If $(m,n)=1$ then $m^{\phi(n)}+n^{\phi(m)}\equiv0\pmod {mn}$

Question

Show that if $(m,n)=1$ then $m^{\phi(n)}+n^{\phi(m)}\equiv0\pmod {mn}$

I tried, $(m,n)=1$ then, Euler theorem we have $m^{\phi (n)}\equiv1\pmod n$ and $n^{\phi (m)}\equiv1\pmod m$. But I could not continue.

Answer 1

You are almost done

Observe that $$m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod m$$

Similarly, $$m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod n$$

$$\implies m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod{\text{ lcm}(m,n)}$$

Answer 2

First, one note: your RHS should be $1$, not $0$.

Given that: you've already shown that $m^{\phi(n)}\equiv 1\pmod n$, and it's easy what $n^{\phi(m)}$ equals $\pmod n$; this allows you to compute $m^{\phi(n)}+n^{\phi(m)}\pmod n$. Do the same $\pmod m$, and then use the Chinese Remainder Theorem (along with the fact that $\mathop{GCD}(m,n)=1$) to derive your result.