Prove that, for distinct primes $p,q$: $$\tag{1} p^{q-1} + q^{p-1} \equiv{1} \pmod{pq}.$$
This is a pretty straight forward problem, I know, and although the standard proof is simpler, I was wondering if there is any flaw in this slightly different one.
Proof: Let $p$ and $q$ be prime numbers, with $p\lt q$. By Fermat’s Little Theorem, since $p$ and $q$ are distinct: $$\begin{align} \tag{2} p^{q-1} \equiv 1 \pmod{q}, \\ \tag{3} q^{p-1} \equiv 1 \pmod{p}.\end{align}$$ By $(2)$ and $(3)$, there are integers $k_1,k_2$ such that $$\begin{align} p^q-p +q^p-q=(k_1 +k_2)pq \\ \implies \tag{4} p^q+q^p\equiv p+q\pmod{pq}.\end{align}$$
Obviously, $pq^{p-1} \equiv qp^{q-1}\equiv 0\pmod{pq},$ and so adding these terms to $(4)$ gives $$\tag{5} (p+q)( p^{q-1}+q^{p-1})\equiv p+q\pmod{pq} .$$
Since $p+q\lt 2q\leq pq $, it follows that $pq$ and $(p+q)$ are coprime, and so, from $(5)$, we can conclude that $$p^{q-1}+q^{p-1} \equiv 1 \pmod{pq}.$$ $\square$
Your prof is correct.
Or, since $p\mid q^{p-1}-1$ and $q\mid p^{q-1}-1$ we have $$pq\mid (q^{p-1}-1)(p^{q-1}-1)$$ so $$pq \mid p^{q-1}q^{p-1}-q^{p-1}-p^{q-1}+1$$ But since $$pq \mid p^{q-1}q^{p-1}$$ we have $$pq\mid -q^{p-1}-p^{q-1}+1$$ and thus a conclusion.