Find a, b, c, Such That a! b! = a! + b! + c!

Date: 12/09/2003 at 06:15:46
From: Peter
Subject: Is there a chance that a! b! = a! + b! + c! ?

Find all triples of nonnegative integers a, b, c such that 

  a! b! = a! + b! + c!

I have no idea how to start.  Would you please give me some hints?

Date: 12/09/2003 at 12:07:24
From: Doctor Rob
Subject: Re: Is there a chance that a! b! = a! + b! +c! ?

Thanks for writing to Ask Dr. Math, Peter!

This may not be the only way, or even the best way, but here is one
way to attack this problem.

Since the equation is symmetric in a and b, we can assume without loss 
of generality that a <= b.  We can see that a = 0 and a = 1 are 
impossible, since 

  b! = 1 + b! + c! 

is impossible.  Also, if a = 2, then 

     2*b! = 2 + b! + c!

  b! - c! = 2

which is impossble.  Thus b >= a >= 3.  Then a! | b!, which implies
that c! = 0 (mod a!), so a! | c!, so a <= c.  Now divide the equation
through by a!:

      b! = 1 + b!/a! + c!/a!

       0 = 1 + 0 + c!/a! (mod b!/a!)

   c!/a! = -1 (mod b!/a!)

That implies that either b!/a! = 1 or else b!/a! does not divide
c!/a!, that is, either b = a or else b > c.

Assume first that b > c.  Then

       a!*b! - a! - b! = c!

   a!*b! - a! - b! + 1 = c! + 1
   
     (a! - 1)*(b! - 1) = c! + 1

Since a >= 3, we must have 5 <= a! - 1, so

   5*(b! - 1) <= (a! - 1)*(b! - 1) = c! + 1 <= (b-1)! + 1
   
         5*b! <=       (b-1)! + 6           <= 3*(b-1)!
  
          5*b <= 3

which is again a contradiction.  Thus b > c is impossible.  That
implies that a = b.

Now we have

   a!^2 = 2*a! + c!,

so a! | c! and a <= c.  But a = c is impossible, since then one would
have

   a!^2 = 3*a!
 
     a! = 3

a contradiction.  Thus a < c, so a + 1 <= c.  Divide through by a!:

   a! = 2 + c!/a!

Now reduce modulo a + 1:

   a! = 2 (mod a + 1)

If a + 1 is prime, then Wilson's Theorem says that

    a! = -1 (mod a + 1)

    -1 = 2 (mod a + 1)

    a + 1 | 3 
 
    a + 1 <= 3

another contradiction.  Thus a + 1 must be composite.  

Now for almost all composite numbers, a + 1 | a!, which would imply
that a + 1 | 2, another contradiction.  Thus a + 1 must be a composite
number with a + 1 not a divisor of a!.  These are few.

I leave the rest to you.

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 12/10/2003 at 01:39:35
From: Peter
Subject: Thank you

Thank you very much!