Prove $(p-1)^{q-1}\equiv (q-1)^{p-1} \pmod{pq}$

Question

Let $p$ and $q$ be $2$ distinct odd prime numbers

Prove $(p-1)^{q-1}\equiv (q-1)^{p-1} \pmod{pq}$

Given :

1. $\gcd(p, q-1) = 1$
2. $\gcd(q, p-1) = 1$

So far I have:

$(p-1)^{q-1} \equiv 1 \pmod{q}$

$(q-1)^{p-1} \equiv 1 \pmod{p}$

using CRT: $(q-1)^{p-1} +(p-1)^{q-1} \pmod{pq} \equiv 1$

Please help, I know I need to use FLT but I'm not sure how to solve it.

Answer 1

Since the statement is not true for when $GCD(p,q-1) \not = 1$ (counterexample using $7,3$). I am assuming that what is "given" is a part of the criteria for the presented identity and not a part of the solution.

By FLT we have $$(p-1)^{q-1} ≡ 1 \mod(q)$$ $$(q-1)^{p-1} ≡ 1 \mod(p)$$

Also by binomial expansion we note that $$ (p-1)^{q-1} = p(\text{some terms here}) + 1^{q-1} \equiv 1 \mod (p)$$ $$(q-1)^{p-1} = q(\text{some terms here}) + 1^{p-1} \equiv 1 \mod (q)$$

Now this means $(p-1)^{q-1} = kpq + 1$ and $(q-1)^{p-1} = kpq + 1$ which implies that $$(p-1)^{q-1} \equiv 1 \equiv (q-1)^{p-1} \mod(pq)$$

Answer 2

$p-1 \equiv - 1 \pmod p$ and $q-1$ is even so $ (p-1)^{q-1} \equiv (-1)^{q-1} \equiv 1 \pmod p$.

Also, by LFT, $(p-1)^{q-1} \equiv 1 \pmod q$.

So, by CRT, $(p-1)^{q-1} \equiv 1 \pmod {pq}$. The same holds for $(q-1)^{p-1}$, by symmetry.