# Prove for any integer $n$, $21n^{22} + 22n^{26} + 34n^{32} \equiv 0 \bmod 77$

I am trying to study for my exam but I'm having trouble with doing some of the questions and any help in figuring them out will be greatly appreciated. Thanks so much!

Prove for any real integer, $21n^{22} + 22n^{26} + 34n^{32} \equiv 0 \bmod 77$.

• Is a real integer the same thing as an integer? – rschwieb Mar 19 '13 at 20:31
• Deconstruct the problem into mod 7 and mod 11 and see if you can show how that equation is true in those cases and use the Chinese Remainder Theorem to get the rest of the proof. – JB King Mar 19 '13 at 20:33

Following JB King's suggestion, we know that $n^6 \equiv 1 \pmod 7$ unless $n \equiv 0 \pmod 7$, in which case we have $0\equiv 0 \pmod 7$. We have $21n^{22}+22n^{26}+34n^{32}\equiv 0+1\cdot n^2-1\cdot n^2\equiv 0 \pmod 7$ See if you can do $\pmod {11}$
• $21\equiv 0, 22\equiv 1, 34 \equiv -1 \pmod 7$ and the exponents come from removing the $6$'s – Ross Millikan Mar 19 '13 at 21:07
$\rm\begin{eqnarray}{\bf Hint} &&\rm mod\ \ \ 7\!:\ \ f(n)\,&\equiv&\,\rm n^{26}-n^{32} &\equiv&\rm\, n^{25}\,(\, n - n^7) &\equiv&\rm\, 0\ \ \ by\ little\ Fermat\\ \rm and, &&\rm mod\ 11\!:\ \ f(n)\,&\equiv&\,\rm n^{32}-n^{22}&\equiv&\rm\, n^{21} (n^{11}\!-n)&\equiv&\rm\, 0\ \ \ by\ little\ Fermat\end{eqnarray}$