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Multiplication of Factorials

  1. Dec 28, 2012 #1
    Given that a, b, and c are positive integers solve the following equation.

    a!b! = a! + b! + c^2

  2. jcsd
  3. Dec 28, 2012 #2
    I found the answer through brute force: a=2, b=3, c=2.

    Not sure if there is a more elegant solution though.
  4. Dec 28, 2012 #3
    vorde, thanks for trying, im getting the same answer too:smile:

    but i'm seeking for method to solve it: how to relate multiplication of 2 factorials and their sums? if we can, then c wouldn't be a problem.

    there should be a way to solve it:rolleyes:
  5. Dec 28, 2012 #4


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    "Brute force" is a method! Please clarify what you are looking for.
  6. Dec 29, 2012 #5
    Doesn't it work for all positive integers c such that [itex]c = \sqrt{a!b! -a! -b!}[/itex]? :biggrin:

    Spit-balling here, we have [itex]a!b! = a! + b! + c^2[/itex]? Doesn't that imply that [itex]\displaystyle a! = 1 + \frac{a! + c^2}{b!} = 1 + \frac{a(a-1)(a-2)(a-3)...}{b(b-1)(b-2)...} + \frac{c^2}{b!} = 1 + \prod_{k = (b+1)}^a k + \frac{c^2}{b!}[/itex]. Don't know where I'm going with that...
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