Number Theory divisibilty How can I check if $$12^{2013} + 7^{2013}$$ is divisible by $19$?
Also, how can I format my questions to allow for squares instead of doing the ^ symbol. 

Another approach by hints:
$\;\bullet\;$ $\forall\;$ prime $\,p\,$ and $\;\forall\,a\in\Bbb Z\;,\;\;a^p=a\pmod p\;$ , and if $\,a\neq 0\pmod p\;$ then $\,a^{p-1}=1\pmod p\,$
$$\bullet\;\;\;\;\;\; 2013=19\cdot105-1\stackrel{\text{arithmetic mod 19}}\implies 12^{2013}+7^{2013}=\left(12^{19}\right)^{106}12^{-1}+\left(7^{19}\right)^{106}7^{-1}=$$
$$=12^5\cdot12^{10}+7^5\cdot7^{10}=(-7)^{15}+7^{15}=0$$
