Prove divisibility: $6\mid 13^n+7^n-2$

Question

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 1

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

Answer 2

Non-induction approach: write $13=6\times 2+1$ and $7=6+1$ then use the binomial theorem.

Edit: Look up the Statement of the Theorem part here. Then, apply it to your case, which gives $$ 13^n=(1+12)^n=1+12\times A,\quad 7^n=(1+6)^n=1+6B $$ for some integers $A$ and $B$. It follows that $13^n+7^n-2=12A+6B$, which of course is divisible by $6$.