The divisibility of $a^p-1$ by $a-1$ and by $(a-1)^2$

Question

I was working my way through some number theoretic proofs and being a newbie am stuck on this problem :

Let a $\geq$ 2 and p be any positive integers , then prove that :

• $(a-1) \mid(a^p - 1) $

• $ (a-1)^2 \mid (a^p - 1) $ iff $ (a-1) \mid p $

My Attempt (Inadequate):

First Part :

• $(a-1) \mid(a^p - 1) $

• We know that $(a^p - 1)$ can be written as $(a-1).(a^{p-1} + \cdots + 1)$

• $\Rightarrow$ $(a^p - 1)$ is of the form (a-1).$\lambda$

• $\Rightarrow$ (a-1) $\mid$ $(a^p - 1)$

Hence , Proved

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Second Part:

I have absolutely no clue and that is why ask for a hint and not a solution

My Question :

I am unable to make any concrete progress . Even a decent hint would be acceptable so that I can build on that ...

Answer 1

Hints: The first can be proven easily by induction on $p$.

The second you can prove by showing (again by induction on $p$) that $$a^p-1\equiv (a-1)p\pmod {(a-1)^2}.$$

(Of course, this second proof could be used to prove the first part.)

Answer 2

Hint: using the binomial theorem, $$a^p-1=\sum_{k=1}^p \binom pk (a-1)^k\ .$$

Answer 3

Here's an idea. $(a-1)k=a^p-1 \\ (a-1)k=a^p-1+a^{p+1}-a^{p+1}$ Play with this.

Answer 4

Let $a-1=d$

So, $a^p-1=(1+d)^p-1$

Using Binomial Expansion, $(1+d)^p-1\equiv1+dp-1\pmod{d^2}$

which will be $\equiv0\pmod{d^2}\iff p\equiv0\pmod d$

Answer 5

Another form

Consider $p(x) =x^p -1 =(x-1)(x^{p-1} + ... + 1)$, now $p(a) = (a-1)(a^{p-1} + ...+1)$ here we have the first step.

Now we consider that $g(x) = x^{p-1} + ...+1$ and take $g(1) = p$, then by remainder theorem we have g(x) = (x-1)q(x) + p, where q(x) is a polynomial and take $$g(a) = (a-1)q(a) + p$$ and how $(a-1)|p$, we have $(a-1)|(a^{p-1} + ...+1)$.

Therore $(a-1)^2|a^p -1$ if $(a-1)|p$

For the only if, is easy with the above fact