If $\gcd(m,n) = 1$, show that $m^{φ(n)} + n^{φ(m)} ≡ 1 \mod {mn}$
The only similar problem I found to this in my notes was "Find all positive integers $n$ such that $φ(n) = 12$." Unfortunately this was no help and I'm really lost, any help is really appreciated.
Since $\gcd(m,n)=1$, you know from Euler-Fermat that $$ m^{\varphi(n)}\equiv 1\pmod{n} $$ and, similarly, $$ n^{\varphi(m)}\equiv 1\pmod{m} $$ Since $n^{\varphi(m)}\equiv 0\pmod{n}$, we also have $$ m^{\varphi(n)}+n^{\varphi(m)}\equiv 1+0\pmod{n} $$ and, similarly, $$ m^{\varphi(n)}+n^{\varphi(m)}\equiv 1+0\pmod{m} $$ Now use the Chinese Remainder Theorem.