My attempt: I rewrote it as $2x^2=5y^2+7. 2x^2$ is always even, so in order for the RHS to be even, this means that $5y^2$ must be odd since an odd number plus $7$ is even.
If I evaluate when y is odd, so if $y=2k+1$ for some integer $k$, I get: $2x^2=20k^2+20k+12$. This is the same as $x^2=10k^2+10k+6$, which implies that $x^2$ is congruent $6$ (mod $10$).
Here, I arrive at an issue because if $x=4$, then I get that $x^2$ is congruent to $6$ (mod $10$), but I am supposed to show that the equation does not have a $6$ (mod $10$) congruency.
Best Answer
Modulo $7$, $2x^2-5y^2=7$ would mean $2x^2+2y^2\equiv0$ or $x^2+y^2\equiv0$ or $x^2\equiv-y^2$.
Now $x^2, y^2\equiv 0, 1, 2, $ or $4 \pmod 7$, so the only solution would be $x^2\equiv y^2\equiv0\pmod7$.
But this means $7|x,y$, which means $49|2x^2-5y^2=7,$ a contradiction.