Let P(x) be a nonconstant polynomial, where all the coefficients are nonnegative integers. Prove that there exist infinitely many positive integers n such that P(n) is composite.

Hint(s):

Remember that if a and b are distinct integers, then P(a) - P(b) is divisible by a - b.

Guest Jun 27, 2021

#1**+1 **

The simplest polynomial is $ax^2 + bx + c$

Let $x = k \cdot c$ for $k \in \mathbb{Z}$

We have \(\begin{align*} ax^2 + bx + c &= a(kc)^2 + b(kc) + c\\ &= ak^2c^2 + bkc + c\\ &= c(ak^2c + bk + 1) \end{align*}\)

Because there is are at least two factors for $ax^2 + bx + c,$ $P(x)$ is composite.

MathProblemSolver101 Jun 27, 2021