What happens to $n^{\phi(p)} \equiv 1$ when $n$ and $p$ are not co-prime?

Question

We know $n^{\phi(p)} \equiv 1$ in the case $n$ and $p$ are co-prime i.e. $ gcd(n,p) = 1$. What is the case when they are not co-prime? What happens to $n^{\phi(p)} \equiv 1$?

Answer 1

It is false: $\,n^{\phi}\equiv 1\pmod{p}\,\Rightarrow\, \color{#c00}n^{\phi}+k\,\color{#c00}p = 1\,\Rightarrow\, \gcd(\color{#c00}{n,p}) = 1,\,$ by $\,\phi \ge 1$

But there do exist such generalizations of the other form of Fermat $\,a^{p}\equiv a\pmod{\!p}\,$ such as below.

Theorem $\ $ Suppose $\ m\in \mathbb N\ $ has the prime factorization $\:m = p_1^{e_{1}}\cdots\:p_k^{e_k}\ $ and suppose for all $\,i,\,$ $\ e\ge e_i\ $ and $\ \phi(p_i^{e_{i}})\mid f.\ $ Then $\ m\mid a^e(a^f-1)\ $ for all $\: a\in \mathbb Z,\,$ i.e. $$\qquad\qquad a^{\large e+f}\equiv a^{\large e}\!\!\!\pmod{\!m}\quad {\rm for\ all}\ \ a\in\Bbb Z$$

See this answer for proofs and examples.