Competition problem — Number theory

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Number theory Difficulty 4.8 AIME Find the answer

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.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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 $2^{5}-1=31$ .

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