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Consecutive Product Square

Problem

We can see that 3 times 4 times 5 times 6 = 360 = 192minus1.

Prove that the product of four consecutive integers is always one less than a perfect square.


Solution

Let Q = n(n + 1)(n + 2)(n + 3) + 1 = n4 + 6n3 + 11n2 + 6n + 1.

If Q is square, it must be of the form, (n2 + an + 1)2.

Expanding (n2 + an + 1)2 = n4 + 2an3 + (a2+2)n2 + 2an + 1, and comparing coefficients, 2a = 6 implies a = 3; checking: a2+2 = 11.
Therefore, Q = (n2 + 3n + 1)2.

Hence, the product of four consecutive integers is always one less than a perfect square.

Problem ID: 184 (01 Nov 2004)     Difficulty: 3 Star

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