HMMT-Nov 2021 guts, Problem 5

A perfect power is an integer $n$ that can be represented as $a^{k}$ for some positive integers $a \geq 1$ and $k \geq 2$. Find the sum of all prime numbers $0<p<50$ such that $p$ is 1 less than a perfect power.

Solution: Solution: First, it is known that $a^{k}-1=(a-1)\left(a^{k-1}+a^{k-2}+\ldots\right)$. This means either $a-1$ or $a^{k-1}+a^{k-2}+\ldots+1$ must be 1 in order for $a^{k}-1$ to be prime. But this only occurs when $a$ is 2 . Thus, the only possible primes are of the form $2^{k}-1$ for some integer $k>1$. One can check that the primes of this form less than 50 are $2^{2}-1=3,2^{3}-1=7$, and $2^{5}-1=31$. $\fbox{41}$.

Answer: 41