}, 2 ( 1 + 27 ) α n + ( 1 − 27 ) α ˉ n = 2 1980 D β m − D ′ β ˉ m , which implies that
C α n − D β m = − C ′ α ˉ n − D ′ β ˉ m . C\alpha^n-D\beta^m=-C'\bar{\alpha}^n-D'\bar{\beta}^m. C α n − D β m = − C ′ α ˉ n − D ′ β ˉ m . Since 0 < α ˉ n < 1 0 < \bar{\alpha}^n < 1 0 < α ˉ n < 1 0 < β ˉ m < 1 0 < \bar{\beta}^m < 1 0 < β ˉ m < 1
∣ C α n − D β m ∣ ≤ ∣ C ′ ∣ α ˉ n + ∣ D ′ ∣ β ˉ m < 196. |C\alpha^n-D\beta^m| \le|C'|\bar{\alpha}^n+|D'|\bar{\beta}^m < 196. ∣ C α n − D β m ∣ ≤ ∣ C ′ ∣ α ˉ n + ∣ D ′ ∣ β ˉ m < 196. Then
∣ C α n D β m − 1 ∣ < 196 D β m < 3 β m . (22) \left|\frac{C\alpha^n}{D\beta^m}-1\right| < \frac{196}{D\beta^m} < \frac{3}{\beta^m}. \tag*{(22)} D β m C α n − 1 < D β m 196 < β m 3 . (22) Now, let us apply Theorem 2.6 with ( γ 1 , b 1 ) = ( C D , 1 ) (\gamma_1,b_1) = \left(\frac{C}{D},1\right) ( γ 1 , b 1 ) = ( D C , 1 ) ( γ 2 , b 2 ) = ( α , n ) (\gamma_2,b_2) = (\alpha,n) ( γ 2 , b 2 ) = ( α , n ) ( γ 3 , b 3 ) = ( β , − m ) (\gamma_3,b_3) = (\beta,-m) ( γ 3 , b 3 ) = ( β , − m )
Λ = C α n D β m − 1. \Lambda= \frac{C\alpha^n}{D\beta^m}-1. Λ = D β m C α n − 1. We claim that Λ ≠ 0 \Lambda\ne0 Λ = 0 n , m ≥ 2 n,m \ge2 n , m ≥ 2 Λ = 0 \Lambda= 0 Λ = 0 n , m ≥ 2 n,m \ge2 n , m ≥ 2 C α n = D β m C\alpha^n = D\beta^m C α n = D β m α = 26 + 5 27 \alpha= 26+5\sqrt{27} α = 26 + 5 27 α n ∈ Q ( 3 ) \alpha^n \in\mathbb{Q}(\sqrt{3}) α n ∈ Q ( 3 ) β = 89 + 12 55 \beta= 89+12\sqrt{55} β = 89 + 12 55 β m ∈ Q ( 55 ) \beta^m \in\mathbb{Q}(\sqrt{55}) β m ∈ Q ( 55 ) C α n = D β m C\alpha^n = D\beta^m C α n = D β m
( 1 + 27 ) α n = D β m 6 55 ∈ Q ( 55 ) , (1+\sqrt{27})\alpha^n = \frac{D\beta^m}{6\sqrt{55}} \in\mathbb{Q}(\sqrt{55}), ( 1 + 27 ) α n = 6 55 D β m ∈ Q ( 55 ) , consequently, we get ( 1 + 27 ) α n ∈ Q ( 3 ) ∩ Q ( 55 ) = Q (1+\sqrt{27})\alpha^n \in\mathbb{Q}(\sqrt{3}) \cap\mathbb{Q}(\sqrt{55}) = \mathbb{Q} ( 1 + 27 ) α n ∈ Q ( 3 ) ∩ Q ( 55 ) = Q
Note that γ 1 , γ 2 , γ 3 ∈ Q ( 3 , 55 ) \gamma_1,\gamma_2,\gamma_3 \in\mathbb{Q}(\sqrt{3},\sqrt{55}) γ 1 , γ 2 , γ 3 ∈ Q ( 3 , 55 ) d = [ Q ( 3 , 55 ) : Q ] = 4 d = [\mathbb{Q}(\sqrt{3},\sqrt{55}):\mathbb{Q}] = 4 d = [ Q ( 3 , 55 ) : Q ] = 4
B = 2 m ≥ max { ∣ b 1 ∣ , ∣ b 2 ∣ , ∣ b 3 ∣ } , B = 2m \ge\max\{|b_1|,|b_2|,|b_3|\}, B = 2 m ≥ max { ∣ b 1 ∣ , ∣ b 2 ∣ , ∣ b 3 ∣ } , A 2 = 8 > 2 log ( α ) ≥ max { D h ( γ 2 ) , ∣ log γ 2 ∣ , 0.16 } A_2 = 8 > 2\log(\alpha) \ge\max\{Dh(\gamma_2),|\log\gamma_2|,0.16\} A 2 = 8 > 2 log ( α ) ≥ max { D h ( γ 2 ) , ∣ log γ 2 ∣ , 0.16 } A 3 = 11 > 2 log ( β ) ≥ max { D h ( γ 3 ) , ∣ log γ 3 ∣ , 0.16 } . A_3 = 11 > 2\log(\beta) \ge\max\{Dh(\gamma_3),|\log\gamma_3|,0.16\}. A 3 = 11 > 2 log ( β ) ≥ max { D h ( γ 3 ) , ∣ log γ 3 ∣ , 0.16 } . Since h ( γ 1 ) ≤ h ( C ) + h ( D ) = 1 2 ( log 51480 + log 684 ) < 9 h(\gamma_1) \le h(C)+h(D) = \frac{1}{2}(\log51480+\log684)<9 h ( γ 1 ) ≤ h ( C ) + h ( D ) = 2 1 ( log 51480 + log 684 ) < 9
A 1 = 36 > max { D h ( γ 1 ) , ∣ log γ 1 ∣ , 0.16 } . A_1 = 36 > \max\{Dh(\gamma_1),|\log\gamma_1|,0.16\}. A 1 = 36 > max { D h ( γ 1 ) , ∣ log γ 1 ∣ , 0.16 } . From Theorem 2.6 and (21), we obtain
exp ( − 1.4 ⋅ 30 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) ) < ∣ Λ ∣ < 3 β m , \exp\left(-1.4\cdot30^6\cdot3^{4.5}\cdot16\cdot(1+\log4)\cdot A_1\cdot A_2\cdot A_3\cdot(1+\log2m)\right)<|\Lambda|<\frac{3}{\beta^m}, exp ( − 1.4 ⋅ 3 0 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) ) < ∣Λ∣ < β m 3 , it follows that
− 1.4 ⋅ 30 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < log 3 − m log β , -1.4\cdot30^6\cdot3^{4.5}\cdot16\cdot(1+\log4)\cdot A_1\cdot A_2\cdot A_3\cdot(1+\log2m)<\log3-m\log\beta, − 1.4 ⋅ 3 0 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < log 3 − m log β , which gives
m log β < log 3 + 1.4 ⋅ 30 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < 2 ⋅ 10 16 ⋅ ( 1 + log 2 m ) . \begin{aligned}
m \log\beta< \log3 + 1.4 \cdot30^{6} \cdot3^{4.5} \cdot16 \cdot(1 + \log4) \cdot A_{1} \cdot A_{2} \cdot A_{3} \cdot(1 + \log2m) \\
&< 2 \cdot10^{16} \cdot(1 + \log2m).
\end{aligned} m log β < log 3 + 1.4 ⋅ 3 0 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < 2 ⋅ 1 0 16 ⋅ ( 1 + log 2 m ) . The above inequality implies that m < 1.6 ⋅ 10 17 m < 1.6 \cdot10^{17} m < 1.6 ⋅ 1 0 17 n n n
Let z = log ( C D ) + n log α − m log β z = \log\left(\frac{C}{D}\right) + n \log\alpha- m \log\beta z = log ( D C ) + n log α − m log β z ≠ 0 z \ne0 z = 0 Λ ≠ 0 \Lambda\ne0 Λ = 0 n , m ≥ 2 n,m \ge2 n , m ≥ 2 z > 0 z > 0 z > 0 (22) , we have
0 < z ≤ e z − 1 < 3 β m . 0 < z \le e^{z} - 1 < \frac{3}{\beta^{m}}. 0 < z ≤ e z − 1 < β m 3 . If z < 0 z < 0 z < 0 (22) , we have
∣ e z − 1 ∣ < 3 β m < 1 2 , for m ≥ 1. \left|e^{z} - 1\right| < \frac{3}{\beta^{m}} < \frac{1}{2}, \qquad\text{for } m \ge1. ∣ e z − 1 ∣ < β m 3 < 2 1 , for m ≥ 1. Since for any nonzero real number x < 0 x < 0 x < 0 ∣ e x − 1 ∣ < 1 2 \left|e^{x} - 1\right| < \frac{1}{2} ∣ e x − 1 ∣ < 2 1 ∣ x ∣ < 2 ∣ e x − 1 ∣ \lvert x\rvert< 2\left|e^{x} - 1\right| ∣ x ∣ < 2 ∣ e x − 1 ∣ m ≥ 1 m \ge1 m ≥ 1
∣ z ∣ < 2 ∣ e z − 1 ∣ < 6 β m . \lvert z\rvert< 2\left|e^{z} - 1\right| < \frac{6}{\beta^{m}}. ∣ z ∣ < 2 ∣ e z − 1 ∣ < β m 6 . From the above inequality, we get
∣ log ( C D ) + n log α − m log β ∣ < 6 β m . \left|\log\left(\frac{C}{D}\right) + n \log\alpha- m \log\beta\right| < \frac{6}{\beta^{m}}. log ( D C ) + n log α − m log β < β m 6 . After multiplying both sides of the above inequality by 1 log β \frac{1}{\log\beta} l o g β 1
∣ log ( C D ) log β + n log α log β − m ∣ < 6 log β ⋅ β m < 2 β m . \left|\frac{\log\left(\frac{C}{D}\right)}{\log\beta} + n\frac{\log\alpha}{\log\beta} - m\right| < \frac{6}{\log\beta\cdot\beta^{m}} < \frac{2}{\beta^{m}}. log β log ( D C ) + n log β log α − m < log β ⋅ β m 6 < β m 2 . Now, let us apply Lemma 2.7 with
u = n , γ = log α log β , v = m , μ = log ( C D ) log β , A = 2 , B = β , w = m . u=n,\qquad\gamma=\frac{\log\alpha}{\log\beta},\qquad v=m,\qquad\mu=\frac{\log\left(\frac{C}{D}\right)}{\log\beta},\qquad A=2,\qquad B=\beta,\qquad w=m. u = n , γ = log β log α , v = m , μ = log β log ( D C ) , A = 2 , B = β , w = m . Since m < 1.6 ⋅ 10 17 m < 1.6 \cdot10^{17} m < 1.6 ⋅ 1 0 17 u = n < 2 m < 3.2 ⋅ 10 17 u=n<2m<3.2 \cdot10^{17} u = n < 2 m < 3.2 ⋅ 1 0 17 M = 3.2 ⋅ 10 17 M=3.2 \cdot10^{17} M = 3.2 ⋅ 1 0 17
p q = p 39 q 39 = 446050159788405153046 585015235298117452451 , \frac{p}{q}=\frac{p_{39}}{q_{39}}=\frac{446050159788405153046}{585015235298117452451}, q p = q 39 p 39 = 585015235298117452451 446050159788405153046 , we get q 39 > 6 M q_{39}>6M q 39 > 6 M
ϵ = ∥ μ q ∥ − M ∥ γ q ∥ = 0.1886 … > 0. \epsilon=\|\mu q\|-M\|\gamma q\|=0.1886\ldots>0. ϵ = ∥ μ q ∥ − M ∥ γ q ∥ = 0.1886 … > 0. Therefore, by Lemma 2.7, we have
m < log ( A q / ϵ ) log B < 10. m < \frac{\log(Aq/\epsilon)}{\log B}<10. m < log B log ( A q / ϵ ) < 10. Therefore, m ≤ 9 m \le9 m ≤ 9 2 ≤ m ≤ 9 2 \le m \le9 2 ≤ m ≤ 9 ξ 1 , 2 ′ = η 2 , 2 ′ = 161 \xi'_{1,2}=\eta'_{2,2}=161 ξ 1 , 2 ′ = η 2 , 2 ′ = 161
Case-4: Since Y 2 ≠ 1 Y_2 \ne1 Y 2 = 1 ξ 1 , n ′ = η 2 , m \xi'_{1,n}=\eta_{2,m} ξ 1 , n ′ = η 2 , m n , m ≥ 2 n,m \ge2 n , m ≥ 2 β n / 2 < ξ 1 , n ′ = ξ 2 , m < β m \beta^{n/2}<\xi'_{1,n}=\xi_{2,m}<\beta^m β n /2 < ξ 1 , n ′ = ξ 2 , m < β m n < 2 m n<2m n < 2 m
( 1 + 27 ) α n + ( 1 − 27 ) α ˉ n 2 = − D ′ β m + D β ˉ m 2 1980 . \frac{(1+\sqrt{27})\alpha^n+(1-\sqrt{27})\bar{\alpha}^n}{2}=\frac{-D'\beta^m+D\bar{\beta}^m}{2\sqrt{1980}}. 2 ( 1 + 27 ) α n + ( 1 − 27 ) α ˉ n = 2 1980 − D ′ β m + D β ˉ m . which implies that
C α n + D ′ β m = − C ′ α ˉ n + D β ˉ m . C\alpha^n+D'\beta^m=-C'\bar{\alpha}^n+D\bar{\beta}^m. C α n + D ′ β m = − C ′ α ˉ n + D β ˉ m . Since 0 < α ˉ n < 1 0<\bar{\alpha}^n<1 0 < α ˉ n < 1 0 < β ˉ m < 1 0<\bar{\beta}^m<1 0 < β ˉ m < 1
∣ C α n + D ′ β m ∣ ≤ ∣ C ′ α ˉ n + D β ˉ m ∣ < 268. |C\alpha^n+D'\beta^m|\le|C'\bar{\alpha}^n+D\bar{\beta}^m|<268. ∣ C α n + D ′ β m ∣ ≤ ∣ C ′ α ˉ n + D β ˉ m ∣ < 268. Then
∣ C α n ( − D ′ ) β m − 1 ∣ < 268 ∣ D ′ ∣ β m < 32 β m . (23) \left|\frac{C\alpha^n}{(-D')\beta^m}-1\right|<\frac{268}{|D'|\beta^m}<\frac{32}{\beta^m}. \tag*{(23)} ( − D ′ ) β m C α n − 1 < ∣ D ′ ∣ β m 268 < β m 32 . (23) Now, let us apply Theorem 2.6 with ( γ 1 , b 1 ) = ( C − D ′ , 1 ) (\gamma_1,b_1)=\left(\frac{C}{-D'},1\right) ( γ 1 , b 1 ) = ( − D ′ C , 1 ) ( γ 2 , b 2 ) = ( α , n ) (\gamma_2,b_2)=(\alpha,n) ( γ 2 , b 2 ) = ( α , n ) ( γ 3 , b 3 ) = ( β , − m ) (\gamma_3,b_3)=(\beta,-m) ( γ 3 , b 3 ) = ( β , − m )
Λ = C α n ( − D ′ ) β m − 1. \Lambda=\frac{C\alpha^n}{(-D')\beta^m}-1. Λ = ( − D ′ ) β m C α n − 1. A similar argument to that used in Case-3 proves that Λ ≠ 0 \Lambda\ne0 Λ = 0 γ 1 , γ 2 , γ 3 ∈ Q ( 3 , 55 ) \gamma_1,\gamma_2,\gamma_3\in\mathbb{Q}(\sqrt{3},\sqrt{55}) γ 1 , γ 2 , γ 3 ∈ Q ( 3 , 55 ) d = [ Q ( 3 , 55 ) : Q ] = 4 d=[\mathbb{Q}(\sqrt{3},\sqrt{55}):\mathbb{Q}]=4 d = [ Q ( 3 , 55 ) : Q ] = 4
B = 2 m ≥ max { ∣ b 1 ∣ , ∣ b 2 ∣ , ∣ b 3 ∣ } , A 2 = 8 > 2 log ( α ) ≥ max { D h ( γ 2 ) , ∣ log γ 2 ∣ , 0.16 } , B=2m\ge\max\{|b_1|,|b_2|,|b_3|\},\qquad A_2=8>2\log(\alpha)\ge\max\{Dh(\gamma_2),|\log\gamma_2|,0.16\}, B = 2 m ≥ max { ∣ b 1 ∣ , ∣ b 2 ∣ , ∣ b 3 ∣ } , A 2 = 8 > 2 log ( α ) ≥ max { D h ( γ 2 ) , ∣ log γ 2 ∣ , 0.16 } , A 3 = 11 > 2 log ( β ) ≥ max { D h ( γ 3 ) , ∣ log γ 3 ∣ , 0.16 } . A_3=11>2\log(\beta)\ge\max\{Dh(\gamma_3),|\log\gamma_3|,0.16\}. A 3 = 11 > 2 log ( β ) ≥ max { D h ( γ 3 ) , ∣ log γ 3 ∣ , 0.16 } . Since h ( D ′ ) = h ( D ) h(D')=h(D) h ( D ′ ) = h ( D ) h ( γ 1 ) ≤ h ( C ) + h ( D ′ ) = h ( C ) + h ( D ) < 9 h(\gamma_1)\le h(C)+h(D')=h(C)+h(D)<9 h ( γ 1 ) ≤ h ( C ) + h ( D ′ ) = h ( C ) + h ( D ) < 9 A 1 = 36 A_1=36 A 1 = 36 Theorem 2.6 and (23) , we obtain
exp ( − 1.4 ⋅ 30 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) ) < ∣ Λ ∣ < 32 β m . \exp\left(-1.4\cdot30^6\cdot3^{4.5}\cdot16\cdot(1+\log4)\cdot A_1\cdot A_2\cdot A_3\cdot(1+\log2m)\right)<|\Lambda|<\frac{32}{\beta^m}. exp ( − 1.4 ⋅ 3 0 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) ) < ∣Λ∣ < β m 32 . it follows that
− 1.4 ⋅ 30 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < log 32 − m log β , -1.4\cdot30^6\cdot3^{4.5}\cdot16\cdot(1+\log4)\cdot A_1\cdot A_2\cdot A_3\cdot(1+\log2m)<\log32-m\log\beta, − 1.4 ⋅ 3 0 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < log 32 − m log β , which gives
m log β < log 32 + 1.4 ⋅ 30 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < 2 ⋅ 10 16 ⋅ ( 1 + log 2 m ) . m\log\beta<\log32+1.4\cdot30^6\cdot3^{4.5}\cdot16\cdot(1+\log4)\cdot A_1\cdot A_2\cdot A_3\cdot(1+\log2m)
<2\cdot10^{16}\cdot(1+\log2m). m log β < log 32 + 1.4 ⋅ 3 0 6 ⋅ 3 4.5 ⋅ 16 ⋅ ( 1 + log 4 ) ⋅ A 1 ⋅ A 2 ⋅ A 3 ⋅ ( 1 + log 2 m ) < 2 ⋅ 1 0 16 ⋅ ( 1 + log 2 m ) . The above inequality implies that m < 1.6 ⋅ 10 17 m<1.6\cdot10^{17} m < 1.6 ⋅ 1 0 17 n n n
Let z = log ( C − D ′ ) + n log α − m log β z=\log\left(\frac{C}{-D'}\right)+n\log\alpha-m\log\beta z = log ( − D ′ C ) + n log α − m log β z ≠ 0 z\ne0 z = 0 Λ ≠ 0 \Lambda\ne0 Λ = 0 n , m ≥ 2 n,m\ge2 n , m ≥ 2 z > 0 z>0 z > 0 (23) , we have
0 < z ≤ e z − 1 < 32 β m . 0<z\le e^z-1<\frac{32}{\beta^m}. 0 < z ≤ e z − 1 < β m 32 . If z < 0 z<0 z < 0 ∣ e z − 1 ∣ < 32 β m < 1 2 |e^z-1|<\frac{32}{\beta^m}<\frac{1}{2} ∣ e z − 1∣ < β m 32 < 2 1 m ≥ 2 m\ge2 m ≥ 2 ∣ z ∣ < 2 ∣ e z − 1 ∣ < 64 β m |z|<2|e^z-1|<\frac{64}{\beta^m} ∣ z ∣ < 2∣ e z − 1∣ < β m 64
∣ log ( C − D ′ ) + n log α − m log β ∣ < 64 β m . \left|\log\left(\frac{C}{-D'}\right)+n\log\alpha-m\log\beta\right|<\frac{64}{\beta^m}. log ( − D ′ C ) + n log α − m log β < β m 64 . After multiplying both sides of the above inequality by 1 log β \frac{1}{\log\beta} l o g β 1
∣ log ( C − D ′ ) log β + n log α log β − m ∣ < 64 log β ⋅ β m < 13 β m . \left|\frac{\log\left(\frac{C}{-D'}\right)}{\log\beta}+n\frac{\log\alpha}{\log\beta}-m\right|<\frac{64}{\log\beta\cdot\beta^m}<\frac{13}{\beta^m}. log β log ( − D ′ C ) + n log β log α − m < log β ⋅ β m 64 < β m 13 . Now, let us apply Lemma 2.7 with
u = n , γ = log α log β , v = m , μ = log ( C − D ′ ) log β , A = 13 , B = β , w = m . u=n,\qquad\gamma=\frac{\log\alpha}{\log\beta},\qquad v=m,\qquad\mu=\frac{\log\left(\frac{C}{-D'}\right)}{\log\beta},\qquad A=13,\qquad B=\beta,\qquad w=m. u = n , γ = log β log α , v = m , μ = log β log ( − D ′ C ) , A = 13 , B = β , w = m . Since m < 1.6 ⋅ 10 17 m<1.6\cdot10^{17} m < 1.6 ⋅ 1 0 17 u = n < 2 m < 3.2 ⋅ 10 17 u=n<2m<3.2\cdot10^{17} u = n < 2 m < 3.2 ⋅ 1 0 17 M = 3.2 ⋅ 10 17 M=3.2\cdot10^{17} M = 3.2 ⋅ 1 0 17
p q = p 37 q 37 = 2487021452540632304 3261842661237276437 , \frac{p}{q}=\frac{p_{37}}{q_{37}}=\frac{2487021452540632304}{3261842661237276437}, q p = q 37 p 37 = 3261842661237276437 2487021452540632304 , we get q 37 > 6 M q_{37}>6M q 37 > 6 M
ϵ = ∥ μ q ∥ − M ∥ γ q ∥ = 0.3987 … > 0. \epsilon=\| \mu q\|-M\|\gamma q\|=0.3987\ldots>0. ϵ = ∥ μ q ∥ − M ∥ γ q ∥ = 0.3987 … > 0. Therefore, by Lemma 2.7, we have
m < log ( A q / ϵ ) log B < 9. m<\frac{\log(Aq/\epsilon)}{\log B}<9. m < log B log ( A q / ϵ ) < 9. Therefore, m ≤ 8 m\le8 m ≤ 8 2 ≤ m ≤ 8 2\le m\le8 2 ≤ m ≤ 8 ξ 1 , n ≠ η 2 , m \xi_{1,n}\ne\eta_{2,m} ξ 1 , n = η 2 , m
Proof of Corollary 1.21. If possible assume that there exists positive integers a ≠ b a\ne b a = b { 4 , 108 , 165 , a , b } \{4,108,165,a,b\} { 4 , 108 , 165 , a , b } D ( 1 ) D(1) D ( 1 ) { 4 , 108 , 165 , a } \{4,108,165,a\} { 4 , 108 , 165 , a } { 4 , 108 , 165 , b } \{4,108,165,b\} { 4 , 108 , 165 , b } D ( 1 ) D(1) D ( 1 ) Theorem 1.20 , we have a = b = 360 a=b=360 a = b = 360
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References
Department of Mathematics,, Indian Institute of Technology Ropar, Punjab, India. Email address: tapasc@iitrpr.ac.in
Department of Mathematics, Indian Institute of Technology Ropar, Punjab, India. Email address: sagar.25maz0008@iitrpr.ac.in, sagarmandal31415@gmail.com
