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I know that the polynomial $f(x)=\frac{1}{2} x^2 + \frac{1}{2}x \in \mathbb{Q}[x] $ has that $f(\mathbb{Z})\subset\mathbb{Z}$.

My question is a bit different: Does there exist a polynomial $f \in \mathbb{R}[x]$ such that $f(\mathbb{Z}) = \mathbb{Q}$? What about a polynomial that $f(\mathbb{Q}) = \mathbb{Z}$?

I suspect that both polynomials do not exist, but I do not know how to prove it.

Noam
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  • Yes, I'll fix it. – Noam Jun 24 '16 at 16:19
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    If $f$ is not constant, then $|f(x)| \to \infty$ as $|x| \to \infty$, so certainly $f(\mathbb Z)$ cannot be $\mathbb Q$. – GEdgar Jun 24 '16 at 16:22
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    The first question is an easy "no" - non-constant polynomials have infinite limits as $n\to\infty$, so $f(\mathbb Z)$ takes only finitely many values in, say, the interval $(0,1)$. – Thomas Andrews Jun 24 '16 at 16:22
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    Next consider that $f(\mathbb Q)$ is dense in $f(\mathbb R)$,which is connected. That rules out $f(\mathbb{Q}) = \mathbb{Z}$ – GEdgar Jun 24 '16 at 16:25
  • Any non-topological argument? – Noam Jun 24 '16 at 16:26
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    @Noam What counts as non-topological? You can easily convert GEdgar's argument into a statement about inequalities: we can assume $f$ is non-constant (otherwise it's trivial), so $f'$ is a non-zero polynomial with finitely many roots. We can thus assume $f'(0) \ne 0$, in which case for large enough $n$, $f(1/n) - f(0)$ can't be an integer. – Erick Wong Jun 24 '16 at 16:36
  • A non-topological argument is an argument that I understand :). Why can't $f(\frac{1}{n})-f(0)$ be an integer? – Noam Jun 24 '16 at 16:43
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    @Noam, it converges to $0$ because of continuity of $f$. However a non-trivial sequence of integers cannot converge. – Hmm. Jun 24 '16 at 16:44

2 Answers2

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there is no polynomial $f$ such that $f (\Bbb{Q}) = \Bbb{Z}$:

Suppose $f(X)=a_nX^n+...+a_1X+a_0$ is such a polynomial, then $f (0) =a_0= n_0\in \Bbb{Z}$, so the polynomial $g_1(X)=\frac{(f-n_0)(X)}{X}(\Bbb{Q})\subset \Bbb{Z} $ and then result $g_1(O)= a_1 = n_1\in \Bbb{Z}$, by repeating this reasoning $f$ to be an element of $\Bbb{Z}[X]$, let $m = lcm(n_i)$ then for suitable m' depending to m , $f (1 / m')\not\in \Bbb{Z}$.

using the same reasoning if $f$ is such that $f (\Bbb{Z}) = \Bbb{Q}$, then $f$ is in $\Bbb{Q}[X]$, and the fiber $f^{-1} (x)$ is empty for all $x \in \Bbb{Q}\cap] f (n), f (n + 1)[$ for suitable $n$.

m.idaya
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1st question: non-constant polynomials have infinite limits as $n\to\infty$, so can talk only finitely many values in the interval $(0,1)$, so no polynomial can take all rational values in $(0,1)$. (by @ThomasAndrews)

2nd question: $f(\mathbb{Q})$ is dense in $f(\mathbb{R})$ which is connected, so we cannot have $f(\mathbb{Q})=\mathbb{Z}$. (by @GEdgar)

If you prefer a non-topological argument, the result obviously holds for $f$ constant. So $f'$ has only finitely many roots. Take $a$ such that $f'(a)\ne0$, then for large enough $n$, $f(a+\frac{1}{n})-f(a)$ cannot be an integer - it must converge to 0, but a non-trivial sequence of integers cannot do that. (by @ErickWong and @Hmm.).

almagest
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