Every $C_4$-free graph on $49$ vertices has a vertex of degree at most $6$

We prove that every graph on $49$ vertices containing no cycle of length four has a vertex of degree at most $6$. Equivalently, $R(C_4,K_{1,42})\le 49$, hence $R(C_4,K_{1,42})=49$, a value left open in the published tables. In the notation of Erdős, the least integer $f(n)$ such that every graph on $n$ vertices with minimum degree at least $f(n)$ contains a $C_4$ satisfies $f(49)=7$, while $f(48)=f(50)=8$; the value $f(48)=8$ comes from an explicit $7$-regular $C_4$-free Cayley graph on $48$ vertices, which is also an explicit lower-bound construction for the tabulated value $R(C_4,K_{1,41})=49$. Hence $f$ is not non-decreasing; to our knowledge no $n$ with $f(n+1)<f(n)$ was known before. Erdős's question whether $f(n+1)\ge f(n)$ for all sufficiently large $n$ (Problem 85 of erdosproblems.com) remains open. The proof combines structural lemmas, valid for every odd $k$, on $C_4$-free graphs with $k^2$ vertices and minimum degree at least $k$, a symmetry reduction for $k=7$, and an exhaustive SAT search carried out by two separately written programs.