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Math Help - Find a prime number

  1. #1
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    Find a prime number

    Prove that if an integer n > 2, then there exist a prime number p such that n < p < n!

    My proof.

    Let p be a prime divisor of n!-1, then  p \leq n!-1 \leq n!.

    If p = n, then p|n!, but that is impossible since p|n!-1.

    Now, how do I prove that p is not < n?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by tttcomrader View Post
    Prove that if an integer n > 2, then there exist a prime number p such that n < p < n!

    My proof.

    Let p be a prime divisor of n!-1, then  p \leq n!-1 \leq n!.

    If p = n, then p|n!, but that is impossible since p|n!-1.

    Now, how do I prove that p is not < n?
    Adapt Euclid's proof.

    Suppose that for some n there is no prime p such that n<p<n!.

    Now consider N=n!-1 this is either a prime or composite the first of
    these possibilities contradicts our assumption, so it must be composite.

    Now let p_1,\ p_2,\ ..,\ p_r be the primes less than n, then clearly p_i \not | N,\ i=1,\ ..,\ r.

    But N is composite, so it has a prime divisor, which therefore must be greater
    than n and less than n! , a contradiction.

    So we conclude that it is not the case that that there are no primes p such
    that n<p<n!, that is for all n there is a prime p such that n<p<n!.

    RonL
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  3. #3
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    the part I don't understand is, for the primes less than n, why don't they divide N?
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  4. #4
    Senior Member ecMathGeek's Avatar
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    For n>2, it's not possible for the divisors of n! to also be the divisors of n!-1 and since all p_i<n divide n!, they can not divide n!-1.
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