# The reducibility of a polynomial in coefficients of integer

This question is from Putnam and Beyond, an example in the section ‘Irreducible Polynomial’:

Let \$P(x)\$ be an nth-degree polynomial with integer coefficients with the property
that \$|P(x)|\$ is a prime number for \$2n + 1\$ distinct integer values of the variable \$x\$.
Prove that \$P(x)\$ is irreducible over \$mathbb{Z}[x]\$.

And the solution:
Assume the contrary and let \$P(x) = Q(x)R(x)\$ with \$Q(x),R(x) ∈ mathbb{Z}[x]\$,
\$Q(x),R(x) = ±1\$. Let \$k = deg(Q(x))\$. Then \$Q(x) = 1\$ at most \$k\$ times and \$Q(x) =
−1\$ at most \$n − k\$ times. Also, \$R(x) = 1\$ at most \$n − k\$ times and \$R(x) = −1\$ at
most \$k\$ times. Consequently, the product \$|Q(x)R(x)|\$ is composite except for at most
\$k + (n − k) + (n − k) + k = 2n\$ values of \$x\$. This contradicts the hypothesis. Hence
\$P(x)\$ is irreducible.

All concept is okay for me except the part “\$Q(x)=-1\$ at most \$n-k\$ times” , because I know \$Q(x)=1\$ at most \$k\$ times is because \$Q(x)\$ will be constant function if \$Q=1\$ at least \$k+1\$ times, but is it related to \$Q(x)=-1\$?

All topic