I assume that the equation was meant for non-negative integers (for negatives it's nothing interesting). It's clear that $ x \geq 1$ and $ z \geq 2$ (mod $ 9$, keep in mind that $ 87 = 3 \cdot 29$). Suppose now that $ y \geq 1$ and $ z \geq 3$. From the binomial expansion it easily follows that $ 27|(9k+1)^l - 1$ iff $ 3|l$. Therefore we must have $ 3|y$ since $ 19 = 18 + 1$, $ 28=27+1$ and $ 27|87^z$. Now we look mod $ 29$. Because $ 29|87$ we have the congruence $ (-1)^x \equiv 19^y \pmod{29}$. It's easy to check that $ 19^y \equiv 1 \pmod{29}$ is equivalent to $ 28|y$ (we have to try $ y=2, 4, 7, 14$) and it follows that in order to have $ 19^y \equiv \pm 1 \pmod{29}$ we must have $ y \equiv 0 \pmod{28}$ or $ y \equiv 14 \pmod{28}$, anyway we have $ 14|y$. So we have obtained that $ 42|y$. In particular $ 19^y \equiv 1 \pmod{7}$. $ 87 \equiv 3 \pmod{7}$ and we check that $ 3^z \equiv -1 \pmod{7}$ is equivalent $ z \equiv 3 \pmod{6}$. Now mod $ 13$. We have $ 2^x \equiv 6^y + 9^z \equiv 6^y + 3^{2z} \pmod{13}$. Since $ 3|z$ and $ 3^3 \equiv 1 \pmod{13}$ it follows that $ 3^{2z} \equiv 1 \pmod{13}$. There are two possibilites: $ y \equiv 6 \pmod{12}$ or $ y \equiv 0 \pmod{12}$. In first case we get that the right side is divisible by $ 13$ which is impossible, and in the second we have $ 2^x \equiv 2 \pmod{13}$ and this is equivalent to $ x \equiv 1 \pmod{12}$. We finish the solution with mod $ 19$. Since $ 3|z$ and $ 11^3 \equiv 1 \pmod{19}$ we have that $ 87^z \equiv 11^z \equiv 1 \pmod{19}$. It follows that $ 28^x \equiv 1 \pmod{19}$. But since $ x \equiv 1 \pmod{6}$ it follows that $ 28^x \in \{28, 28^7, 28^{13}\} \pmod{19}$ but this numbers give residues: $ 9, 4, 6$ which is a contradiction. We are left with the cases $ y=1$ and $ z=2$. Let's start with the first one. From the above reasoning the relation $ 3|z$ still holds (mod $ 7$). So let $ z=3k$, then $ 28^x = 87^{3k}+1 = (87^{k}+1)(87^{2k} - 87^{k} + 1)$. It's easy to see that $ 87^{k} + 1$ and $ 87^{2k}-87^{k}+1$ are coprime, so $ 87^{k} + 1 = 4^x$ and $ 87^{2k} - 87^{k}+1 = 7^x$ because the second number is bigger. But: $ 87^{2k} - 87^{k} + 1 = (4^x - 1)^2 - (4^x-1) + 1 = 4^{2x} - 3 \cdot 4^x + 3 = 7^x$, which is a contradiction mod $ 8$, since obviously $ x \geq 2$. Now consider the case $ z=2$. We still have that $ 14|y$ (mod $ 29$), so in particular $ 2|y$. But that's a contradiction mod $ 4$ since the right side is congruent to $ 2 \pmod{4}$. So the equation doesn't have a solution.

Actually, the equation $ 28^x = 19^y + 87^z$ is not solvable modulo $ 889$.

maxal wrote: Actually, the equation $ 28^x = 19^y + 87^z$ is not solvable modulo $ 889$. Mmm. That is quite interesting. How did you work it out? Only trial? I don't think, there must be something tricky.

maxal wrote: Actually, the equation $ 28^x = 19^y + 87^z$ is not solvable modulo $ 889$. Could you tell me how in the world did you manage to work that out, please ?