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 $a^4-b^4=c^4$.

If $gcd(a,b)>1$, then there is a prime number $p$ dividing $gcd(a,b)$, i.e., $p|a$ and $p|b$.
Since $a^4-b^4=c^4$, we have $p|c$. Thus $p|gcd(a,b,c)$. But $gcd(a,b,c)=1$, a contradiction.
