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

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How do x and y fit in? – bmm6o Dec 9 '12 at 1:39
up vote 1 down vote accepted

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

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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.

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