Let $p$ and $q$ be distinct primes. Prove that
$$ p^{q−1} + q^{p−1} \equiv 1 \pmod {pq} $$
We can use result of the previous exercise: if $\gcd (m, n) = 1$, then
$$ m^{\phi(n)} + n^{\phi(m)} \equiv 1 \pmod {mn} $$
Since $p$ and $q$ are prime, we have $\gcd(p,q)=1$, and so
$$ p^{\phi(q)} + q^{\phi(p)} \equiv 1 \pmod {pq} $$
Using $\phi(q)=q-1$ and $\phi(p)=p-1$, we conclude
$$ p^{q-1} + q^{p-1} \equiv 1 \pmod {pq} $$