### 18th Putnam 1958

Problem B2

Let n be a positive integer. Prove that n(n + 1)(n + 2)(n + 3) cannot be a square or a cube.

Solution

(n+1)(n+2) = n2 + 3n + 2 and n(n+3) = n2 + 3n. So their product is (n2 + 3n + 1)2 - 1. Hence n(n + 1)(n + 2)(n + 3) is 1 less than a square, so it cannot be a square.

One of n+1, n+2 must be odd. Suppose it is n+1. Then n+1 has no factor in common with n(n + 2)(n + 3), so n(n + 2)(n + 3) = n3 + 5n2 + 6n must be a cube. But (n + 1)3 = n3 + 3n2 + 3n + 1 < n3 + 5n2 + 6n < n3 + 6n2 + 12n + 8 = (n + 2)3, so n3 + 5n2 + 6n cannot be a cube.

Similarly, suppose n + 2 is odd. Then it has no factor in common with n(n + 1)(n + 3) = n2 + 4n2 + 3n, so n2 + 4n2 + 3n must be a cube. But for n ≥ 2, (n + 1)3 < n2 + 4n2 + 3n < (n + 2)3, so n2 + 4n2 + 3n cannot be a cube for n ≥ 2. The case n = 1 is checked by inspection: 24 is not a cube.
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