Question

Proof that the product of any 4consecutive integers is one less than a perfect square

Answer 1

Let the four consecutive numbers be n, (n + 1), (n + 2) and (n + 3)
Product of four numbers when added to 1 = n(n + 1)(n + 2)(n + 3) + 1
= n(n + 3)(n + 1)(n + 2) + 1
= (n2 + 3n)(n2 + 3n + 2) + 1
Let y = n2 + 3n
⇒ y(y + 2) + 1 = y2 + 2y + 1 = (y + 1)2
(y + 1)2 = (n2 + 3n + 1)2
Hence, the product of four consecutive numbers when added to 1 is a perfect square.

Answer 2

Well you can write the equation as this
(Every 2 after n is square)
(n2 + 3n) (n2 + 3n + 2) = (n2 + 3n + 1 − 1) (n2 + 3n + 1 + 1) = (n2 + 3n + 1)2−12
If you know you have to make this equation into an perfect square you need to get 2 same things and -1