<s>[INST] Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$ 
Let $x,y$ be positive integers satisfying $2x^2-y^2 = 1$. Prove that if $x \equiv 5 \pmod{10}$, then $y \equiv 0 \pmod{7}$.

I wasn't sure how to use the fact that $x,y$ are positive integers satisfying $2x^2-y^2 = 1$. We could use the theory of Pell's equations to find the solutions, but that would get complicated. Is there a simpler way?
 [/INST] The solutions of this Pell equation are 
$$ \pmatrix{x_n\cr y_n\cr} = \pmatrix{ 3 & 2\cr 4 & 3\cr}^n \pmatrix{1\cr 1\cr} $$
for nonnegative integers $n$.  If $M = \pmatrix{3 & 2\cr 4 & 3\cr}$, we have
$M^6 \equiv I \mod 70$. Thus $(x_n, y_n) \mod 70$ (and therefore mod $10$ and mod $7$) is periodic with period $6$.  We find that $x_n \equiv 5 \mod 10$ for $n = 1$ and $4$, and therefore whenever $n \equiv 1$ or $4 \mod 6$.  For those $n$ we find $y_n \equiv 0 \mod 7$.
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