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Math Help - hard algebra question

  1. #1
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    Post Factorial that are one less than a Square

    hey guyz..will u please help me on getting the required of this problem and the solution..i tried but i think i really need your help..

    "Are the integers n and x (with n>7) such that n!=x^2-1?

    By n! we mean that product of the integers from 1 to n. It is known that 4!+1=25=5^2, 5!+1=121=11^2, and 7!+1=5041=71^2 .
    Last edited by fireboy25; August 21st 2013 at 03:43 AM. Reason: changed the title
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  2. #2
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    Re: hard algebra question

    Hey fireboy25

    It should be false but to prove it, you need to have all prime factors to have even power where x^2 = n! + 1. If you can prove that this is not the case then you're done.

    Basically find a value of n! where n is prime and check the solution.
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  3. #3
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    Re: hard algebra question

    ..thank you for your suggestion man ..actually, this question was only given to us by our teacher and does not explained to us anymore..what should i do?..will you please give some informations..
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  4. #4
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    Re: hard algebra question

    If n! = x^2 - 1 then n! = (x+1)(x-1). If n! has a prime factor then it means that the result should be n! = prime*(prime-1).

    Use this to pick a prime > 7 and show a contradiction if you assume that n! = (x+1)(x-1).

    Remember that prime factors can't be factorized further which means that if a prime exists on the LHS, it must also be on the RHS and both sides are simplified as best as they can be (in terms of linear factors).
    Thanks from fireboy25
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  5. #5
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    Re: hard algebra question

    This is Brocard's Problem. It is unsolved as of 2012.
    Thanks from johng and fireboy25
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