TITLE: How to prove that $n^7\equiv n^3\mod40,\forall n\in\mathbb{Z}$
QUESTION [8 upvotes]: I have a problem when I try to induction on $\mathbb{Z}$. I don't know how to solve when I try $n-1$ or $n+1$.

REPLY [2 votes]: It's special case $\ p^i = 2^3,\ q^j = 5,\ \phi = 4,\ k=3\ $ of the Euler-Fermat generalization below.
Theorem $\ \  n^{\large k+\phi}\equiv n^{\large k}\pmod{p^i q^j}\ \, $ assuming $ \ \color{#0a0}{\phi(p^i),\phi(q^j)\mid \phi},\, $ $\, \color{#90f}{k\ge i,j},\,\ p\ne q.\ $ Proof:
$ p\nmid n\,\Rightarrow\, {\rm mod\ }p^i\!:\  n^{ \phi}\equiv 1\,\Rightarrow\, n^{k + \phi}\equiv n^k,\, $ by $\  n^{\Large \color{#0a0}\phi}\! = (n^{\color{#0a0}{\Large \phi(p^{ i})}})^{\large \color{#0a0}m}\overset{\color{blue}{\rm E}}\equiv 1^{\large m}\!\equiv 1\,$ by $\,\rm\color{blue}{E}=$ Euler.
$\color{#c00}{p\mid n}\,\Rightarrow\, {\rm mod\ }p^i\!:\  n^k\equiv 0\,\equiv\, n^{k + \phi}\ $ by $\ n^k = n^{k-i} \color{#c00}n^i = n^{k-i} (\color{#c00}{mp})^i$ and $\,\color{#90f}{k\ge i}.$
So $\ p^i\mid n^{k+\phi}\!-n^k.\,$ By symmetry $\,q^j\,$ divides it too, so their lcm $ = p^iq^j\,$ divides it too. $\ $ QED
Remark $\ $ Obviously the proof immediately extends to an arbitrary number of primes. This leads the way to Carmichael's Lambda function, a generalization of Euler's phi function.