Let f (x) be a nonconstant element of Z[x]. Prove that f (x) takes on infinitely many values in Z.

This is a homework problem so no need to give me an immediate answer.

My general plan has been to try to prove that f(x) cannot have both an upper bound and a lower bound.

One strategy I tried was using induction on the degree of f. The base case for $f(x) = a_0 + a_1x$ is very doable to show that it cannot have both upper & lower bounds. It's when you get to $deg(f) = n$ that things start to get tricky.

The chapter of the book I'm working with, Contemporary Abstract Algebra, 8th edition, has the theorem for the division algorithm for polynomials, and as a corollary we see that f(a) is the remainder of f(x) when dividing by (x - a). So I am assuming they want me to try using that, but I'm not sure.

I feel as though I could brute force this problem by

1. Starting with f(b) = lower bound and f(c) = upper bound
2. Checking if the leading term has an even/odd power
3. Checking if the leading coefficient is positive/negative

Once all those are fixed, I would have to come up with a clever value for $b$ or $c$, and then use that to show that we can't have both an upper bound and a lower bound.

What I would really love is if there is a more elegant solution using the image of the polynomial f(Z) and showing some contradiction if it's finite. But nothing I've come across fits that.

• But induction doesn't seems so hard, every $n+1$ degree polynomial can be written as $b+x(\sum_{i=0}^na_ix^i)$, the constant doesn't matter, so you have unbounded expression times $x$, so unless $x\to0$ you have $n+1$ degree also unbounded – Holo May 12 '18 at 22:10

If $f(X)\in\Bbb{Z}[X]$ takes on only finitely many values in $\Bbb{Z}$, then it must take one of these values (say $c$) infinitely many times. But then $f(X)-c$ has infinitely many zeroes in $\Bbb{Z}$...