If there is a solution to Fermat’s Last Theorem, \(A^n+B^n=C^n\), where \(n>2\), and integers \(x\), \(y\), \(z\) are greater than 1, then there is a solution where \(x\), \(y\), and \(z\) are coprime.
Proof:
By our hypothesis, there is a solution \(A=a\), \(B=b\), and \(C=c\). We proceed in two cases: \(gcd(a,b,c)=1\) or \(gcd(a,b,c)\neq 1\).
Case 1: \(gcd(a,b,c)=1\)
In this case, if \(gcd(a,b,c)=1\), they are coprime and the case is complete.
Case 2: \(gcd(a,b,c)\neq 1\)
Assume the \(gcd(a,b,c)=d\) where \(d\) is an integer greater than 1. It follows: \[a=dj\qquad b=dk\qquad c = dl\quad\text{for } j,k,l\in\mathbb{Z}\] Let us substitute these values into our equation. It goes: \[a^n+b^n=c^n\] \[(dj)^n+(dk)^n=(dl)^n\] By the laws of exponents we get: \[d^nj^n+d^nk^n=d^nl^n\] Factoring out \(d^n\) we see: \[d^n(j^n+k^n)=d^nl^n\] Dividing by \(d^n\) on both sides of the equation we get: \[j^n+k^n=l^n\] Since \(d\) is the greatest common divisor of \(a\), \(b\), and \(c\), it follows that \(j\), \(k\), and \(l\) must be coprime. Hence, we have found a solution to Fermat’s Last Theorem where \(x\), \(y\), \(z\) are coprime.
Q.E.D.