TITLE: Even degree rational polynomials must attain some rational values twice?
QUESTION [9 upvotes]: If $P$ is a polynomial, with rational coefficients and of even degree, do there exist arbitrarily large rational $y$ such that $P(x)=y$ has two roots in the rationals? 
The tricky bit, I suppose, is to show that polynomials without an axis of symmetry, which cannot be described as $Q(ax^2+bx+c)$, fit this criterion. But I have no idea if it is true at all.

REPLY [10 votes]: I don't think the question is an easy one, and I don't have an answer, but I suspect the answer is "no, not necessarily".

As a test case, consider the polynomial
$$P(x) = x^4 + 2x\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\,$$
Based on limited testing, I don't think there is any rational number $y$ such that $P(x) = y$ has more than one rational root.

To prove the statement 
\begin{align*}
&\text{There is no rational number}\;y\;\text{such that the equation}
\qquad\;\;\;\;\;\\[4pt]
&\qquad P(x) = y\\[4pt]
&\text{has more then one rational root}\\[4pt]
\end{align*}
it would suffice to prove the statement
\begin{align*}
&\text{There is no rational number}\;a\;\text{such that the polynomial}\qquad\;\;\,\\[4pt]
&\qquad Q(x) = \frac{P(x)-P(a)}{x-a}\\[4pt]
&\text{has a rational root}\\[4pt]
\end{align*}
Thus, for the case $P(x) = x^4 + 2x$, after dividing out the common factor of $(x-a)$, we get
$$Q(x) = x^3+ax^2+a^2x+(a^3+2)\qquad\qquad\qquad\qquad\;\,$$
So perhaps continue by trying to prove that there is no rational number $a$ such that the cubic polynomial $Q(x)$, as defined above, has a rational root.