# 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.

• Also $6^2 \equiv 6 \pmod {10}$ so you have $x \equiv 4,6 \pmod {10}$. You need another modulus or something else. Jul 5, 2019 at 2:01
• I suspected that, but I have no clue how to obtain it. Jul 5, 2019 at 2:02
• mod $7$ it's $2(x^2+y^2)\equiv0\implies x,y\equiv0$ Jul 5, 2019 at 2:13

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.

• also $x^2\equiv-y^2\pmod7\implies x^2\equiv y^2\equiv0\pmod 7$ because $-1$ is not a quadratic residue modulo $7$ Jul 5, 2019 at 2:44

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

• so there are also no integer solutions to $2x^2-5y^2=1$ or more generally none to $2x^2-5y^2=n,$ where $n\equiv\pm1\pmod8$ Jul 7, 2019 at 2:23

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.

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.

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");
}