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. 

Proposition : $a+b$ divides $a^m+b^m$ if $m$ is odd
Some proofs :
$1:$ Let $a+b=c,$
$a^m+b^m=a^m+(c-a)^m\equiv a^m+(-a)^m\pmod c\equiv \begin{cases} 2a^m &\mbox{if } m \text{ is even }  \\
0 & \mbox{if } m \text{ is odd } \end{cases}\pmod c $
$2:$ If $m$ is odd, $a^m+b^m=a^m-(-b)^m$ is divisible by $a-(-b)=a+b$
as $\frac{A^r-B^r}{A-B}=A^{r-1}+A^{r-2}B+A^{r-3}B^2+\cdots+A^2B^{r-3}+AB^{r-2}+B^{r-1}$ which is an integer if $A,B$ are integers and integer $r\ge0$
$3:$ Inductive  proof:
$\underbrace{a^{2n+3}+b^{2n+3}}=a^2\underbrace{(a^{2n+1}+b^{2n+1})}-b^{2n+1}(a^2-b^2)\equiv a^2\underbrace{(a^{2n+1}+b^{2n+1})}\pmod {(a+b)}$
So, $a^{2(n+1)+1}+b^{2(n+1)+1}$ will be divisible by $a+b$ if $(a^{2n+1}+b^{2n+1})$ is divisible by $a+b$
Now clearly,$(a^{2n+1}+b^{2n+1})$ is divisible by $a+b$ for $n=0,1$
Hence the proposition will hold for all positive integer $n$ (By induction)
