resolve in N...

the equation :

$<br /> x^2-3y^2+4z^2=0<br />$

good luck

sorry, it's not in N but in Z
there isn't any difference ^^

First, we obviously have the trivial solution $(x,y,z)=(0,0,0)$

Now suppose there's another solution $(x,y,z)$

If $x$ and $z$ are not multiples of 3 then $x^2+4z^2\equiv{5}(\bmod.3)$ a contradiction.

Again if only one of them ( x or z) is multiple of 3 we get a contradiction ( try it).

So both, $x$ and $z$ are multiples of 3. And thus $3y^2$ is multiple of 9, so it must be that $y$ is multiple of 3. Now set $x=3x'$ ; $y=3y'$ ; $z=3z'$

We get: $x'^2-3y'^2+4z'^2=0$ (dividing by 9)

And it follows by infinite descent that there can be no other solution.

yes good, i post an other equation ^^