$5$ divides one of $x,y,z$ in $x^2+y^2=z^2$? 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$$∣$$xy$.
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 mod $5$? Not that $5$$∣$$xy$, but that $5$ divides one of $x,y,z$?
I believe I have 1) down: 
We note that $3$ is a prime, so $3∣xy⟺3∣x$ or $3∣y$. Suppose to the contrary that $3∤x$, and $3∤y$, 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.
But I am having trouble with 2). Can anyone show a solution for 2)?

For the second part since $5$ is prime, if $5 \mid xy \implies 5 \mid x$ or $5\mid y$. For the first part, we have: $x = 2mn, y = m^2-n^2=(m-n)(m+n)$. Thus if $3 \nmid x$, then $3 \nmid m, 3 \nmid n$. Thus if $m = n \pmod 3$, then $m-n = 0 \pmod 3$, otherwise say $m = 1\pmod 3, n = 2\pmod 3 \implies m+n = 0 \pmod 3$. 
