## Consecutive Product Square

#### Problem

We can see that 3 4 5 6 = 360 = 19^{2}1.

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 = `n`^{4} + 6`n`^{3} + 11`n`^{2} + 6`n` + 1.

If Q is square, it must be of the form, (`n`^{2} + `a``n` + 1)^{2}.

Expanding (`n`^{2} + `a``n` + 1)^{2} = `n`^{4} + 2`an`^{3} + (`a`^{2}+2)`n`^{2} + 2`an` + 1, and comparing coefficients, 2`a` = 6 `a` = 3; checking: `a`^{2}+2 = 11.

Therefore, Q = (`n`^{2} + 3`n` + 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