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Factorial-Square Sum (Posted on 2014-03-23) Difficulty: 3 of 5
Determine all possible triplets (X,Y,Z) of positive integers that satisfy:

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

Prove that there are no others.

No Solution Yet Submitted by K Sengupta    
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re: more | Comment 5 of 7 |
(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
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