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Can anybody prove that the following equation is true?

$7^n + 9^n \equiv 0 \pmod {11}\quad\text{where}\quad n\equiv 5 \pmod{10}$

Thanks in advance.

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Hint $\rm\: mod\ 11\!:\ 9^{\!\:5+10j}\!+7^{\:\!5+10k}\! \equiv (3^2)^{5+10j}\! + (-2^2)^{5+10k}\!\equiv (3^{10})^{1+2j}\!+(-2^{10})^{1+2k}\!\equiv 1 - 1 $

Remark $\ $ Thus it is just a special case of the fact that for a prime $\rm\:p = 4\:k+3$

$\rm mod\ p\!:\ (a^2)^{2k+1}+(-b^2)^{2k+1}\equiv\: a^{p-1} - b^{p-1}\equiv 1 - 1\ \ \ for\ \ a,b\not\equiv 0$

The innate structure will become clearer when you learn about the group structure of squares and quadratic reciprocity.

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    Got it! Thanks a lot both of you.2012-04-29
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Hint: Express $n$ as $n=10m+5$, then show $7^{10}\equiv 9^{10}\equiv 1$ mod 11.

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    Most textbooks and lectures are very weak at explicitly pointing out relations between abstractions and prior-known concrete manifestations. It is essential that students learn how to efficiently navigate such abstract-concrete hierarchies. I often emphasize examples of such, because it can help to greatly aid students in the process of learning how to comprehend abstractions. Perhaps I emphasize it too much at times, often pointing out generalizations, and, conversely, various interesting specializations. But better too much than too little - better to be overwhelmed by beauty than blind.2012-04-29