Pythagorean Triple divisible by $5$

Question

Show that, if x, y and z are integers such that $x^2+y^2 = z^2$,then at least one of $x,y,z$ is divisible by $5$. I was able to show that at least one of $x$ or $y$ is divisible by $2$. Can someone please show me how to prove it for $5$?

Answer 1

Hint: $x^2 = 0, 1 \text{ or } 4 \pmod 5$, for all $x$.

Answer 2

The only squares in $Z_5$ are $0$, $1$, and $4$.

Assume that none of $x$, $y$ and $z$ is divisible by 5. Then, we also have $x^2$, $y^2$ and $z^2$ non-divisible by 5. This means that the possible values of $x^2$ + $y^2$ - $z^2$ ($mod$ $5$) are:

$$(1 + 1 - 1 = 1);(1 + 1 - 4 = -2 = 3);(1 + 4 - 1 = 4);(1 + 4 - 4 = 1);(4 + 1 - 1 = 4);(4 + 1 - 4 = 1);(4 + 4 - 1 = 7 = 2); (4 + 4 - 4 = 4)$$

$0$ is none of the above values. A contradiction.