$a^x + b^y = c^z \Rightarrow a^{x-2} + b^{y-2} = 0 $ (mod c) if $x,y,z > 2$ and $a,b,c$ are pairwise coprime? Let $a,b,c,x,y,z$ be positive integers such that $x,y,z > 2$ and $a$,$b$,$c$ are pairwise coprime.
Suppose it is given that $a^x + b^y = c^z$ (i.e $a^x + b^y \equiv 0 (\text{mod } c)$), then is it true that $a^{x-2} + b^{y-2} \equiv 0 (\text{mod } c)$ ? If not, could you give a counterexample?
I tried to looking manually for counterexamples in special equations like $x^n + y^n = z^{n+1}$ and $x^n + y^{n+1} = z^{n+2}$, but there weren't any for these special equations.
Source: This question was asked in one of the FB groups I am in.
 A: This might be a trick question. Note the Beal Conjecture says that $A^x + B^y = C^z$ has no solution

in positive integers $A, B, C, x, y, z$ with $A, B,$ and $C$ being pairwise coprime and all of $x, y, z$ being greater than $2$

There's been extensive computer checks done so far and a US million dollar prize for its proof or disproof is on offer, so I don't believe your FB group would know of any such values existing.
However, assume hypothetically the Beal Conjecture is wrong and there are such values. Then
$$a^x + b^y \equiv 0 \pmod c \tag{1}\label{eq1}$$
Without knowing what the specific values of $a,b,c,x,y$ and $z$ are, it's not possible to definitively prove or disprove that
$$a^{x-2} + b^{y-2} \equiv 0 \pmod c \tag{2}\label{eq2}$$
is always true. Nonetheless, given that \eqref{eq1} holds, the following provides an analysis on some of the conditions required for \eqref{eq2} to be true. In particular, it requires that
$$a^x + a^2b^{y - 2} \equiv 0 \pmod c \\
-b^y + a^2b^{y - 2} \equiv 0 \pmod c \\
a^2 \equiv b^2 \pmod c \\
a \equiv \pm b \pmod c \tag{3}\label{eq3}$$
For $a \equiv b$, you get
$$b^{x-2} + b^{y-2} \equiv 0 \pmod c \tag{4}\label{eq4}$$
while for $a \equiv -b$, you get
$$(-b)^{x-2} + b^{y-2} \equiv 0 \pmod c \tag{5}\label{eq5}$$
If $x = y$, then \eqref{eq4} becomes $2b^{x-2} \equiv 0 \pmod c$ which is only true if $c = 2$, and likewise with \eqref{eq5} if $x$ is even. If $x$ is odd, then \eqref{eq5} will always be true. If $x \neq y$, then WLOG, let $y \gt x$, say $y = x + n$ for $n \gt 0$. Then after dividing by $b^{x-2}$, \eqref{eq4}, and \eqref{eq5} if $x$ is even, become
$$1 + b^n \equiv 0 \pmod c \tag{6}\label{eq6}$$
while for $x$ odd, \eqref{eq5} becomes
$$-1 + b^n \equiv 0 \pmod c \tag{7}\label{eq7}$$
Both such cases are possible for various values of $b, n$ and $c$.
Overall, the required conditions for \eqref{eq2} to be true are relatively stringent, so I believe it's unlikely for it to always be true, but I don't see any way to rule out this possibility.
