Show that $3\mid xy$ in $x^2 + y^2 = z^2$ Part 1:

If $x$, $y$, and $z$ are positive integers for which $gcd(x, y, z) = 1$ and $x^2 + y^2 = z^2$, show that $3\mid xy$

Part 2:

Now again if $x$, $y$, and $z$ are positive integers for which $gcd(x,y,z) = 1$ and $x^2 + y^2 = z^2$, can a similar result to the one in (part 1) be said modulo 5? Not that $5|xy$, but one of $x,y,z$?
Note that $3$ is a prime, so $3 \mid xy \iff 3\mid x$ or $3\mid y$. Suppose to the contrary that $3 \nmid x$, and $3\nmid y$, then:$x = 1,2$ mod $3$ and $y = 1,2$ mod $3$. Then:

$x^2+y^2 = 2$ mod $3$, and $z^2 = 0,1$ mod $3$. Contradiction, proving the claim.
Primitive Pythagorean triplets can be characterized precisely.


  If $x^2+y^2=z^2$, where $x,y,z \in \mathbb{Z}$, and $\gcd(x,y,z) =1$, then $x=m^2-n^2$, $y=2mn$ and $z=m^2+n^2$, where $\gcd(m,n) = 1$ and are of opposite parity.


From the above, we can immediately conclude that $3$ divides $xy$. Similarly, we can conclude that $4$ divides $xy$ and $5$ divides $xyz$. Look here for more properties on Pythagorean triplets.