mathschallenge.net

Problem

We can see that 3 4 5 6 = 360 = 19²1.

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 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.