Why do we need an even number between $x$ and $y$ in Pythagorean triple where $x^2+y^2=z^2$ and $x$, $y$ and $z$ are relatively prime?

Question

I needed to know about Pythagorean triple in details. So I started searching in Google.

Suppose that $x$, $y$ and $z$ are relatively prime and $x^2 + y^2 = z^2$. Then I've found an interesting fact that, among $x, y$, and $z$, only $x$ or $y$ can be even. $z$ can never be.

I searched for this and haven't found the reason behind this. Would anybody please tell me the reason behind this?

Answer 1

This may be overkill.

if $z^2 = x^2+ y^2$ is even if and only if $x , y $ are both same parity and if $x,y,z $ are relatively prime that parity must be odd.

Let $x =2a+1$ and $y=2b+1$ and $z=2c $ and $(2a+1)^2 + (2b+1)^2 =4c^2$ so $4a^2 + 4a +4b^2 + 4b +2 = 4c^2$

So, $a^2 + a + b^2 + b + \frac 12 = c^2$. Hence, $a,b,c $ can not all be integers. So no integer pythagorean triple $x,y,z $ exists where $x,y$ are both odd.

And as $x,y$ can't be both even and relative prime to $z $ (which would also be even) primitive pythagorean triples must be such $x,y$ are opposite parity and $z $ is odd.

No triples, primitive or otherwise, can have $x,y $ both odd. $x,y,z $ can all be even, but such a triple is, of course, not primitive.

Answer 2

Since $x,y$ and $z$ are relatively prime, $x$ and $y$ cannot both be even, for then $z$ would be even as well. So we just have to rule out the possibility that $x$ and $y$ are both odd.

To do so, recall that the only squares mod $4$ are $0$ and $1$. Hence if $x$ and $y$ are odd, then $x^2+y^2\equiv 2$ (mod $4$), but on the other hand $z^2\equiv 0,1$ (mod $4$), so we cannot have $x^2+y^2=z^2$ in this case.

Therefore exactly one of $x,y$ is odd, and then $z$ must be odd as well.