$5^{2012}+1$ is divisible with $313$ Prove that: $\displaystyle5^{2012}+1$ is divisible with $313$. 
What I try and what I know: 
$313$ is prime
and I try use the following formula : 
$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\ldots\pm(-1)^{n}b^{n-1})$$ 
but still nothing. this problem can be solved using a elementary proof because I found it a mathematical magazine for children with the age of 14. 
 A: $$ \frac {5^{2012} + 1}{626} =\frac {5^{2012} + 1}{5^{4} + 1 }  = \frac {(5^{4})^{503} + 1}{(5^{4}) + 1 } = \frac {a^{n} + b ^n }{a+b}
$$
Where $ a = 5^4  and\ $ $ b =1 \ and\ $ $ n=503 (odd)$  
Concept : $ {a^{n} + b ^n } $ is always a multiple of $ a+b\ $ when n is odd
.
so this division gives remainder $0 $.
As $ 626 = 313 *2 $  It should also be divisible by 313 .
hence $ \frac {5^{2012} + 1}{313}  $ gives remainder $ 0 $. 
so its divisible by 313 .
Hence proved
A: $5^4=625\equiv -1\pmod {313}$ as $626=2\cdot313$
So, $5^{2012}=(5^4)^{503}\equiv (-1)^{503}\pmod {313}\equiv-1$

Alternatively, $5^4=625=313\cdot2-1$
So,  $5^{2012}=(5^4)^{503}=(313\cdot2-1)^{503}=(313\cdot2)^{513}+\binom {513}1(313\cdot2)^{512}(-1)^1+\cdots+\binom {513}{512}(313\cdot2)(-1)^{512}-1$
Observe that all the except the last is divisible by $313$
So, the remainder i.e.,  $5^{2012} \mod {313}$ is $-1$
A: $5^{2012}+1=(5^4)^{503}-(-1)^{503}$, thus $5^4-(-1)|5^{20212}+1$ Finally we note that $313|5^4-(-1)$. This is because $5^4-(-1)=626=2(313)$
A: Just use the fact that $a+b$ divides $a^n+b^n$, if $n$ is odd.
Here $5^4+1=626$ divides $(5^4)^{503}+1^{503}=5^{2012}+1$.
