Determine all possible pairs (x,y) of non-negative integers that satisfy this equation:
(x!+1)(y!+1)= (x+y)!
Assume both x>=2 and y>=2. Then the left side of the equation will be odd and the right side will be even. This is a contradiction, so at least one of x and y is less than 2.
Without loss of generality, assume x<2. Then for either x=0 or x=0 we have x!=1 and the equation reduces to 2*(y!+1) = (y+1)! Now assume y>=3. Then working mod 3 the left side reduces to 2 and the right side reduces to 0. This is a contradiction. so y is less than 3.
This leaves exactly two cases to check, y=0 or y=1 making y!=1 or y=2 making y!=2. By direct evaluation only the last case works. Then (x,y)=(0,2) or (1,2) are solutions. Because x and y are symmetric then their reversals also work and (x,y)=(2,0) or (2,1) are also solutions.
The possible pairs (x,y) of non-negative integers that satisfy the given equation are (0,2), (1,2), (2,0), and (2,1).
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