Prove that if $x^2+y^2 = z^2$ then $x$ or $y$ is even

Question

I am having trouble proving this. I feel that proof by contradiction would be the best method, although I quickly got stuck after $x=(2k+1), y=(2j+1)$. I expanded so that $4j^2+4k^2+4j+4k+2=z^2$ but I don't know where to go from here.

Answer 1

HINT: You have $x^2+y^2$ in the form $4n+2$. Can the square of an odd integer have that form? What about the square of an even integer?

Answer 2

If $z$ is odd, then either $x$ is odd or $y$ is odd but not both. Suppose $z$ is even and both $x$ and $y$ are odd. Then $z^{2} = 4k + 2$ for some integer $k \geq 1.$ Since $z$ is even, we have $4 \mid z^{2},$ but $4 \nmid 4k + 2,$ a contradiction.

Answer 3

If $z$ is odd, it is of the form $z = 2k + 1$ where $k$ is a natural number. $$\text{Then }z^2 = (2k+1)^2 + 4(k^2 + k) + 1 = x^2 + y^2,\text{ which must also be odd.}$$ Because of this, exactly one of $x^2$ or $y^2$, (or similarly, $x, y$) is odd, leaving the other to be even.

If $z$ is even, it is of the form $z = 2k$ where $k$ is a natural number $$\text{Then }z^2 = (2k)^2 + 1 = x^2 + y^2,\text{ which must also be even.}$$ Because of this, either both of $x^2$ and $y^2$, (or similarly $x,y$) are even. $$\text{Since } (2l+1)^2 + (2j + 1)^2 = 4l^2 + 4l + 4j^2 + 4j + 2 \neq 4k^2$$ due to the $2$ term.