The exponent of $p$ in $x^{p-1}-1$: Yamada's constant 283 lowered to 69

Result

For a prime $p$ and an integer $x$ coprime to $p$, the exponent $v_p(x^{p-1}-1)$ equals 1 unless $p$ is a Wieferich prime to base $x$. The trivial bound $v_p(x^{p-1}-1)<(p-1)\log x/\log p$ comes from $p^v\leq x^{p-1}-1$. The theorems of Bugeaud–Laurent [1] on linear forms in two $p$-adic logarithms assume $\log A_i\geq\log p$ and give nothing better here. Yamada [3] removed this obstacle and obtained a non-trivial bound, with the constant 283.

The corollary follows with $(x,y)=(q,2)$ and $(x,y)=(2,3)$. Theorem 4 of [3], on $v_p(\sigma(q^c))$, is deduced from Theorem 2 of [3] by an inequality of Nagell, valid for $p\notin\{2,q\}$ (and $p\ne2,3$ for $\sigma(2^c)$); in that range it therefore holds with 69 in place of 283.

	constant	additive term
Yamada 2010 [3], Theorem 2	283	4
Theorem 1	69	4

Theorem 1 gives nothing when $\log p\leq69\log y\log(xy)/\log x$, since the trivial bound is then better. For $y=2$ and $x$ large it becomes non-trivial as soon as $p>2^{69(1+o(1))}$, against $p>2^{283(1+o(1))}$ in [3].

A rearrangement lemma

Fix integers $K,L \ge1$, put $N=KL$ and $l_j=\lfloor(j-1)/K \rfloor$ for $1 \le j \le N$: each value $l \in\{0,\ldots,L-1\}$ is taken exactly $K$ times.

Proof. Put $\bar l=(L-1)/2$. Since $\sum_i(l_i-\bar l)=K\sum_{l=0}^{L-1}(l-\bar l)=0$,

whose absolute value is at most $\frac{R-1}{2}\sum_i|l_i-\bar l|=\frac{R-1}{2}K\sum_{l=0}^{L-1}|l-\frac{L-1}{2}|=\frac{R-1}{2}K\lfloor L^2/4\rfloor$. In the configuration indicated, the increasing matching gives equality.

Lemma 10 of [1] bounds the same quantity by $\frac{NL(R+g-1)}{8}-\frac{gN^2L}{24(S+g-1)}$, for pairs $(r_j,s_j)$ in a single class $m_1r+m_2s\equiv c\pmod g$. Lemma 6 of [3] replaces it by $G_Y=NL(R-1)/4$. Lemma 3 gives $G=NL(R-1)/8$ for $L$ even and $G=NL(R-1)(1-L^{-2})/8$ for $L$ odd, with no hypothesis on the $r_j$.

The improved criterion

For $L\ge1$ put

Proof. This is the proof of Theorem 1 of [1], in the case $\mathbb{Q}_p$: $D=e=f=1$, $t=0$ because $p$ is odd, and $u=0$ because $\gcd(b_1,b_2,p)=1$. Only the arithmetic lower bound changes. Assume $v_p(\Lambda)\ge N-\frac{1}{2}$. The matrix $M$ formed by the columns $(r,s)$, $0\le r<R$, $0\le s<S$, $m_1r+m_2s\equiv c_1+c_2$, with entries $(r b_2+s b_1)\alpha_1^{lr}\alpha_2^{ls}$ (rows $(k,l)$, $0\le k<K$, $0\le l<L$) has rank $N$ by the zero lemma ([1], Lemmas 6 and 7, which use only (1)–(2)). Let $\Delta\ne0$ be a minor of order $N$. The analytic upper bound ([1], Lemma 8) gives

It uses the decomposition $\alpha_i=\zeta^{m_i}\theta_i$, $\theta_i\in1+p\mathbb{Z}_p$, and the fact that all columns lie in one class. It does not use the condition $\gcd(m_1,m_2,g)=1$, which serves only in Lemma 10 of [1].

Arithmetic lower bound. Index the columns of $\Delta$ by $(r_j,s_j)$, $1\le j\le N$, and the rows by $(k_i,l_i)$, with $l_i=\lfloor(i-1)/K\rfloor$. The determinant is $\Delta=P(\alpha_1,\alpha_2)$ with

By Lemma 3, applied to the $r_j$ and then to the $s_j$, all exponents of $X$ (resp. $Y$) lie in $[M_1-G_1,M_1+G_1]$ (resp. $[M_2-G_2,M_2+G_2]$), with $G_1=\frac{(R-1)K\lfloor L^2/4\rfloor}{2}$ and $G_2=\frac{(S-1)K\lfloor L^2/4\rfloor}{2}$. Write $\Delta=\alpha_1^{U_1}\alpha_2^{U_2}\widetilde{P}(\alpha_1,\alpha_2)$, where $U_1$ (resp. $U_2$) is the least exponent of $X$ (resp. $Y$) among the monomials with non-zero coefficient, $\widetilde{P}\in\mathbb{Z}[X,Y]$ has degrees at most $2G_1$ and $2G_2$, and $|\widetilde{P}|=|P|$ for the sup norm on the torus $|X|=|Y|=1$. By Lemma 6 of [2],

Since $\widetilde{P}(\alpha_1,\alpha_2)$ is a non-zero integer with the same $p$-adic valuation as $\Delta$, and since the maximum principle on the polydisc, applied to $X^{2G_1}Y^{2G_2}\widetilde{P}(1/X,1/Y)$ with $|\alpha_i|\ge1$, gives $|\widetilde{P}(\alpha_1,\alpha_2)|\le|\widetilde{P}|\,|\alpha_1|^{2G_1}|\alpha_2|^{2G_2}$,

Now $4G_1/N=\lambda_L(R-1)$ and $4G_2/N=\lambda_L(S-1)$. Comparing the two bounds for $v_p(\Delta)$ and multiplying by $2/N$ gives the negation of (3). Hence $v_p(\Lambda)<N-\frac{1}{2}$, that is $v_p(\Lambda)\le N-1$. $\square$

Proof of Theorem 1

Reduction

Put $P=\log p$. If $P\le69\log y\log(xy)/\log x$, the trivial bound $v_p(x^{p-1}-1)<(p-1)\log x/P$ implies the theorem. From now on assume

In particular $p>2^{69}$ is odd. We apply Theorem 4 with

$(\alpha_1,\alpha_2)=(xy,y),\qquad b_1=b_2=g=p-1,\qquad\Lambda=(xy)^{p-1}-y^{p-1}=y^{p-1}(x^{p-1}-1),$

so that $v_p(\Lambda)=v_p(x^{p-1}-1)$. Write $a_1=\log(xy)/P$, $a_2=\log y/P$, $\Gamma=ga_1a_2$, and take

Then $R_1S_1>gL$ and $R_2S_2>g(K-1)L$. The classes modulo $g$ partition the boxes, so there is a class $c_1$ (resp. $c_2$) containing at least $L$ (resp. more than $(K-1)L$) points of the box $R_1\times S_1$ (resp. $R_2\times S_2$). The numbers $\alpha_1^r\alpha_2^s$ are pairwise distinct by multiplicative independence, whence (1). Since $xy\ge6$ and $y\ge2$, under (4) we have $\Gamma\ge(e^{69\log2}-1)\log2\log6/(69\log2)^2>3.2\cdot10^{17}$.

Degenerate case

If $(r,s) \mapsto(p-1)(r+s)$ is not injective on the class $c_2$ of the box $R_2 \times S_2$, two points $(r_1,s_1) \ne(r_2,s_2)$ of this class satisfy $r_1+s_1=r_2+s_2$, whence $(m_1-m_2)(r_1-r_2) \equiv0 \pmod{g}$. Since $\zeta^{m_1-m_2} \equiv x$, the order $g_0$ of $x$ modulo $p$ divides $r_1-r_2$, so $g_0<R_2$. We have $v_p(x^{p-1}-1)=v_p(x^{g_0}-1)$ because $(p-1)/g_0$ is coprime to $p$, and $p^v \le x^{g_0}-1$. Hence

which is stronger than the theorem.

Non-degenerate case

Then (2) holds, and it remains to check (3).

Proof. We have $\log j! \ge\int_1^j \log t\,dt=j\log j-j+1$ and $\sum_{j=1}^n j\log j \ge\int_1^n t\log t\,dt$. With $n=K-1$,

Multiply by $\frac{2}{n(n+1)}$ and bound term by term. □

Proof. We have $R+S-2\le(1+\sqrt{K-1})\sqrt{gL}(a_1+a_2)/\sqrt{a_1a_2}$ and $gLa_1a_2=L\Gamma\le K/k$.

Lemma 5 then gives, as in Lemma 13 of [1],

with $\eta(K)=\log\frac{\sqrt{K}(1+\sqrt{K-1})}{K-1}+\frac{\log(K-1)}{K}$. Since $K>kL\Gamma>4\cdot10^{18}$, we have $\eta(K)<10^{-8}$. We conclude with $g<p$ and $1/a_1+1/a_2=P\left(\frac{1}{\log xy}+\frac{1}{\log y}\right)$. □

Proof. Put $E=\log\left(\frac{1}{\log xy}+\frac{1}{\log y}\right)+\kappa$. If $\log P_0+E\ge1$, the function $P\mapsto(\log P+E)/P$ decreases on $[P_0,\infty)$ and its bound is attained at $P_0$. Otherwise its supremum on $\mathbb{R}_{>0}$ is $e^{E-1}<-\log P_0\le1/(69\log2)$, which is smaller than the bound sought. In the first case, with $u=\log y$, $w=\log x$ and $t=u/w$,

so $(\log P_0+E)/P_0=h(t)/(69u)$ with $h(t)=(A+\log(1+2t))/(1+t)$ and $A=\log69+\kappa>4.5$. We have $h(0)=A$, and

Hence $h(t)\le A$, and $u\ge\log2$ gives the bound. □

Verification of (3). We have

and likewise for $(S-1)\log|\alpha_2|$. Also $K-1\le kL\Gamma$, $K\ge kL\Gamma$, $\log b\le\beta^*P$, and $N\le kL^2\Gamma+L$. It therefore suffices that

With $L=5$, $k=69/25$ and $\lambda_5=12/5$:

and the last two terms are less than $2\cdot10^{-8}$ since $\Gamma>3.2\cdot10^{17}$ and $P>47$. The margin is 0.2201.

Theorem 4 gives $v_p(\Lambda)\le KL-1\le kL^2\Gamma+L-1=69\Gamma+4$, and $\Gamma=(p-1)\log y\log(xy)/P^2$.

Remarks

Remark 8 (choice of $L$). The same chain, with $\lambda_L$ and the corresponding $\beta^*$, gives the minimal constants:

$L$	3	4	5	6	7	8	9
coefficient $\lambda_L$	75.27	70.09	68.29	84.44	95.04	116.30	132.85
coefficient $L$ of [3]	337.75	263.13	282.79	327.95	387.46	457.97	538.12

The choice $L=5$ is the best in the table. The choice $L=4$ gives 71 with additive term 3.

Remark 9 (the constant 283). Equation numbers refer to the published version of [3]. In [3 (24)], the intermediate bound $\log b\le\log p+\log\log p$ does not hold for $y=2$. It would require $\log\left(\frac{1}{\log2x}+\frac{1}{\log2}\right)+\frac{3}{2}-\log2-\frac{1}{2}\log11.32\le0$, whereas the left-hand side is positive for $3\le x\le12\,311\,938$, $x$ not a power of 2 (it equals 0.287 at $x=3$). With only the reduction $p>2^{283}$ of [3], one finds at $(x,y)=(3,2)$ and $\log p=283\log2$ that $\log b/\log p$ can reach $1.02837>1.027$, and (21) of [3] is no longer satisfied. Statement (6) of [3] nevertheless remains true. On the one hand, Lemma 3 gives $G\le NL(R-1)/8\le G_Y$ with no hypothesis; Theorem 1 of [3] therefore holds even when $\gcd(m_1,m_2,p-1)\ne1$, a case its Lemma 6 does not reach. On the other hand, the chain above, run with the coefficient $L$ of [3] and the exact range (4), gives the minimal constant $282.79\le283$ at $L=5$.

Remark 10 (additive term). The analytic bound used here (Lemma 8 of [1]) contains the loss $N\log N/\log p$. Expanding the powers $\theta^\ell z$ as a binomial series in $\theta-1$ instead of the exponential series gives $v_p(\Delta)\ge NK(L-1)/2$ with no loss as soon as $v_p(\Lambda)\ge N-1$ [4 Corollary 3]; the conclusion of Theorem 4 becomes $v_p(\Lambda)\le KL-2$, and the additive term 4 of Theorem 1 can be replaced by 3.

Remark 11 (large $p$). Stewart [5 proof of Lemma 8, (41)], using the refinements of Yu [6], proves, for an integer $a\ge2$ and $p\nmid a$ larger than an effectively computable but unspecified constant,

with $n=p-1$, this bound is asymptotically better than bounds of the form $p/(\log p)^2$, those of [3] as well as Theorem 1. For $y=2$, computation shows that it beats Theorem 1 only for $\log p$ larger than about $1.1\cdot10^4$ (and earlier if $y$ is larger). Theorem 1, like that of [3], is explicit and holds for every $p$.

Remark 12 (scope). Lemma 3 and Theorem 4 are proved here. The ingredients used as they stand are the zero lemma and the analytic bound (Lemmas 6 to 8 of [1]) and the bound $|\mathcal{P}|\le N^{N/2}b^{(K-1)N/2}$ (Lemma 6 of [2]).

Verification

The ancillary file VERIF_FERMAT_69.py (Python and numpy, no argument, about ten seconds) checks:

• Lemma 3, by brute force on 15 120 configurations, together with its equality case;

• Lemma 5 for $2 \le K \le20\,000$;

• the bound $\beta^*$, against a grid of pairs $(x,y)$;

• the final margin 0.2201;

• condition (3) at 40 explicit primes of up to 1521 bits, with the integer parameters built in decimal arithmetic with precision proportional to the size of $p$;

• the bound $\eta(K) < 10^{-8}$ for $K \ge k_L \cdot3.2 \cdot10^{17}$;

• the calibration, which recovers $282.79 \le283$ with the coefficient of [3], and the table of constants.

Acknowledgements. Technical assistance: Claude (Anthropic).