For $\text {HCF}(x,y,z)=1$ and $x^2+y^2=2z^2$, prove the following Let $x,y,z$ be positive integers such that $\text {HCF}(x,y,z)=1$ and $x^2+y^2=2z^2$. Prove the following 
$1.$ $3$ divides $x+y$ or $3$ divides $x-y$
$2.$ $5$ divides $z(x^2-y^2)$
From $x^2+y^2=2z^2$,  it is clear that $x$ and $y$ will be both odd or both even.
If both are even then $z$ will be odd so that $\text {HCF}(x,y,z)=1$ holds true but I think $x^2+y^2=2z^2$ could give more information which I am not able to observe and hence not able to prove the given statements. Could someone give some hint to proceed.

b) By Fermat little theorem we have if $5\nmid a$ then $a^4 \equiv_5 1$ thus $a^2 \equiv_5 \pm1$.
If $5|z$ we are done, if $5\nmid 5$ then $2z^2\equiv_5 \pm 2$ so $x^2 \equiv_5 y^2 \equiv_5 1$ or  $x^2 \equiv_5 y^2 \equiv_5 -1$ thus $5\mid x^2-y^2$. And we are done.
a) The same procedure can be done with mod $3$ of $3\nmid z$. If $3\mid z$ then $3\mid x^2+y^2$ then $3\mid x$ and $3\mid y$ and we are done.
