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I am struggling to show that $a^{ϕ(n)+1}≡ a \pmod p$, when $n$ is a product of distinct odd primes. Also, $p$ is prime.

First, I considered a case where $n$ is simply the product of one odd prime; that is, $n = p$.

If I let $n = p$, it's easy:

$$a^{ϕ(p)+1} ≡a^{(p-1)+1} ≡ a^{p}≡ a\pmod p$$. This is certainly true by Fermat's Little Theorem.

I struggle, however, to see this relationship if we let $n = pq$, where $q$ is an odd prime and $p \nmid q$.

For example, if we let $n = pq$, we have:

$$a^{ϕ(pq)+1} ≡ a^{ϕ(p)ϕ(q)+1} ≡ a^{(p-1)(q-1)+1} ≡ a^{(pq-p-q+1) +1}$$

I have no idea how to argue that $ a^{(pq-p-q+1) +1}$ is congruent to $a$ (mod $p$), assuming that's even what you're supposed to do.

I also am trying to figure out the case where $n = pqr$, where $r$ is another distinct odd prime, etc. My goal, by figuring out all these cases is to figure out how I can then generalize it when n is equal to the product of any number of distinct odd primes.

Hopefully my question makes sense. I appreciate any insight I can receive.

Bill Dubuque
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John Coltrane
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    If $a$ is a multiple of $p$ it is easy, if not then we have $a^\varphi(n)\equiv 1 \bmod p$ because $\varphi(n)$ is a multiple of $\varphi(p)$ and euler's theorem. – Asinomás Jun 30 '21 at 21:00
  • As is in the linked proofs $\bmod p!:\ a^{1+(p-1)k}\equiv a,$ is clear if $,a\equiv 0,,$ else by Fermat $,\color{#c00}{a^{p-1}\equiv {\bf 1}}\Rightarrow a^{1+(\color{#c00}{p-1})k} \equiv a(\color{#c00}{a^{p-1}})^k\equiv a(\color{#c00}{\bf 1})^k\equiv a,$ by Congruence Power and Product rules. – Bill Dubuque Jun 30 '21 at 21:17
  • Thank you, @BillDubuque. That answers my question perfectly. – John Coltrane Jun 30 '21 at 22:19

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