Determine all possible pairs (x,y) of non-negative integers that satisfy this equation:
(x!+1)(y!+1)= (x+y)!
(In reply to
Another solution by Brian Smith)
But if x=0, y=2,
(x!+1)(y!+1) = (x+y)! becomes
(0!+1)(2!+1) = (0+2)!
(2) * (3) = 2! but 6 is not 2
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Posted by Larry
on 2023-09-07 12:40:32 |