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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?

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    $\begingroup$ These numbers aren't that big. Can you use a computer to solve the problem? Or even a calculator? $\endgroup$ – Mastrem Nov 9 '16 at 15:15
  • $\begingroup$ $27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \mod 5$. $\endgroup$ – Santiago Nov 9 '16 at 15:15
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    $\begingroup$ Well I've used Wolfram to check if it's indeed divisible (it is), but the proof requires algebraic manipulation, no more no less. Tnx @Santiago, that helps, onto 7 now. $\endgroup$ – nullpotent Nov 9 '16 at 15:19
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    $\begingroup$ @iccthedral: notice that $10887 - 10152 = 735 = 7 \cdot 105$. $\endgroup$ – Santiago Nov 9 '16 at 15:23
  • $\begingroup$ @Santiago All is clear now, thank you very much. $\endgroup$ – nullpotent Nov 9 '16 at 15:25
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Hint $\ $ Exploit innate symmetry using the following observation

$$\begin{eqnarray}a\equiv c,&&\ b\equiv d\!\!\pmod{m}\\ a\equiv d,&&\ b\equiv c\!\!\pmod{k}\end{eqnarray}\, \Rightarrow\,\ e := a^n+b^n-c^n-d^n\equiv 0\!\! \pmod{{\rm lcm}(m,k))}$$

because $\ \ {\rm mod}\ m\!:\ a\equiv c\,\Rightarrow a^n\equiv c^n,\ \ b\equiv d\,\Rightarrow\ b^n\equiv d^n\,\Rightarrow\,e\equiv 0\, $ by Congruence Power Rule.

Similarly $\, {\rm mod}\ k\!:\,\ e\equiv 0.\ $ Thus $\ m,k\mid e\,\Rightarrow\,{\rm lcm}(m,k)\mid e$.

Remark $\ $ More generally, bringing to the fore the innate symmetry yields:

Assume $ \{a,b\}\equiv \{c,d\}\ {\rm mod}\ m,k\ $ and $\,f(x,y)\,$ is symmetric $\,f(x,y)=f(y,x).$

Then $\,\ \ f(a,b)\equiv f(c,d)\ {\rm mod}\ m,k\,$ so also mod $\,{\rm lcm}(m,k).$

OP has $\, f(x,y) = x^n + y^n.$

Remark $\ $ See here and here for some explicit examples.

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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 $0 \bmod m$.

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    $\begingroup$ lhf, what does the matrix represent? $\endgroup$ – nullpotent Nov 9 '16 at 15:29
  • $\begingroup$ Haha, this is too clever to be used as my own proof, professor knows my wits :) Thank you nonetheless, I love the construction. $\endgroup$ – nullpotent Nov 9 '16 at 15:34
  • $\begingroup$ @iccthedral It's a classic exploitation of symmetry - I explain it further in my answer (esp. the linked example). Once you see the innate symmetry it's not "too clever" but somewhat trivial. $\endgroup$ – Bill Dubuque Nov 9 '16 at 15:42

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