On generalizing a corollary of Fermat's little theorem (Q2278753)

Fermat's little theorem states that, for every prime \(p\nmid a\), it holds that \(a^{p-1}\equiv 1\pmod{p}\). This implies that \(a^p\equiv p\pmod{p}\) for every prime \(p\) and integer \(a\), even if \(p\mid a\). Fermat's little theorem can be naturally extended to the so-called Euler's theorem which states that, for every prime \(n\) coprime to \(a\), it holds that \(a^{\phi(n)}\equiv 1 \pmod{n}\), where \(\phi\) is the well-know Euler's totient function. Unlike in the case of Fermat's theorem, Euler's theorem does not imply that \(a^{\phi(n)+1}\equiv n \pmod{n}\) for every pair of integers \(a\) and \(n\). This paper is devoted to investigate this situation. The author reaches the following result. Given \(n\) and \(n>1\) be integers, let us denote by \(n_a\) the largest factor of \(n\) that contains all the primes in the factorization of \(a\), and those primes only. Then, \(a^{\phi(n)+1}\equiv a\pmod{n}\) if and only if \(n_a\mid a\). As a corollary, the following result if obtained. Given an integer \(n>1\), there exists \(k>1\) such that \(a^k\equiv a\pmod{n}\) for every \(a\) if and only if \(n\) is square-free. Moreover, \(k=\phi(n)+1\).