elementary number theory - Prove divisibility: $6mid 13^n+7^n-2$

We have the following proposition: $$P(n): 13^n+7^n-2\vdots 6$$.

• Prove $P(n)$ in two ways. I know that one of them is mathematical induction. I don't know many things about the other one, I know it's something from modular arithmetic.


• If we had $p_n = 13^n+7^n-2$ with $n\in\Bbb N^*$, how should we calculate the rest of $p_n:6$?

Answer

use the following facts
$13\equiv 1 \mod 6$ and
$7\equiv 1 \mod 6$
yes you can write $13=2\cdot 6+1$ this means the remainder is $1$ and the same for $7$, $7=6+1$
see here

http://en.wikipedia.org/wiki/Modulo_operation
we use this in our math circle in Leipzig