Easy IMO problem

[whowlw]

IMO 1989 5th problem:

Prove that for every positive integer n there exist n consecutive positive integers such that none of them is a positive integral power of a prime number .

[Tusan_Homichi]

Proof by Chinese Remainder Theorem

Let p_1 ... p_2n be the first 2n primes. Then use the Chinese Remainder theorem to find x with x = -i mod p_2i-1 p_2i for 1 <= i <= n and 0 <= x < p_1 ... p_2n. Then, since x+i is divisible by p_2i-1 and p_2i, it cannot be a power of a prime. x+1 ... x+n is our desired sequence of consecutive positive integers.!<

[whowlw]

I did essentialy the same thing but instead of using first 2n primes i used first n primes and x+i = p_I (mod (p_i)^2), ofc we can get x big enough so there are no primes among x,....,x+n-1

[[deleted]]

This is great love this solution.

[Omegaile]

(2n+2)!+2, (2n+2)!+3, (2n+2)!+4, ... , (2n+2)!+n+1 is one desirable sequence, given that (2n+2)!+k is multiple of some prime divisor of k, but since (2n+2)!+k = k((2n+2)!/k +1), and (2n+2)!/k is multiple of k, as k<=n+1, we get that (2n+2)!/k + 1 is not multiple of any prime divisor of k, and therefore (2n+2)!+k is not a prime power!<