, 2013

Write $p^4 - q^6 = r^n$ for some prime $r$ and some positive integer $n$. The expression $p^4 - q^6$ can be factored as $p^4 - q^6 = (p^2 - q^3)(p^2 + q^3)$. Since $p^4 - q^6 > 0$, the primes $p$ and $q$ cannot be equal and are therefore relatively prime. Let $d$ be the greatest common divisor of the numbers $p^2 - q^3$ and $p^2 + q^3$. Then $d$ divides $2p^2$ and $2q^3$. Since $p$ and $q$ are relatively prime, it follows that $d$ can be at most $2$.

If $d = 2$, then $r = 2$ as well as $p^2 - q^3 = 2$ and $p^2 + q^3 = 2^{n-1}$. Adding the equations and dividing by $2$ we get $p^2 = 1 + 2^{n-2}$, or $2^{n-2} = p^2 - 1 = (p-1)(p+1)$. The numbers $p-1$ and $p+1$ have to be powers of $2$, but this is only possible if $p=3$. Now, we get a contradiction with $q^3 = 7$.

So, $d=1$. Since $p^2 - q^3$ and $p^2 + q^3$ are relatively prime and their product is a power of a prime, we have $p^2 - q^3 = 1$, or $p^2 = q^3 + 1 = (q+1)(q^2 - q + 1)$. Since $q$ is prime, we have $q+1 > 2$ and $q^2 - q + 1 > 2$, so neither of the factors $q+1$ and $q^2 - q + 1$ is equal to $1$. If their product is to be the square of a prime, the two factors have to be equal. The equality $q+1 = q^2 - q + 1$ implies $q = 2$ and from $p^2 = 1+q^3$ we then obtain $p=3$. Finally, we check that $3^4 - 2^6 = 17$ is indeed a power of a prime. The only solution is $p=3, q=2$.