Solvability of a variant Fermat equation If
$p$
is an odd prime,
are there any pairwise coprime integers
$a, b, c > 0$
such that
$$(a-b+c)^{p} + (a+b-c)^{p} = (a+b+c)^{p}$$
[a proof without Fermat's last theorem]
 A: Fermat's Last Theorem states that there are no non-trivial (i.e. $x,y,z\neq 0$) positive integer solutions to $x^n+y^n=z^n$ for $n\geq 3$.  The first odd prime is 3, so there are no solutions to your equation.
The only possibilities are for at least one of $a-b+c$, $a+b-c$, or $a+b+c$ to be non-positive.  If one is zero, all must be zero, but $a+b+c>0$ since $a,b,c>0$.  
Thus, we need only check the case when some of them are negative. Since $a,b,c>0$, it follows that $a+b+c>0$.  Thus, with $x=a-b+c$, $y=a+b-c$ and $z=a+b+c$, we know that $z^p>0$.  Thus, it cannot be that both $x,y<0$ as $x^p,y^p<0$ and $x^p+y^p<0<z^p$.  Thus, only one of $x$ or $y$ may be negative.  Suppose without $x<0$.  Then $y^p=(-x)^p+z^p$ contradicts Fermat's Last Theorem.  Similarly, if $y<0$, then $x^p=(-y)^p+z^p$, again giving us a contradiction.
For this reason, there can be no solutions.
A: The goal of dealing with the variant Fermat equation without solving the Fermat problem is not achievable. Hayden has shown that if the variant equation has a solution with $a$, $b$, $c$ positive, then the original Fermat equation has a non-trivial solution. We show that if the variant equation has no positive solution, then the Fermat equation has no non-trivial solution. 
Suppose to the contrary that $x^p+y^p=z^p$, with $x$, $y$, and $z$ positive.
Set $a-b+c=2x$, $a+b-c=2y$, $a+b+c=2z$. Then $a=x+y$, $b=z-x$  and $c=z-y$. If they happen not to be pairwise relatively prime, divide each by $\gcd(a,b)$.
