Maths Olympiad Prep

Library / /90 of 348

Number theory Difficulty 4.8 AIME Find the answer

A perfect power is an integer nn that can be represented as aka^{k} for some positive integers a1a \geq 1 and k2k \geq 2. Find the sum of all prime numbers 0<p<500<p<50 such that pp is 1 less than a perfect power.

A number or a short expression. Spacing and $ signs are ignored.

Solution

First, it is known that ak1=(a1)(ak1+ak2+)a^{k}-1=(a-1)\left(a^{k-1}+a^{k-2}+\ldots\right). This means either a1a-1 or ak1+ak2++1a^{k-1}+a^{k-2}+\ldots+1 must be 1 in order for ak1a^{k}-1 to be prime. But this only occurs when aa is 2 . Thus, the only possible primes are of the form 2k12^{k}-1 for some integer k>1k>1. One can check that the primes of this form less than 50 are 221=3,231=72^{2}-1=3,2^{3}-1=7, and 251=312^{5}-1=31.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.