1
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Also $6^2 \equiv 6 \pmod {10}$ so you have $x \equiv 4,6 \pmod {10}$. You need another modulus or something else. $\endgroup$ Jul 5, 2019 at 2:01
  • $\begingroup$ I suspected that, but I have no clue how to obtain it. $\endgroup$
    – FoiledIt24
    Jul 5, 2019 at 2:02
  • 2
    $\begingroup$ mod $7$ it's $2(x^2+y^2)\equiv0\implies x,y\equiv0$ $\endgroup$ Jul 5, 2019 at 2:13

5 Answers 5

6
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ also $x^2\equiv-y^2\pmod7\implies x^2\equiv y^2\equiv0\pmod 7$ because $-1$ is not a quadratic residue modulo $7$ $\endgroup$ Jul 5, 2019 at 2:44
3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ so there are also no integer solutions to $2x^2-5y^2=1$ or more generally none to $2x^2-5y^2=n, $ where $n\equiv\pm1\pmod8$ $\endgroup$ Jul 7, 2019 at 2:23
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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");
}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.