The product of four consecutive integers, starting with n, is n(n+1)(n+2)(n+3).

Note that n(n+3) = n^2 + 3n

and (n+1)(n+2) = n^2 + 3n + 2

So if we let x = n(n+3), we see that the product of the four conseuctive integers starting with n is of the form x(x+2).

x(x+2) = x^2 + 2x = x^2 + 2x + 1 - 1 = (x+1)^2 - 1

Thus the product of four consective integers is always one less than a square.