1989 IMO Problems/Problem 5
Problem
Prove that for each positive integer there exist consecutive positive integers none of which is an integral power of a prime number.
Solution 1
There are at most 'true' powers in the set . So when gives the amount of 'true' powers we get that .
Since also , we get that . Now assume that there is no 'gap' of lenght at least into the set of 'true' powers and the primes. Then this would give that for all in contrary to the above (at east this proves a bit more).
Edit: to elementarize the part: Look . Then all numbers in the residue classes are not primes (except the smallest representants sometimes). So when there wouldn't exist a gap of length , there has to be a 'true' power in each of these gaps of the prime numbers, so at least one power each numbers, again contradicting .
This solution was posted and copyrighted by ZetaX. The original thread for this problem can be found here: [1]
Solution 2
By the Chinese Remainder Theorem, there exists such that: where are distinct primes. The consecutive numbers each have at least two prime factors, so none of them can be expressed as an integral power of a prime.
Solution 3
OG, If , then select any composite number If , then the consectutive number , give the desired consecutive numbers as each of the numbers are divisible by at least 2 primes. Proof: Each of the numbers are , Now since , Thus , which clearly has atleast 2 prime divisors. Hence DONE, OG
See Also
1989 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |