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Thread: Another number theory

  1. #1
    Junior Member
    Feb 2013

    Another number theory

    If n is equal to or greater than 3 then prove that there's atleast one prime number between n and n!.
    Help would be appreciated.
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  2. #2
    MHF Contributor
    Nov 2010

    Re: Another number theory

    Use a standard induction argument. Show that it is true for n=3. Assume it is true for all values up to n. Prove it is true for n+1.

    So, 3<5<3!=6, so it is obviously true for n=3. Next, assume it is true up to n. How would you show it is true for n+1? There are two cases. Either n+1 is prime or there exists some prime number n+1 < p < n!. If it is the latter case, then you are done. So, suppose n+1 is prime. Let's consider the factors of n!. Since n+1 is prime, it is not a factor of n!, so any factor of n! that is greater than n is also greater than n+1. There are at most n! - (n+1) factors that are greater than n and at most n-1 factors that are greater than 1 and less than or equal to n. So, for (n+1)!, its factors that are greater than n+1 are the factors of n! that are greater than n+1, those same factors times n+1, and also the factors of n! that are no greater than n multiplied by n+1. So, there are at most 2(n!-(n+1)) + (n-1) factors of (n+1)! that are greater than n+1. All that remains is to show that (n+1)! - (n+1) > 2(n!-(n+1)) + (n-1).

    This was only a very rough outline of what you need to prove. You would need to flesh it out.

    Edit: Reading through this, this is not a terribly good method. Instead, you should probably use Euler's totient.

    \varphi(n) = n\prod_{p|n}\left(1-\dfrac{1}{p}\right)

    If n+1 is prime, \varphi((n+1)!) = n\varphi(n!). If \gcd((n+1)!,k)=1 and k>(n+1) then k has a prime factor greater than or equal to n+1. So, you just need to show that \varphi(n!)\ge 2 for all n\ge 3. By assumption, n+1 is prime, so it is relatively prime to n!. Hence 1 and n+1 are both relatively prime to n! making \varphi(n!)\ge 2. This means there are at least 2n relatively prime numbers to (n+1)!. For every n\ge 3, 2n>n+1 implying that at least one of the numbers relatively prime to (n+1)! is greater than n+1.
    Last edited by SlipEternal; Oct 17th 2013 at 09:40 PM.
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