Problem 1711

Solution:
We consider two cases.

Case 1. Let $(a, b) = (a, c) = 1$. Then $(a, b c) = 1$ and hence there are integers $u$ and $v$ such that $u a + v b c = 1$. This means that $a$ divides $-v b c + 1$. If $k \geq 1$ is a positive integer such that $a$ divides $-v - k$, then $a$ divides $k b c + 1$, i.e. $k b c + 1 = a t$. Hence setting $x = 2^{t}$, $y = 2^{k c}$ and $z = 2^{k b}$ we have that
$$y^{b} + z^{c} = 2^{k b c} + 2^{k b c} = 2^{k b c + 1} = \left(2^{t}\right)^{a} = x^{a}$$

Case 2. Let $(c, a) = (c, b) = 1$. Then $(c, a b) = 1$ and as above we find a positive integer $k$ such that $c$ divides $k a b + 1$, i.e., $k a b + 1 = c t$. Hence setting $x = 2\left(2^{a} - 1\right)^{k b}$, $y = \left(2^{a} - 1\right)^{k a}$ and $z = \left(2^{a} - 1\right)^{t}$ one has that
$$x^{a} - y^{b} = 2^{a}\left(2^{a} - 1\right)^{k a b} - \left(2^{a} - 1\right)^{k a b} = \left(2^{a} - 1\right)^{k a b + 1} = z^{c}$$