1. ## diophantine equation problem

Solve: $\displaystyle x^2 - 7y^2 = 3z^2$

I set it congruent (mod 3) to reduce it to $\displaystyle x^2 - y^2 \equiv 0 (mod 3)$

This does not fit the sum of two squares theorem and I can not seem to find any difference of two squares theorem. Also, I can not make $\displaystyle -y^2 \equiv a^2 (mod 3)$ for any a, so I can not make it sum of two squares that way either.
The answer is supposed to be x=y=z=0, which is obvious. But setting it congruent (mod 3) seems to give me additional answers. How do I go about solving this problem? Thanks

2. Hello,
Originally Posted by Proggy
Solve: $\displaystyle x^2 - 7y^2 = 3z^2$

I set it congruent (mod 3) to reduce it to $\displaystyle x^2 - y^2 \equiv 0 (mod 3)$

This does not fit the sum of two squares theorem and I can not seem to find any difference of two squares theorem. Also, I can not make $\displaystyle -y^2 \equiv a^2 (mod 3)$ for any a, so I can not make it sum of two squares that way either.
The answer is supposed to be x=y=z=0, which is obvious. But setting it congruent (mod 3) seems to give me additional answers. How do I go about solving this problem? Thanks
Set it congruent (mod 4) (in my opinion, it's not good making disappear a variable, but my knowledge in diophantine equations is not wide enough to be certain)

$\displaystyle -7 \equiv 1 \bmod 4$ and $\displaystyle 3 \equiv -1 \bmod 4$.

The equation becomes : $\displaystyle x^2+y^2=-z^2 \implies x^2+y^2+z^2=0$

This means that $\displaystyle x \equiv y \equiv z \equiv 0 \bmod 4$

$\displaystyle x=y=z=0$ is obviously a solution.

Now, if they're all $\displaystyle \neq 0$, let's say $\displaystyle x=4x'$, $\displaystyle y=4y'$ and $\displaystyle z=4z'$, the equation is :
$\displaystyle 16x'^2-7*16y'^2=16*3z'^2 \implies x'^2-7y'^2=3z'^2$ and we're back to the same equation...