Suppose by contradiction that $P(x) \neq 0$ for all integer $x$. For each $(x,y)$ such that $xP(x) = yP(y)$, we have \[ a (x^4 - y^4) + b (x^3 - y^3) + c (x^2 - y^2) + d (x - y) = 0, \] that is, simplifying by $x-y \neq 0$ and letting $k=x+y$ \[ a k(2 x^2 - 2kx + k^2) + b (x^2 - kx + k^2) + c k + d = 0, \] thus \[ x^2 (2ak+b) - kx(2ak+b) + P(k) = 0. \] If $2ak+b=0$, then we get $P(k)=0$, that is a contradiction. If $2ak+b \neq 0$, \[ x=\frac{k(2ak+b)\pm\sqrt{k^2 (2ak+b)^2 - 4 (2ak+b) P(k)}}{2(2ak+b)}, \] that is \[ 2x=k \pm \sqrt{k^2 - 4 \frac{P(k)}{2ak+b}}, \] thus $4 \frac{P(k)}{2ak+b}$ is an integer. We deduce that $\frac{8 a^2 P(k)}{2ak+b}$ also is. But it is equal to \begin{eqnarray*} & & \frac{8 a^3 k^3 + 8 a^2 b k^2 + 8 a^2 c k + 8 a^2 d}{2ak+b} \\ &=& \frac{(4 a^2 k^2 + 2ab k + 4ac - b^2)(2ak+b) + (b^3 - 4abc + 8a^2 d)}{2ak+b} \\ &=& (4a^2 k^2 + 2abk + 4ac - b^2) + \frac{b^3 - 4abc + 8da^2}{2ak +b}. \end{eqnarray*} Thus for all $(x,y)$ such that $xP(x) = yP(y)$, $2a(x+y)+b$ divides $b^3 - 4abc + 8da^2$, which is a constant. $(x+y)$ thus takes only a finite number of values. Since we can compute $x$ and $y$ in terms of $x+y$ (in fact each value of $x+y$ gives two possible values of $x$), we get that there is a finite number of $(x,y)$ verifying the condition $xP(x) = yP(y)$. It also gives a contradiction.

As in the previous solution, we have \[a(x^{4}-y^{4})+b(x^{3}-y^{3})+c(x^{2}-y^{2})+d(x-y) = 0. \] Since $x \neq y$, we may divide by $(x-y)$ to obtain \[a(x^{3}+x^{2}y+xy^{2}+y^{3})+b(x^{2}+xy+y^{2})+c(x+y)+d = 0, \] or, using the notation $k = x+y$, \[P(k) = xy (2ak+b). \] Let $P(x)$ have roots $\lambda_{1}\le \lambda_{2}\le \lambda_{3}$. We note that the function $xP(x)$ is bounded on the interval $\Big[ \min(\lambda_{1}, 0) , \max(\lambda_{3}, 0 ) \Big]$ and injective on each of the intervals $(-\infty , \min(\lambda_{1}, 0)) , \; (\max(\lambda_{3}, 0), \infty)$. Because each value of $xP(x)$ yields at most finitely many solutions $x,y$, there are only finitely many solutions $x,y$ such that $x$ and $y$ have the same sign. Clearly, if $P(k) \neq 0$, then each value of $k$ yields at most one solution. But for arbitrarily large values of $k$, $P(k)$ and $(2ak+b)$ have the same sign, making our equation $P(k) = xy(2ak+b)$ impossible if $x$ and $y$ have different signs. Thus there must be some integer $k$ such that $P(k) = 0$. (Indeed, if $P(k) = 2ak+b = 0$, then any pair of integers with sum $k$ will work.)

My solution starts the same as previous ones but it is different in the later part. Maybe someone will find that interesting. So, suppose that $P(x)$ satisfy given conditions but it doesn't have integer root. Therefore, it is irreducible in the $\mathbb{Z}[x]$. As before, we reach the form $P(\alpha)=\beta(2a\alpha+b)=\beta Q(\alpha)$ where $\alpha=x+y$ , $\beta=xy$ and $Q(x)=2ax+b$. We will prove that this equation has solution only for finitely many $\alpha$. Indeed, suppose on the contrary. From the fact that $P$ is irreducible and $Q$ is a of lower degree than $P$ we conclude that $P, Q$ are relatively prime in $\mathbb{Z}[x]$. So, there exist $f, g \in \mathbb{Z}[x]$ such that: $f(x)P(x)+g(x)Q(x) = C$ for some non-zero integer constant $C$. From putting $x=\alpha$ and relation $Q(\alpha)|P(\alpha)$ we obtain that $Q(\alpha)|C$. Since $Q$ is non constant we see that $C$ must have infinitely many divisors, but that's impossible because $C \not = 0$. So, our previous equation has solution for only finitely many $\alpha$. But then it has also solution for finitely many $\beta$ (since $\beta$ is linear in terms of $\alpha$). For given $\alpha, \beta$ there is at most one pair of integers $(x,y)$ satisfying $x+y=\alpha, xy=\beta$, so our equation has only finitely many solutions in pairs of integer $(x, y)$. That's contradiction with assumption.

TomciO wrote: but it doesn't have integer root. Therefore, it is irreducible in the $\mathbb{Z}[x]$. This is wrong. $8x^{3}-1=(2x-1)(4x^{2}+2x+1)$ has no integer root, but is reducible in $\mathbb{Z}[x]$. Here I showed $xP(x)=Q((x-{t\over 2})^{2})$ for some $Q(x)\in\mathbb{R}[x]$ and $t\in\mathbb{Z}$. Then $0=0\cdot P(0)=Q(({t\over 2})^{2})=tP(t)$. If $t\not =0$, then $P(t)=0$ with $t\in\mathbb{Z}$. If $t=0$, then $Q(0)=0$ and $x^{2}|Q(x^{2})=xP(x)$ and $P(0)=0$.

Sorry for reviving an old topic toumaf wrote: If $2ak+b=0$, then we get $P(k)=0$, that is a contradiction. For this to be a contradiction , we need $\frac{b}{2a}$ to be an integer. How is that so?

For $ xP(x)=yP(y) $, suppose, $ x=\frac {u+v}{2} $ and $ y=\frac {u-v}{2} $ where $ u,v $ are integers. Simplifying the equation we get that $ v^2(2au+b)=-(2au^3+3bu^2+4cu+4d) $. f $ u=\frac {-b}{2a} $, then $ f(\frac {-b}{2a})=0 $. Otherwise note that dividing $ 2au^3+3bu^2+4cu+4d $ by $ 2au+b $ we get that the remainder(which is $ 8a^3f(\frac {-b}{2a}) $) is divisible by $ 2au+b $ for infinitely many $ u $. So $ f(-\frac {b}{2a})=0 $. So $ xP(x)=(-\frac {b}{2a}-x)P(-\frac {b}{2a}-x) $. So if $ y\ne -\frac {b}{2a}-x $ for all $ x $, then note that the value of $ xP(x) $ is equal for three values of $ x $, but this is possible only within a bounded region, but $ x $ is not bounded. So $ y=-\frac {b}{2a}-x $ after the bound. So $ \frac {b}{2a} $ is integer. So done.

Sahil I don't really understand what's bothering you we have set $k=x+y$ so that is an integer value regardless of $a$ and $b$ there is no need for RSM's proof.

Am I missing something? Let $Q(x)=xP(x)$ Consider $Q(a-x)-Q(x)$. We can see that for $a$ with large enough magnitude, this will have no integer roots. Therefore, for some $a, Q(a-x)-Q(x)=0$ as a polynomial, and so $P(a)=0$.

First of all, don't use $a$, since that already is the leading coefficient of $P$. So, replacing with $\alpha$, even if your claim that for large enough magnitude of $\alpha$ you have the polynomial $Q(\alpha - x) - Q(x)$ to have no integer root, all you do is rule out the pairs $u\neq v$ with $u+v=\alpha$ for which $Q(u) = Q(v)$. But there exist infinitely many pairs $u\neq v$ with $u+v<\alpha$, so there is room for infinitely often having $Q(u) = Q(v)$. What are you trying to do there?

There do indeed exist infinitely many such pairs, but if we bound $a$ it follows by polynomial identity that for a specific value of $a$, $Q(x)=Q(a-x)$ is true for all $x$. Then either $a=0$, in which case $Q$ is even and so $0$ is a double root and must be a root of $P$, or $a \neq 0$ and $a$ is a root of $P$.

Ok. So we agree that if for all those infinitely many pairs $(u,v)$ with $u\neq v$ for which $uP(u) = vP(v)$ we had the set $\{\alpha = u+v \mid u\neq v, uP(u)=vP(v)\}$ to be bounded, then for some $\alpha$ we would have infinitely many pairs $(u,v)$ with $u+v=\alpha$, and then $R_{\alpha}(x) = Q(\alpha-x) - Q(x)$ would be identically null. It is your claim relative to that set needing to be bounded that is unsustained as yet. In fact $R_{\alpha}(x) = Q(\alpha-x) - Q(x)$ always has an integer root for even $\alpha$ (albeit being related to $u=v$, which is not allowed).

For $a$ large enough, suppose $Q(a-x)-Q(x)$ has a root $a/2+r$ where $r$ is nonzero. Then $Q(a/2+r)=Q(a/2-r)$. For $a$ large enough magnitude, the polynomial$ R(x)=Q(a/2+x)$ will have $x^3$ and $ x$ terms with the same sign. Therefore, $R(x)-R(-x)$ will have no roots other than at $x=0$.

This problem is solved if you think to reduce the polynomia to degree $2$ by setting $x+y=k$. There is really nothing much else to it. And without this trick I don't think there is a nice solution is there?

https://artofproblemsolving.com/community/c6h1231860_generalization_of_sl_2002_a3 This is the best solution!

Hopefully this is right. The equation $xP(x) = yP(y)$ rewrites as \[a(x^4-y^4) + b(x^3-y^3) + c(x^2-y^2) + d(x-y) = 0.\]Since $x\neq y$, we can divide out $x-y$ to get \[a(x^3+x^2y+xy^2 + y^3) + b(x^2+xy+y^2) + c(x+y) + d = 0.\]Now let $s = x + y$, $p=xy$. Then \[a(s^3 - 2sp) + b(s^2 - p) + cs + d = 0,\]or $p(2sa+b) = P(s)$. Thus $2as + b$ divides $P(s)$ for infinitely many pairs $(x,y)$. But this means $2as + b$ divides $P(s)$ for infinitely many integer values of $s$. Write $P(x) = (2ax + b)Q(x) + R$ for some rational $R$ and rational polynomial $Q(x)$ . Scaling up by some positive integer $N$, we may assume that $Q(x)$ and $R$ are integers at the expense of dividing into $NP(x)$ instead. Hence $2as + b$ divides $R$ for infinitely many integers $s$, implying $R = 0$. (Here we use $a\neq 0$.) Thus $2ax + b$ divides $P(x)$ as polynomials, ergo $P(x)$ has a rational root (again using $a\neq 0$). Thus $P(x)$ is reducible in $\mathbb Q[x]$, so by Gauss's Lemma it's also reducible in $\mathbb Z[x]$. In turn, $P$ has an integer root.

We more generally show the assertion when $P$ is replaced by any polynomial of odd degree. Lemma: Given an non-constant even degree polynomial $Q(x)$ such that $Q(x) = Q(y)$ holds for infinitely many pairs $(x,y)$ of integers with $x \ne y$. Then $Q(x) \equiv Q(c-x)$ for some constant $c \in \mathbb Z$. Proof: WLOG assume leading coefficient of $Q$ is positive, otherwise change $Q \to -Q$. First choose a constant $M,K,N \in \mathbb Z_{> 0}$ such that both sequences $$ P(M),P(M+1),P(M+2),\ldots ~~;~~ P(-M),P(-M-1),P(-M,-2),\ldots $$are strictly increasing ; and $$Q(x) - Q(N-x) >0 ~~ \forall ~ x \ge K \qquad , \qquad Q(x) - Q(-N-X) < 0 ~~ \forall ~ x \ge K$$By changing $M \to \max(M,K)$, WLOG $M \ge K$. Let set $$ S_1 = \{M,M+1,M+2,\ldots\} ~~,~~ S_2 = \{-M,-M-1,-M-2,\ldots\} $$Basically, all elements of $Q(S_1),Q(S_2)$ are distinct. Now for infinitely such $x,y$ with $Q(x) = Q(y)$, we have that one of $x,y$ is in $S_1 + K$ while other is in $S_2$. Now for any $x \in S_1 + K$, We know the element $y \in S_2$ for which $Q(x) = Q(y)$ must satisfy $$ y \in (-N-x,N-x) $$Basically, $x+y$ can take finitely many values only. So there is a constant $c$ such that for infinitely many pairs $x,y$ we have $x+y = c$ and $Q(x) = Q(y) = Q(c-x)$. More particularly, the polynomial $$Q(x) - Q(c-x)$$has infinitely many roots, and hence is identically $0$. This proves our Lemma. $\square$ Our Lemma for $Q(x) \equiv xP(x)$ implies for some $c \in \mathbb Z$ we have $$ xP(x) \equiv (c-x)P(c-x) $$If $c \ne 0$, then $x=c$ just gives $P(c) = 0$. Otherwise for $c=0$ we have $$ xP(x) \equiv -xP(-x) \qquad \text{or} \qquad P \text{ is odd} $$But then $P(0) = 0$, completing the proof. $\blacksquare$

Because the polynomial $Q(x)=xP(x)$ is bounded on the interval containing its turning points, for $xP(x)=yP(y)$ to hold infinitely often we need $x$ and $y$ to lie on the "left branch" and the "right branch" of $Q$ infinitely often. If $y=-x+k$, then $k$ must be bounded both below and above for size reasons, else the absolute value of the $x^2$ term in $Q(y)$ is greater than $|b|$, and thus $Q(y)-Q(x)$ is nonzero for large enough $x$. Then by infinite pigeonhole, we must pick some value of $k$ infinitely many times, in which case $0 \equiv xP(x)=yP(y) \equiv kP(k) \pmod{x}$ for infinitely many $x$. But then $kP(k)=0$, so we're done unless $k=0$. If $k=0$, then $Q(x)$ is odd and thus is of the form $ax^3+cx^2$, hence $x=0$ is a root of $P$. $\blacksquare$ This proof works for any odd degree $P$

Suppose otherwise. Let $x+y=s$. Since $x-y\neq 0$, for infinitely many $x$, $y$, the following is zero: \[\frac{xP(x)-yP(y)}{x-y}=as(s^2-2xy)+b(s^2-xy)+cs+d\]Therefore, we have $P(s)=xy(2as+b)$. If $s$ is made constant, then $xy$ is determined, so there are finitely many $(x,y)$ that work per $s$. Therefore, it is true for infinitely many $s$. Thus, $|P(s)|=|xy||(2as+b)|\le \tfrac{s^2}{4}|2as+b|$. For large values of $s$ the left will overpower the right due to a larger leading term. Contradiction.

Let $Q(x) = x P(x) = ax^4 + bx^3 + cx^2 + dx$. WLOG $a > 0$ since we can swap $Q$ with $-Q$. Note that there must be infinitely many integers $x,y$ with opposite sign satisfying $Q(x) = Q(y)$ (as $Q$ is injective over the large positives and the large negatives). Letting $p = x + y$, we have \[ \frac{Q(x) -Q(y)}{x-y} = a \cdot p (p^2 - 2xy) + b \cdot (p^2 - xy) + c \cdot p + d = P(p) - xy (2ap - b)\]Hence there are infinitely many integers $x,y$ with opposite sign satisfying $P(p) = xy(2ap - b)$. Notice that $xy$ is fixed for a given value of $p$ (as if $2ap - b = 0$ then $P(p) = 0$, absurd), so there can be at most one solution for each $p$. Now if we take $p$ sufficiently large, then $P(p)$ and $2ap - b$ are greater than $0$, which contradicts the fact that $x,y$ are opposite signs.

Notice $x+y$ is bounded (? seems like this is debated on and I haven't proved it either) and takes on finitely many values, thus $xP(x)=(c-x)P(c-x)$ occurs somewhere. Taking $x=c$ we find $cP(c)=0$ thus $P(c)=0$ or $c=0$. If $c=0$ then $P(0)=0$ as the constant term becomes $0$.

Good size exercise but not very interesting otherwise. Let $Q(x) = xP(x)$ be a quartic polynomial. Assume without loss of generality that $a_3$ is positive. The main claim is the following: Claim: There exists a real number $c$ such that $Q(x+c) = Q(-x)$ for all real numbers $x$. Proof: Fix a large positive integer $x$ to be determined such that there exists a pair $(x, y)$ with $xP(x) = yP(y)$. First, set $(x, y) = (x+c, -x)$ with $x > 0$ fixed, and note that $(-x-c, x)$ is also a valid pair. I claim that $c = \frac{-a_2}{2a_3}$ is a constant. Assume $f(x+c) > f(x)$ (and thus $c>0$), otherwise replace $x$ by $-x$. We can also assume that $f$ is strictly increasing and convex after $x$ by taking $x$ sufficiently large. Then, notice that \[f(-x)=f(x+c) \geq f(x) + cf'(x) = f(x) + c\left(4 a_3 x^3 + 3 a_2 x^2 + 2a_1 x + a_0\right).\]On the other hand, $f(-x) - f(x) = -2a_2 x^3 - 2a_0x$ by definition. So $-2a_2 \geq 4 a_3 c$ by taking $x$ to be large enough (leading terms must compare). So we obtain $c \leq \frac{-a_2}{2a_3}$ by dividing through by $4a_3$, which is positive. On the other hand, because $c$ is bounded by a constant, $f(x+c)-f(-x)$ has bounded coefficients and thus must have its leading $x^3$ term equal to zero. Evaluating yields $c = \frac{-a_2}{2a_3}$ precisely. So now $Q(x+c) = Q(-x)$ for this fixed $c$ for all sufficiently large $x$ part of a pair, and it follows that $Q(x+c) = Q(-x)$ for all real numbers $x$. $\blacksquare$ So now $Q(c/2+x) = Q(c/2-x)$ is an even polynomial; but $Q(0) = 0$ is given, so $Q(c) = 0$ as well. Clearly $c$ is an integer, so it is our desired integer root of $P$.