International Mathematical Olympiad

Let $M = (p+q)^{p-q}(p-q)^{p+q} - 1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$.
By eliminating the term $-1$ in the numerator,
$$\begin{align*} (p+q)^{p+q}(p-q)^{p-q}-1 &\equiv (p+q)^{p-q}(p-q)^{p+q}-1 \quad (\bmod M) \\ (p+q)^{2q} &\equiv (p-q)^{2q} \quad (\bmod M) \tag{1}\\ \left((p+q) \cdot (p-q)^{-1}\right)^{2q} &\equiv 1 \quad (\bmod M) \tag{2} \end{align*}$$

Case 1: $q \geqslant 5$.
Consider an arbitrary prime divisor $r$ of $M$. Notice that $M$ is odd, so $r \geqslant 3$. By (2), the multiplicative order of $\left((p+q) \cdot (p-q)^{-1}\right)$ modulo $r$ is a divisor of the exponent $2q$ in (2), so it can be $1, 2, q$ or $2q$.
By Fermat's theorem, the order divides $r-1$. So, if the order is $q$ or $2q$ then $r \equiv 1 (\bmod q)$. If the order is $1$ or $2$ then $r \mid (p+q)^2 - (p-q)^2 = 4pq$, so $r = p$ or $r = q$. The case $r = p$ is not possible, because, by applying Fermat's theorem,
$$M = (p+q)^{p-q}(p-q)^{p+q} - 1 \equiv q^{p-q}(-q)^{p+q} - 1 = \left(q^2\right)^p - 1 \equiv q^2 - 1 = (q+1)(q-1) \pmod{p}$$
and the last factors $q-1$ and $q+1$ are less than $p$ and thus $p \nmid M$. Hence, all prime divisors of $M$ are either $q$ or of the form $kq+1$; it follows that all positive divisors of $M$ are congruent to $0$ or $1$ modulo $q$.
Now notice that
$$M = \left((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}} - 1\right)\left((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}} + 1\right)$$
is the product of two consecutive positive odd numbers; both should be congruent to $0$ or $1$ modulo $q$. But this is impossible by the assumption $q \geqslant 5$. So, there is no solution in Case 1.

Case 2: $q = 2$.
By (1), we have $M \mid (p+q)^{2q} - (p-q)^{2q} = (p+2)^4 - (p-2)^4$, so
$$\begin{gathered} (p+2)^{p-2}(p-2)^{p+2} - 1 = M \leqslant (p+2)^4 - (p-2)^4 \leqslant (p+2)^4 - 1, \\ (p+2)^{p-6}(p-2)^{p+2} \leqslant 1. \end{gathered}$$
If $p \geqslant 7$ then the left-hand side is obviously greater than $1$. For $p = 5$ we have $(p+2)^{p-6}(p-2)^{p+2} = 7^{-1} \cdot 3^7$ which is also too large.
There remains only one candidate, $p = 3$, which provides a solution:
$$\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1} = \frac{5^5 \cdot 1^1 - 1}{5^1 \cdot 1^5 - 1} = \frac{3124}{4} = 781.$$
So in Case 2 the only solution is $(p, q) = (3, 2)$.

Case 3: $q = 3$.
Similarly to Case 2, we have
$$M \left\lvert\, (p+q)^{2q} - (p-q)^{2q} = 64 \cdot \left(\left(\frac{p+3}{2}\right)^6 - \left(\frac{p-3}{2}\right)^6\right) .\right.$$
Since $M$ is odd, we conclude that
$$M \left\lvert\, \left(\frac{p+3}{2}\right)^6 - \left(\frac{p-3}{2}\right)^6\right.$$
and
$$\begin{gathered} (p+3)^{p-3}(p-3)^{p+3} - 1 = M \leqslant \left(\frac{p+3}{2}\right)^6 - \left(\frac{p-3}{2}\right)^6 \leqslant \left(\frac{p+3}{2}\right)^6 - 1, \\ 64(p+3)^{p-9}(p-3)^{p+3} \leqslant 1. \end{gathered}$$
If $p \geqslant 11$ then the left-hand side is obviously greater than $1$. If $p = 7$ then the left-hand side is $64 \cdot 10^{-2} \cdot 4^{10} > 1$. If $p = 5$ then the left-hand side is $64 \cdot 8^{-4} \cdot 2^8 = 2^2 > 1$. Therefore, there is no solution in Case 3.