Continuing my study on basic number theory and particularly the properties of pairwise relatively coprime numbers (with my interest being in gaining a basic understanding Euler's proof of $x^3+y^3=z^3$), suppose we have the following situation: (the following obviously has nothing to do with Euler's proof directly, just a problem i came up with to get a better grip on congruence and properties of pairwise coprime numbers)
Let $a,b,c$ satisfy $(a+c)^3+(b+c)^3=(a+b+c)^3$, where $a+c,b+c,a+b+c$ are pairwise relatively coprime integers. Can we conclude $(a,b)=1$?. Or what, if anything, can we conclude about the pairwise relationship between $a,b,c$?
Consider the more general case of pairwise coprime integers $x,y,z$ satisfying $x^3+y^3=z^3$ Then suppose a prime $p$ divides $z-x$ and $z-y$. Then $x^n\equiv z^n\,\equiv x^n+y^n\,\pmod p$ and so $y$ is divisible by p. However if that is true and $z-y$ is divisible by p, then $z$ must also be divisible by p (since $(z,y)=1$) contradicting the original hypothesis. Hence $(z-x,z-y)=1$.
Assuming that's correct can we simply substitute $(x,y,z)=(a+c,b+c,a+b+c)$ all pairwise coprime and conclude $(a,b)=1$?
The reason i ask is that if i use this proof directly i run into trouble with the logic. For example the first step would involve, effectively, assuming a prime $p$ divides $a$ and $b$. Then
$$a+c \equiv a+b+c\pmod p \,\Rightarrow (a+c)^n\equiv (a+b+c)^n\,\equiv (a+c)^n+(b+c)^n\,\pmod p$$
and so $b+c$ is divisible by p, but how would you continue? Unlike the previous example I can't see how to utilise the hypothesis on $a$.
I suppose we could run the argument again, focusing on $a+c$ and conclude that it too must be divisible by $p$. Then if $a+c$ and $b+c$ are divisible by p, and $a$ and $b$ are divisible by $p$, then this contradicts $(a+c,b+c,a+b+c)=1$. However, is this reasoning circular?
Best Answer
Let the prime $p$ divide $a$ and $b$. Then modulo $p$, $$ c^3+c^3\equiv c^3\pmod p.$$ Hence $p$ divides $c$ and then $p$ divides all of $a+c,b+c,a+b+c$, a contradiction.