jmc

Question

If x is a positive integer and x(x + 1)(x + 2)(x + 3) + 1 = 379^2, compute x.

Accepted Answer

We have that x(x + 3) = x^2 + 3x and (x + 1)(x + 2) = x^2 + 3x + 2. So, let y = x^2 + 3x + 1. Then align* x(x + 1)(x + 2)(x + 3) + 1 &= (x^2 + 3x)(x^2 + 3x + 2) + 1 &= (y - 1)(y + 1) + 1 &= y^2 - 1 + 1 &= y^2. align*So, y = 379, or x^2 + 3x + 1 = 379. Then x^2 + 3x - 378 = 0, which factors as (x - 18)(x + 21) = 0. Hence, x = 18.