the equation :
$\displaystyle
x^2-3y^2+4z^2=0
$
good luck
sorry, it's not in N but in Z
there isn't any difference ^^
First, we obviously have the trivial solution $\displaystyle (x,y,z)=(0,0,0)$
Now suppose there's another solution $\displaystyle (x,y,z)$
If $\displaystyle x$ and $\displaystyle z$ are not multiples of 3 then $\displaystyle x^2+4z^2\equiv{5}(\bmod.3)$ a contradiction.
Again if only one of them ( x or z) is multiple of 3 we get a contradiction ( try it).
So both, $\displaystyle x$ and $\displaystyle z$ are multiples of 3. And thus $\displaystyle 3y^2$ is multiple of 9, so it must be that $\displaystyle y$ is multiple of 3. Now set $\displaystyle x=3x'$; $\displaystyle y=3y'$; $\displaystyle z=3z'$
We get: $\displaystyle x'^2-3y'^2+4z'^2=0$ (dividing by 9)
And it follows by infinite descent that there can be no other solution.