TITLE: how to prove $m^{\phi(n)}+n^{\phi(m)}\equiv 1 \pmod{mn}$ where m and n are relatively prime?
QUESTION [5 upvotes]: I do not know how to prove this question: $m^{\phi(n)}+n^{\phi(m)}\equiv 1 \pmod {mn}$ where $m$ and $n$ are relatively prime.
Can anyone help?

REPLY [3 votes]: Since $\gcd(m, n) = 1$, so by Euler's theorem
$$m^{\phi(n)} \equiv 1 \mod n$$
and
$$n^{\phi(m)} \equiv 1 \mod m$$
But
$$m^{\phi(n)} \equiv 0 \mod m$$
and
$$n^{\phi(m)} \equiv 0 \mod n$$
Thus,
$$m^{\phi(n)} + n^{\phi(m)} \equiv (1 + 0) \mod n  \equiv 1 \mod n$$
and
$$m^{\phi(n)} + n^{\phi(m)} \equiv (1 + 0) \mod m  \equiv 1 \mod m$$
Hence,
$$m^{\phi(n)} + n^{\phi(m)} \equiv 1 \mod mn$$