# resolve in N...

• Dec 21st 2008, 11:52 PM
guigui51250
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 ^^
• Dec 22nd 2008, 02:28 AM
PaulRS
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.
• Dec 22nd 2008, 03:04 AM
guigui51250
yes good, i post an other equation ^^