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, $$\begin{array}{rl}(p+q{)}^{p+q}(p-q{)}^{p-q}-1& \equiv (p+q{)}^{p-q}(p-q{)}^{p+q}-1(\mathrm{mod}M)\\ (p+q{)}^{2q}& \equiv (p-q{)}^{2q}(\mathrm{mod}M)\\ {\left((p+q)\cdot (p-q{)}^{-1}\right)}^{2q}& \equiv 1(\mathrm{mod}M)\end{array}\text{\hspace{1em}(1)(2)}$$

Case 1: $q⩾5$. Consider an arbitrary prime divisor $r$ of $M$. Notice that $M$ is odd, so $r⩾3$. 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 $r\equiv 1(\mathrm{mod}q)$. 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, $$M=(p+q{)}^{p-q}(p-q{)}^{p+q}-1\equiv q^{p-q}(-q{)}^{p+q}-1={\left(q^2\right)}^p-1\equiv q^2-1=(q+1)(q-1)(\mathrm{mod}p)$$ 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 $q⩾5$. 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 $$\begin{array}{c}(p+2{)}^{p-2}(p-2{)}^{p+2}-1=M⩽(p+2{)}^4-(p-2{)}^4⩽(p+2{)}^4-1,\\ (p+2{)}^{p-6}(p-2{)}^{p+2}⩽1.\end{array}$$ If $p⩾7$ 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 $$M|(p+q{)}^{2q}-(p-q{)}^{2q}=64\cdot \left({\left(\frac{p+3}{2}\right)}^6-{\left(\frac{p-3}{2}\right)}^6\right).$$ Since $M$ is odd, we conclude that $$M|{\left(\frac{p+3}{2}\right)}^6-{\left(\frac{p-3}{2}\right)}^6$$ and $$\begin{array}{c}(p+3{)}^{p-3}(p-3{)}^{p+3}-1=M⩽{\left(\frac{p+3}{2}\right)}^6-{\left(\frac{p-3}{2}\right)}^6⩽{\left(\frac{p+3}{2}\right)}^6-1,\\ 64(p+3{)}^{p-9}(p-3{)}^{p+3}⩽1.\end{array}$$ If $p⩾11$ 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.