We consider the equation (x+1)...(x+k) = (y+1)...(y+mk). If m = 1, the equation implies that x = y. If m = 2, the authors showed that x = 7, y = 0, k = 3 is the only solution in integers x >= 0, y >= 0, k >= 2. In this paper, we prove that (x+1)...(x+k) = (y+1)...(y+3k) and (x+1)...(x+k) = (y+1)...(y+4k) have no solution in integers x >= 0, y >= 0, and k >= 2.
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