TITLE: $P(x)=P(y)$ has infinitely many integer solutions
QUESTION [17 upvotes]: Determine all polynomials $P(x)$ with integer coefficients such that $P(x)=P(y)$ has infinitely many integer solutions in integer $x$ and $y$ with $x \neq y$.

Choose $P(x)=a_n(x-k)^{2n}+a_{n-1}(x-k)^{2n-2}+...+a_0$, with integers $k,n,a_n,a_{n-1}, ...,a_0$ Then $P(x)=P(2k-x)$ for every integer $x \neq k$, thus satisfy the condition.
Now, I have a gut feeling that if $P(x)=P(y)$ and $x+y=M$, then for every integer $x$, $P(x)=P(M-x)$, but I cannot analyze any futher.
So my question is: Are there any other types of polynomials $P(x)$ that satisfy the orange question above?
(Any answers or comments will be appreciated!) 
(If this question should be closed or off topic, please let me know. If this site cannot answer this question, let me know, I will delete this question immediately)

REPLY [11 votes]: Here's a more abstract proof:
If $P(x) - P(y) =0$ but $x \neq y$ then $ (P(x) - P(y) )/(x-y)=0$. Because this is a polynomial of degree $n-1$, when this identity is satisfied its leading terms $ (x^{n} - y^{n} )/ (x-y)$ must be equal to minus its remaining terms, and thus must be $O( \max(x,y)^{n-2})$. 
Now if $n$ is odd, $ (x^{n} - y^{n} )/ (x-y)$ is a homogeneous polynomial of degree $n-1$ with no nozero real roots. Because it has no nonzero real roots, its absolute value has some minimum value $C$ on the boundary of the unit square. Then homogeneity gives $ | (x^{n} - y^{n} )/ (x-y)| \geq C \max(x,y)^{n-1}$ and thus can be $O( \max(x,y)^{n-2})$ for only finitely many $x,y$.
If $n$ is even, $(x^n-y^n)/(x^2-y^2)$ is a homogeneous polynomial of degree $n-2$ with no real roots. By the same logic, we have $| (x^{n} - y^{n} )/ (x^2-y^2)| \geq C \max(x,y)^{n-2}$. Thus if $| (x^{n} - y^{n} )/ (x-y)| = O ( \max(x,y)^{n-2})$ then $C |x+y| = O(1)$. So this can happen for only finitely many values of $|x+y|$.
Thus if it happens infinitely often, it happens infinitely often for one particular value of $x+y$, say $M$, thus $P(x) - P(M-x)$ vanishes for infinitely many $x$ and thus is zero.