If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that... If $x$,$y$ are odd integers $a$,$b$ and $c$ such that $a^4-b^4=c^4$, argue that $\gcd(a,b,c)=1$ implies that $\gcd(a,b)=1$
What I know:
A Pythagorean Triple is a triple of positive integers $a$,$b$,$c$ such that $a^2+b^2=c^2$
A Primitive Pythagorean Triple is a Pythagorean triple $a$,$b$,$c$ with the constraint that $\gcd(a,b)=1$, which implies $\gcd(a,c)=1$ and $\gcd(b,c)=1$.
I'm not sure how to use this, since we are dealing with $a^4-b^4=c^4$.

Hint:  Suppose to the contrary that $a$ and $b$ are not relatively prime. Let $p$ a common prime divisor of $a$ and $b$. Show that $p$ must divide $c$.
Remark: We do not need anything about Pythagorean triples to solve this problem. And we do not need to use the fact that $a$ and $b$ are odd.
But if we do want to borrow facts about Pythagorean triples, we can note that $a^4 -b^4=c^4$ if and only if $(b^2)^2 +(c^2)^2=(a^2)^2$, so $(b^2, c^2, a^2)$ is a Pythagorean triple. Then knowledge about triples becomes useful in the further analysis of $a^4-b^4=c^4$.  
