Find all the positive natural solutions of $x^2+y^2=3z^2$.

Question

Find all the positive natural solutions of $x^2+y^2=3z^2$. I guess it has something to do with Pythagorean triples, but I don't know how to relate it properly. Suggestions, hints, or any sort of assisting is welcome.

Answer 1

The only solution to this equation is $(0,0,0)$ prove it by vieta jumping.

• Suppose that there is a solution such that $x^2+y^2=3z^2$ with $x+y$ is minimal and $x\neq 0, y\neq 0$
• By considering all cases modulo $3$ prove that $3$ divides both $x$ and $y$ conclude that $3$ divides $z$ and $(\frac{x}{3},\frac{y}{3},\frac{z}{3})$ is a solution and gives you a small value of $\frac{x}{3}+\frac{y}{3}<x+y$ which is a contradiction.

so $x=y=0$

Answer 2

Soppuse that the triple (x,y,z) is a solution to the equation in positive integers, WLoG suppose that $gcd(x,y,z)=1$
Obviously both $x$ and $y$ are divisible by $3$, that's enough to take the equation in mod $3$, so the LHS of the equation is divisible by $9$ and so $z$ is divisible by $3$ Which means $3|(x,y,z)$, a contradiction. So there is no triple of positive integers $(x,y,z)$ satisfiying the equation.

Answer 3

Hint Reduce the equation modulo $3$ to give $$x^2 + y^2 \equiv 0 \bmod 3.$$ Now, $u^2$ is congruent to $0$ or $1$ modulo $3$ when $u$ is, resp. isn't, a multiple of $3$, so what can you conclude about $x$ and $y$?