The product of 4 consecutive positive numbers is always 1 less than a perfect square. Examples:

1*2*3*4 = 5^2 -1

2*3*4*5 = 11^2 - 1

3*4*5*6 = 19^2 - 1

Actually, in trying to prove it directly, I found that it works for all integers, not just positive numbers.

Well, I found that the general form is:

x^4 + 6x^3 + 11x^2 + 6x = y^2 -1, where x is any integer and y is any whole number.

Doing the steps:

sqrt(x^4 + 6x^3 + 11x^2 + 6x + 1) is what we're trying to find.

Assuming that equation has an correct polynomial solution, we're looking for a polynomial in the following fashion:

(ax^2 + bx +c)^2, because of a degree four polynomial.

Because of the coefficients of the first and last terms, we have verified that our square root polynomial must be in the form:

x^2 + ax + 1

Squaring that then solving, we find that our square root polynomial is:

x^2 + 3x + 1.

Any polynomial is an integer for all integers.

Therefore, we can definitively say that for all integers x and whole numbers y,

x(x + 1)(x + 2)(x + 3) = y^2 - 1, or:

The product of 4 consecutive integers is always 1 less than a perfect square.