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. 
 A: 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)
A: 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 $\,a^{p-1}=1\pmod p\,$
$$\bullet\;\;\;\;\;\; 2013=19\cdot105-1\stackrel{\text{arithmetic mod 19}}\implies 12^{2013}+7^{2013}=\left(12^{19}\right)^{106}12^{-1}+\left(7^{19}\right)^{106}7^{-1}=$$
$$=12^5\cdot12^{10}+7^5\cdot7^{10}=(-7)^{15}+7^{15}=0$$
A: Since $12\equiv -7\pmod{19}$, we have
$$12^{2013}+7^{2013}\equiv (-7)^{2013}+7^{2013}\equiv 0\pmod{19}.$$
A: First we prove two theorems that we will use

If $a,\;b,\;c\;\in \mathbb{N}$ with $a\neq0$ and $x,y\in\mathbb{N} $such that $a\mid b$ and $a\mid c$, then, $a\mid bx+cy$;

Show:$a\mid b$ and $a\mid c$ implies that there $m,n\in\mathbb{N}$ such that $b=a\cdot m$ and $c=a\cdot n$;$$bx+cy=am\cdot x+an\cdot y=a(mx+ny)\Longrightarrow a\mid bx+cy \;\;\;\Box$$Now we have to prove the theorem we use to solve this issue, and in this test we use the theorem proved above

Are $a,b,n\in\mathbb{N}$ with $a+b\neq0$ we have $a+b\mid a^{2n+1}+b^{2n+1}$ 

Show: Have demonstrated induction $n=0$ the statement is true because $a+b\mid a^{2\cdot 0+1}+b^{2\cdot0+1}\Longrightarrow a+b\mid a+b$; hypothesis: $a+b\mid a^{2n+1}+b^{2n+1}$;$$a^{2(n+1)+1}+b^{2(n+1)+1}=a^{2n+3}+b^{2n+3}=a^{2n+1+2}+b^{2n+1+2}=a^2a^{2n+1}+b^2b^{2n+1}=$$$$=a^2a^{2n+1}+b^2b^{2n+1}\underbrace{-b^2a^{2n+1}+b^2a^{2n+1}}_{=0}=a^2a^{2n+1}-b^2a^{2n+1}+b^2b^{2n+1}+b^2a^{2n+1}=$$$$=a^{2n+1}(a^2-b^2)+b^2(a^{2n+1}+b^{2n+1})$$we know that $a+b\mid a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$; our hypothesis gives us that $a+b\mid a^{2n+1}+b^{2n+1}$ and the theorem shown above assures us that $a+b\mid a^{2n+1}(a^2-b^2)+b^2(a^{2n+1}+b^{2n+1})$ thus showing that the theorem is valid for $n +1$, then, goes for any $n\in\mathbb{N}\;\;\Box$

Question: Show that $19\mid 12^{2013}+7^{2013}$

Using the theorem proved above, we have easily that $19=12+7$ and $$12+7\mid 12^{2013}+7^{2013}$$because $2013=2\cdot1006+1$ and $1006\in\mathbb{N}$
