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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, $21n^{22} + 22n^{26} + 34n^{32} \equiv 0 \bmod 77$.

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    $\begingroup$ Is a real integer the same thing as an integer? $\endgroup$ – rschwieb Mar 19 '13 at 20:31
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    $\begingroup$ 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. $\endgroup$ – JB King Mar 19 '13 at 20:33
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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 $\pmod {11}$

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  • $\begingroup$ I'm sorry, I'm still confused as how you arrived from 21n^22+22n^26+34n^32 to 0+1⋅n^2−1⋅n^2≡0(mod7) $\endgroup$ – Michael Mar 19 '13 at 21:03
  • $\begingroup$ $21\equiv 0, 22\equiv 1, 34 \equiv -1 \pmod 7$ and the exponents come from removing the $6$'s $\endgroup$ – Ross Millikan Mar 19 '13 at 21:07
  • $\begingroup$ Thank you! So in order to do (mod 11) I would use n^10≡1 (mod 11) and the result would look like 10*n2+0+1*n^2 ≡0 (mod 11)? $\endgroup$ – Michael Mar 19 '13 at 21:16
  • $\begingroup$ That is correct $\endgroup$ – Ross Millikan Mar 19 '13 at 21:17
  • $\begingroup$ Thanks so much! But how would I approach this proof now? $\endgroup$ – Michael Mar 19 '13 at 21:26
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$\rm\begin{eqnarray}{\bf Hint} &&\rm mod\ \ \ 7\!:\ \ f(n)\,&\equiv&\,\rm n^{26}-n^{32} &\equiv&\rm\, n^{25}\,(\, n - n^7) &\equiv&\rm\, 0\ \ \ by\ little\ Fermat\\ \rm and, &&\rm mod\ 11\!:\ \ f(n)\,&\equiv&\,\rm n^{32}-n^{22}&\equiv&\rm\, n^{21} (n^{11}\!-n)&\equiv&\rm\, 0\ \ \ by\ little\ Fermat\end{eqnarray}$

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