A perfect power is an integer n that can be represented as ak for some positive integers a≥1 and k≥2. Find the sum of all prime numbers 0<p<50 such that p is 1 less than a perfect power.
A number or a short expression. Spacing and $ signs are ignored.
Solution
First, it is known that ak−1=(a−1)(ak−1+ak−2+…). This means either a−1 or ak−1+ak−2+…+1 must be 1 in order for ak−1 to be prime. But this only occurs when a is 2 . Thus, the only possible primes are of the form 2k−1 for some integer k>1. One can check that the primes of this form less than 50 are 22−1=3,23−1=7, and 25−1=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.