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Some Factorials Sum Power (Posted on 2008-12-21) Difficulty: 3 of 5
Determine all possible pair(s) (P, Q) of positive integers that satisfy this equation.

                                        P! + Q! = PQ

where P and Q are not necessarily distinct.

See The Solution Submitted by K Sengupta    
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Some Thoughts computer exploration | Comment 1 of 4

The only values found for p + q totaling 1015 or less are (2,2) and (2,3).

I have no proof that higher values of p+q won't lead to other solutions.

For any given total of p+q, the left side of the equation exceeds the right hand side when p and q are most different; the right side exceeds the left side when p and q are more nearly equal. Thus in the continuous case (with factorials replaced by a modified gamma function and integers by reals) there are many solutions--in fact two for any given value of p+q. To the point in this puzzle, I don't know if some of these solutions, at higher values of p+q, might happen to involve only integers.

   10   for T=2 to 1015
   20    for P=1 to T-1
   30      Q=T-P
   40      A=!(P)+!(Q)
   50      B=P^Q
   60      if A=B then print P,Q
   70    next
   80   next

(If P+Q = T were allowed to be 1016, the program overflows trying to take the factorial of 1015.)

  Posted by Charlie on 2008-12-21 15:18:29
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