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 $n^{\phi(p)} \equiv 1$?
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2Is $p$ a prime? – Randall Mar 11 '19 at 02:58
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3congruent modulo what? – J. W. Tanner Mar 11 '19 at 02:58
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1Carmichael Numbers – JavaMan Mar 11 '19 at 03:05
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@J.W.Tanner modulo $p$. – YFP Mar 11 '19 at 04:31
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2If $p$ is prime and $n$ and $p$ are not co-prime, then $n \equiv 0 \pmod p$ – J. W. Tanner Mar 11 '19 at 04:33
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Even, if $p$ is not prime, if $n$ and $p$ are not coprime, there is a prime number $q$ dividing both $n$ and $p$ , then $n^k\equiv 1\mod p$ (with $k\ge 1$) would imply $n^k\equiv 1\mod q$ , hence $q\mid n^k-1$ which contradicts to $q\mid n$. So, the congruence cannot be satisfied when $n$ and $p$ are not coprime. – Peter Mar 11 '19 at 11:36
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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.
Bill Dubuque
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