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.
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.
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.
Hint: Express $n$ as $n=10m+5$, then show $7^{10}\equiv 9^{10}\equiv 1$ mod 11.