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Proof that n! is not a power of an integer.

  1. Jun 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove that n! for n>1 cannot be a square or cube or any other power of an integer.
    Hint: There is always a prime between n/2 and n if n>3
    3. The attempt at a solution
    Lets assume for contradiction that [itex] n!=x^r [/itex] where x and r are natural numbers
    and n>3 , so there is some prime p that is between n/2 and n. well this occurs once in prime factorization of n!. Now x has some prime factorization and in order for [itex] x^r [/itex] to be equal to n! the prime p must be in x's prime factorization, but if r>1 then p will have r copies in
    [itex] x^r [/itex] so this is a contradiction because numbers have unique prime factorizations so
    n! is never a power of an integer for r>1.
     
  2. jcsd
  3. Jun 4, 2013 #2

    Dick

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    Sounds just fine to me. Any question?
     
  4. Jun 4, 2013 #3
    ok thanks, just wanted to make sure it worked
     
  5. Jun 4, 2013 #4
    Hi cragar, yes it looks good, but you are mixing n and n! here and there so it is a bit confusing, you might want to rephrase it.
    The original statement is probably at the origin of it, but when they say
    "Hint: There is always a prime between n/2 and n if n>3", it's not the same n that was used at the beginning of the problem, and your proof, although correct in the spirit (I think) would benefit from using different variable names or to be more explicit about when you talk about n or n!
    they are freely intermixed and it could be considered as just wrong
     
  6. Jun 4, 2013 #5

    haruspex

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    Looks like the same n to me.
     
  7. Jun 4, 2013 #6

    Dick

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    Sure it's the same n. Knowing there's a prime between n!/2 and n! is pretty useless.
     
  8. Jun 5, 2013 #7
    Hi guys, of course it is! I am the one mixing n! and n :)
    Cheers...
     
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