# Proof of equation in modular arithmetic

Can anybody prove that the following equation is true?

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

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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. –  Nick Apr 29 '12 at 16:17
Hint: Express $n$ as $n=10m+5$, then show $7^{10}\equiv 9^{10}\equiv 1$ mod 11.
That's the easy part. But you give no hint why $7^5 + 9^5\equiv 0\pmod{11}.\:$ For that, see my answer. –  Bill Dubuque Apr 29 '12 at 15:36
Strange, both steps are equally easy. I think it would be rather unlikely for one who proves the given hint to get stuck on the part you mention. This hint is easily reapplied to general problems of the same type: eliminating $m$ is the hard part, then reducing the constant residue is easy. –  rschwieb Apr 29 '12 at 15:41