Show that $19\mid 12^{2013}+7^{2013}$ Show that $$19\mid 12^{2013}+7^{2013}$$$$$$No use of modular arithmetic, have not come this content.
 A: Note that $12\equiv-7\pmod{19}$, so that
$$
12^{2013}+7^{2013}\equiv(-7)^{2013}+7^{2013}\equiv-7^{2013}+7^{2013}\equiv0\pmod{19}.
$$
Hence $19$ divides $12^{2013}+7^{2013}$.

Edit: The OP has changed the question, asking to not use modular arithmetic. 
Here, we can use
$$
12^{2013}+7^{2013}=12^{2013}-(-7)^{2013},
$$
along with the following fact:
$$
\frac{a^n-b^n}{a-b}=a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+a^2b^{n-3}+ab^{n-2}+b^{n-1}.
$$
This tells us that
$$
12^{2013}-(-7)^{2013}=M\cdot (12-(-7))=19M,
$$
where
$$
M:=\sum_{k=0}^{2012}12^k\cdot(-7)^{2012-k}\in\mathbb{Z}.
$$
A: Using 
$$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-...-ab^{2n-1}+b^{2n})$$
you get
$$12^{2013}+7^{2013}=(12+7) \left(12^{2012}-12^{2011}\cdot 7+12^{2010}\cdot 7^2-...+7^{2012}  \right)$$
P.S. Alternately, you can use the binomial theorem.
$$12^{2013}+7^{2013}=(19-7)^{2013}+7^{2013}$$
$$(19-7)^{2013} =\sum_{k=0}^{2013} \binom{2013}{k}19^{2013-k}(-1)^k7^k$$
Now, use the fact that for $1 \leq k \leq 2013$ the number $\binom{2013}{k}19^{2013-k}(-1)^k7^k$ is divisible by $19$. Hence
$$(19-7)^{2013} =\mbox{multiple of 19}+  \binom{2013}{2013}19^{0}(-1)^{2013}7^{2013}$$
What I am doing in this second solution is just hidden modular arithmetic ;)
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 $1006\in\mathbb{N}$
