If $n \mid (a-1)^k$ then $n \mid(a^{n-1} + a^{n-2} + ...+ a + 1) $ and $n,k,a \in N_{\geq2} $

Question

I know that $\nu_{p}(n) \leq \nu_{p}((a-1)^k)$ and $\nu_{p}(a^n-1^n) = \nu_{p}(a-1) + \nu_{p}(n)$

But I'm stuck on this..

Also I was wondering If there is kind of relation between n and k.

Answer 1

You're nearly done, since you've already figured this out,

$$v_p(a^n-1)=v_p(a-1)+v_p(n)$$

You can use the additivity of the p-adic valuation to rewrite it as,

$$v_p(a^n-1)-v_p(a-1)=v_p(n)$$

$$v_p\left(\frac{a^n-1}{a-1}\right)=v_p(n)$$

Now expand the geometric series,

$$v_p\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)=v_p(n)$$

Since this is true for every prime dividing n, it makes n a divisor. There is also a slight adjustment you must make in order to account for when $p=2$. When using the Lifting the Exponent lemma there is an extra $v_2\left(\frac{a+1}{2}\right)$ to account for in the initial equation on the right hand side. Make sure you convince yourself why this can't cause problems.