Determine all possible triplets (X,Y,Z) of positive integers that satisfy:

                          X!*Y! = X! + Y! + Z²

Prove that there are no others.

(In reply to more by Ady TZIDON)

(n,1,1)  for any n?

5!*1! = 120, but 5!+1!+1^2 = 122

The same holds for (1,n,1)

Also, for (0,n,1) and (n,0,1) the same inequality holds, as well as using a zero, which is not a positive integer.

Posted by Charlie on 2014-03-23 10:27:33