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Homework Help: Find all solutions in positive integers

  1. Dec 29, 2009 #1
    Find all solutions in positive integers a; b; c to the equation
    a!b! = a! + b! + c!

    I have rearranged and got (a!-1)(b!-1) = c!+1

    And the only solutions I can find are a=3 b=3 c=4 but I can't be sure that they are the only ones. How would I go about finding other solutions?

    I have tried b=1 to 4 but after that the numbers start getting a bit big
  2. jcsd
  3. Dec 29, 2009 #2
    Re: Factorials

    It's clear that we without loss of generality can assume [itex]a \leq b \leq c[/itex]. Consider the equation modulo b!. Then you get
    [tex]a! \equiv a! + c! \equiv 0 \pmod {b!}[/tex]
    so [itex]b! \, | \, a![/itex] and therefore you must have [itex]a=b[/itex]. Now you can substitute and see [itex]c! = a!(a!-2)[/itex] which you can prove has no solution for a > 3 by noting that if a+1 is prime, then a+1 divides c!, but not a!(a!-2) which is a contradiction; and if a+1=pn is composite with p the smallest possible prime then p and n divide a! so if p and n are not equal a+1=pn|a! and if they are equal, then [itex]a+1=p^2[/itex] so [itex]p,2p \leq a[/itex] for a>3 which proves a+1 | a!. Together this shows [itex]a! \equiv 0 \pmod {a+1}[/itex] whenever a>3 and since [itex]a+1 | a!-2[/itex] we have:
    [tex]2 \equiv a! \equiv 0 \pmod {a+1}[/tex]
  4. Dec 30, 2009 #3
    Re: Factorials

    Isn't a! + 1 + c! equivalent to 0 (mod b!) though?
  5. Dec 30, 2009 #4
    Re: Factorials

    Sorry...Just thought about it...I think I get it now.

    Thanks for your help
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