# The diophantine equation $x^2+y^2=3z^2$

I tried to solve this question but without success:

Find all the integer solutions of the equation: $x^2+y^2=3z^2$

I know that if the sum of two squares is divided by $3$ then the two numbers are divided by $3$, hence if $(x,y,z)$ is a solution then $x=3a,y=3b$. I have $3a^2+3b^2=z^2$ and that implies $$\left(\frac{a}{z}\right)^2+\left(\frac{b}{z}\right)^2=\frac{1}{3}$$ so I need to find the rational solutions of the equation $u^2+v^2=\frac{1}{3}$ and I think that there are no solutions for that because $\frac{1}{3}$ doesn't have a rational root, but I dont know how to explain it.

Thanks

How about using infinite descent ?

As for any integer $a\equiv0,\pm1\pmod4\implies a^2\equiv0,1$

We have $x^2+y^2\equiv0\pmod3$

So, $3|(x,y)$

Let $x=3X,y=3Y$

and so on

Let $(x,y)=d$ and $\dfrac xX=\dfrac yY=d$

$$\implies d^2(X^2+Y^2)=3z^2\implies d|z,$$ $z=dZ$(say)

$$\implies X^2+Y^2=3Z^2$$

Now $X^2+Y^2\equiv1,2\pmod4$ as $X,Y$ both can not be even

$$Z^2\equiv0,1\pmod4\implies3Z^2\equiv0,3\pmod4$$

• Thank you! but I don't understand why cant X,Y both be even? – Ben Apr 2 '15 at 9:57
• @Ben, If both are even, the GCD will divide 2 – lab bhattacharjee Apr 2 '15 at 10:01