Prove that the diophantine equation $2x^2-5y^2=7$ has no integer solutions.

Question

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.

Answer 1

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.

Answer 2

If odd, $$ 2 x^2 - 5 y^2 \equiv 3,5 \pmod 8 $$

Answer 3

you are almost there. Since $x^2=2[5k(k+1)+3]$ then $x$ is even which makes $x^2 \equiv 0 \pmod{4}$, Whereas $5k(k+1)+3$ is odd and $2[5k(k+1)+3]$ is not multiple of 4.

Answer 4

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.

Answer 5

The equation has no solutions mod 8. https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=0658726a8cc6ca5c19be08d83ee7e8f3

fn main() {
    let p = 8;
    for x in 0..p { 
    for y in 0..p {
        if 2*x*x % p == (7 + 5*y*y) % p { 
            println!("{} {} {}", x, y, p);
        }
    }}
    println!("Done");
}