Prove $27195^8 - 10887^8 + 10152^8\equiv 0\pmod{26460}$ So far I've tried:
1) $26460 = 2^2 * 3^3 * 5 * 7^2$
2) $10152 = 2^3 * 3^3 * 47$
3) $27195 - 10887 = 16308 = 2^2 * 3^3 * 151$ (I know $a^8 - b^8\equiv0\pmod{a - b}$)
Therefore I conclude that $27195^8 - 10887^8 + 10152^8$ is divisible by $2^2 * 3^3$, as well is 1).
But what about $5 * 7^2$ part? 

Let $a=27195, b=10887, c=10152$ and consider this:
$$
\matrix{
m       &a \bmod m      &b \bmod m      &c \bmod m      \\
2^2     &3      &3      &0      \\
3^3     &6      &6      &0      \\
5\hphantom{^1}       &0      &2      &2      \\
7^2     &0      &9      &9      \\
}
$$
Note how for each $m$, we have $b \equiv a$ or $b \equiv c \bmod m$, and the other is $0 \bmod m$.
Therefore, the terms in $27195^8 - 10887^8 + 10152^8$ reduce to $0 \bmod m$.
