Let $a > 2$ and $n ≥ 1$ be any integers. Prove that $a − 1 ∣ a^n − 1$.
Is this a problem involving induction or on using mods? I thought it was induction but would I need to have two base cases, ($a=3$ and $n=1$). If I do, how would I proceed with more than one base case? Or is it not related to induction at all and there is some other way?
Any help would be highly appreciated!
We have the identity
$$ a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a^2+a^1+1).$$ So, $(a-1)|(a^n-1)$.
This is not true. Since $a-1|a-1$ we have $a-1|n(a-1)= an-n$ so $$ a-1| (an-1)-(an-n) = n-1$$
So if $n= a+1$ we get $a-1|a$, which is true only if $a-1=1$, thus if $a=2$.
Ah, this is completely different question and this is true since we have a factorization:
$$a^n-1 = (a-1)(a^{n-1}+a^{n-2}+...+a^2+a+1)$$
