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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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Some Thoughts computer exploration | Comment 4 of 7 |

10   for T=2 to 999999
20    for X=1 to int(T/2)
30      Y=T-X
40      Fx=!(X):Fy=!(Y)
50      Z2=Fx*Fy-Fx-Fy
55      if Z2>=0 then
60      :Z=int(sqrt(Z2)+0.5)
70      :if Z*Z=Z2 then print X;Y;Z
80    next
90   next

which stopped by overflow when the total of X and Y reached 1004, so higher totals weren't checked

finds only

(2,2,0) and (2,3,2)

The former contains a zero and so doesn't fulfill the "positive" requirement.

The latter solution also leads to (3,2,2); the program looked only for X<=Y, but of course the result is symmetrical.

  Posted by Charlie on 2014-03-23 10:19:54
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