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Could you give me any hint how to prove that if $f\in\mathbb{Q}[X]$ and $f(\mathbb{Q})=\mathbb{Q}$ then degree of $f$ is equal to $1$? I have tried to prove it by using Bézout's theorem but I guess it is not the best way to do it... Thanks in advance.

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    I changed$\mbox{deg}f$to \deg f. One of the differences is clearly visible: $\mbox{deg}f$, $\deg f$. One should NEVER NEVER use \mbox for that kind of purpose. Please see [this](http://en.wikipedia.org/wiki/Wikipedia:Manual_of_Style/Mathematics/Why_you_should_never_use_%5Cmbox_within_Wikipedia). That essay does _not_ apply _only_ to Wikipedia.2011-10-28

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Since there already is one complete answer:

Suppose that $f$ has degree $k\ge 2$. Without loss of generality we may assume that $f$ has integer coefficients. Let $p$ be a prime that is not a factor of the leading coefficient, and assume there is an $x$ such that $f(x)=1/p$. Let us say that $x=m/n$ in lowest terms. Then $f(x)=\frac{a_km^k+a_{k-1}m^{k-1}n+\cdots+a_0n^k}{n^k}$ For this to equal $1/p$, clearly $p$ must divide $n^k$ and thus also $n$ itself. But then actually $p^k$ divides the denominator, so $p^{k-1}$ must divide the numerator. All terms except the first are multiples of $p$, so $p$ must also divide $a_km^k$. However neither $a_k$ nor $m^k$ is divisible by $p$, and since $p$ was prime this is a contradiction. Thus $1/p \not\in f(\mathbb Q)$.

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    @André Nicolas I also thought to use the Rational Roots Theorem but unfortunately quickly gave up this idea, Thanks.2011-10-28
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Presumably the solution below is not the intended one. Perhaps you can translate the idea into one more algebraic in spirit.

Suppose that $f$ has degree $\ge 2$. Without loss of generality we may assume that $f$ has positive lead coefficient, and integer coefficients. Note that $f$ must have odd degree, else $f$ misses all large enough negative rationals.

Suppose that the lead coefficient of $f$ is $a$, and the constant term is $c$. Finally, let $p$ be a very large prime.

We look at the equation $f(x)=p+c$, and show that it has no rational roots, meaning that $f(\mathbb{Q})$ misses $p+c$. By the Rational Roots Theorem, any such root has to be of the form $\pm 1/d$ or $\pm p/d$, where $d$ is a divisor of $a$. The negative possibilities can be discarded, as can $1/d$ (too small), so any rational solution has to be $\ge p/a$. This is impossible, for when $p$ is huge, $f(p/a)$ is much larger than $p$ if $f$ has degree $\ge 2$.