# If $f$ is at least degree $1$, then $f(n)$ cannot be prime for all n

I can't figure it out. Can you give me some advice? Let $f$ be a polynomial in $\mathbb{Z}[X]$ of degree at least 1. Prove that $f(n)$ cannot be a prime for each $n \in \mathbb{Z}$. I tried induction on the degree, but that didn't work. What can I do now?

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Suppose $f(n)$ is a prime for every integer $n$. In particular, $f(0)=p$ for some prime $p$, and so the constant term of $f$ is $p$. Now consider $f(kp)$ for $k$ an integer. $f(kp)$ must be divisible by $p$, but must also be a prime, so we must have $f(kp)=p$ for every $k$. But then $f$ takes the value $p$ infinitely many times, so must be the constant polynomial $p$.
@Kevin - If $f$ takes a vaule $a$ infinitely many times then $f-a$ takes the value $0$ infinitely many times. But this contradicts the fundamental theorem of algebra – Belgi Oct 15 '12 at 11:10
Ok, so, if I get it: suppose $g(k)=f(k \cdot a)-a$, then $deg(g)=deg(f)$, but $g(k)=0 \forall k \in \mathbb{Z}$, zo $g(k)=...(x+2)(x+1)(x)(x-1)(x-2)... \Rightarrow deg(g)=\infty$? – Kevin Oct 15 '12 at 12:20
@Belgi Every nonzero polynomial with coefficients in a ring R has no more roots than its degree iff R is an integral domain (i.e. $\rm\:ab = 0\:\Rightarrow\: a=0\ \ or\ \ b=0).\:$ Conceptually, this has little to do with the fundamental theorem of algebra. – Bill Dubuque Oct 15 '12 at 13:03