<s>[INST] Prove that the diophantine equation $2x^2-5y^2=7$ has no integer solutions. 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.    
 [/INST] We can write the equation in this form too:
$$2(x^2+y^2) = 7(y^2+1)$$
This means that $7 | x^2+y^2$, also its easy to show that $\gcd(x, y) = 1$, becouse if $\gcd(x, y) = d > 1$, then $d | 7$. Now suppose that $d = 7$, then for $x=7a$ and $y=7b$:
$$2x^2 - 5y^2 = 2\times 49 \times a^2 - 5 \times 49 \times b^2 = 49(2a^2 - 5b^2) = 7 $$
Its contradiction. So $d=1$.
But $x^2+y^2 = 7k$ has no solution where $x$ and $y$ are coprime, . So the given diophantine equation has no solutions. 
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