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?

If $(x,y)$ is a solution of $2x^2-y^2=1$ such that $x>5$, $y>0$ and $x\equiv 5(10)$ then
$$(x',y')=(99 x-70 y,-140 x+99 y)$$
is also a solution with $0<x'<x$ , $x' \equiv 5(10)$, and $y' \equiv y(7)$
The process of repeatedly applying this transformation to a solution $(x,y)$ will stop at the solution $(5,7)$ thereby proving the result
It easy to check that $(x',y')$ is a solution. Next
$$x\sqrt{2}-y=\frac{1}{x\sqrt{2}+y}<\frac{1}{5\sqrt{2}}$$
gives, after some algebra,
$$10 y> 10\sqrt{2}x-\sqrt{2} > 14x$$
the last inequality is true since $x\geq  10$
Therefore $99x-70y<x$
Also, $x\sqrt{2}-y>0$ gives $99x-70y>0$
Finally,
$$99 x-70 y\equiv 99*5\equiv 5(10)$$
and
$$-140 x+99 y\equiv y(7)$$
