Product of four consecutive integers

Let the first number be 'n', then we can write the product P

P = n(n + 1)(n + 2)(n + 3)

Expanding:
P = (n^2 + n)(n^2 + 5n + 6)
P = (n^4 + 6n^3 + 11n^2 + 6n)

It's convenient to set P + 1

P + 1 = n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2

For example, if the product P = 120, then P + 1 = 120 + 1 = 121 = 11^2
Then,
(n^2 + 3n + 1)^2 = 11^2 => n^2 + 3n + 1 = 11 or
n^2 + 3n = n(n + 3) = 10 = 2 * 5
n = 2

The four consecutive numbers are 2, 3, 4 and 5

Say that you wanted to multiply 14, 15, 16, 17

(14 * 15 * 16 * 17) + 1 = (14^2 + 3*14 + 1)^2 = (196 + 42 + 1)^2 = 239^2 = 57121

Notice:

The product of four consecutive positive integers cannot be a square.

Moreover, it is always 1 less than a square of an integer.

P = n(n + 1)(n + 2)(n + 3)

Observe that (n + 1)(n + 2)− n(n + 3) = 2
and define A =(n + 1)(n + 2) − 1
so that (n + 1)(n + 2)= A + 1 and n(n + 3) = A − 1
implying that
P = A^2 - 1