The primitive solutions of the equation $x^2+3y^2=z^2$

Question

I am working on the following exercise:

Consider the Diophantine equation $x^2+3y^2=z^2$. Show that for each primitive solution $(x,y,z)$ to this equation holds

1. $x$ has to be odd.
2. If $y$ is odd then $z$ is even.

I think that 2) is clear. It is enough to recall the arithmetic tables of $Z/2Z$. But I do not know how to do 1.). Could you give me a hint?

Answer 1

The equation $x^2+3y^2=z^2$ cannot have solutions with $x$, $y$, and $z$ all odd. For similar reasons, if two of the variables in a solution are even, then so is the third. So primitive solutions must have exactly one variable even and the other two odd. If, in a primitive solution, $x$ were even, then $y$ and $z$ would be odd, and we would have $3\equiv1$ mod $4$, since $y^2\equiv z^2\equiv1$ mod $4$. So $x$ must be odd, leaving exactly one of $y$ and $z$ to be even, thus proving both 1. and 2. together.

Answer 2

Hint

Look at the equation modulo $8$. Use the fact that $y^2 \equiv 0,1,4 \pmod{8}$.

Answer 3

Suppose $x$ is even. The $y^2\equiv z^2\pmod 2$ and hence $y\equiv z\pmod 2$. For a primitive solution, this means that $y,z$ must both be odd.

Answer 4

Two types, with $u,v$ coprime integers, not both odd. First one also $u \neq 0 \pmod 3$ $$ x = | u^2 - 3 v^2 |, \; y = 2|uv|, \; z = u^2 + 3 v^2 $$

$$ x = | u^2 + 4uv + v^2 |, \; y = |u^2-v^2|, \; z = 2u^2 + 2uv + 2 v^2 $$