a)First we prove this Lemma:If $x,y,z$ is a pythogoran triple then atleast one of $x,y,z$ is divisible by 5. Proof:The reduced residue class $\pmod 5$ i.e ${F_5}=(0,1,-1)$.Now it is easy to see from this that no permutation of $\pm1$ and $\pm 1$ can be congruent to $1,-1$ so one of $x,y,z$ is divisible by 5. Now if $x$ or $y$ or $z$ would be divisible by 5 then we would be done so we just check the case where $5|a^2,5|b^2,5|c^2$. So FTSOC,we will assume that $y^2 \equiv -1 \pmod 5$(it doesn't really matter which of the $x,y,z$ we choose since the equations are cyclic so we can interchange $x$ to $y$ and $y$ to $z$ and by the same procedure we will get a contradiction.). Now we have $x^2 \equiv 1 \pmod 5$ and also since $y^2+z^2=b^2$ we have $z^2 \equiv -1 \pmod 5$ which will be a contradiction to the last relation. Also if $y^2 \equiv 1 \pmod 5$ then $x^2 \equiv -1 \pmod 5,z \equiv -1 \pmod 5$ which is again a contradiction to the last equation. Hence proved.

Guys i've been trying to solve this problem and i ran into something strange: if i sum up the three equations i get $2(x^2+y^2+z^2)=a^2+b^2+c^2$ $a$, $b$ and $c$ are the greatest terms of three pythagorean triples, so they are in the form $m^2+n^2$ where $m$ and $n$ are two integers, one of them even and the other odd. That means that they must be odd. But since $a$, $b$ and $c$ are odd, the sum of their squares is odd too, so we reach a contradiction. In fact the left hand side of the equation is even while the right side surely isn't! Have i made any mistake?

Actually, $a,b,c$ are of the form $k(m^2+n^2)$ for some integer $k$.

uh my guilt, sorry.