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

Question

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

Answer 1

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

Answer 2

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

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

Answer 3

There actually is one solution: $(x,y,z)=(0,0,0)$. What you want to prove is that that's the only solution in integers. And you were off to a good start, showing that $x^2+y^2=3z^2$ implies $3\mid x,y$, so that the equation to solve becomes $3a^2+3b^2=z^2$ with $x=3a$ and $y=3b$.

The next step is to notice that we now have $3\mid z$, so that the equation now becomes $a^2+b^2=3c^2$ with $z=3c$. But this is the original equation back again, just with different, smaller numbers. The "smaller" is what's important: if $|x|+|y|+|z|\not=0$, then $0\not=|a|+|b|+|c|\lt|x|+|y|+|z|$. This gives us an "infinite descent" of solutions, each with smaller positive sum of absolute values than its predecessor, which is impossible, so we conclude there are no solutions except for $(0,0,0)$.