tastymath75025 wrote: Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$holdsfor all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$. Typo. There should be a space between "holds" and "for."

First of all, rewrite the equation as \[xP(x) + yP(y) + zP(z) = xyz(P(x-y) + P(y-z) + P(z-x)).\] Now fix an $x \neq 0$ and let $y = \tfrac{x+z}{2xz - 1}$. Since $y, 2xz - 1 \neq 0$ for arbitrarily small but nonzero $z$, the equation is valid as long as $z \in (-\varepsilon, \varepsilon) \setminus \{0\}$ for some $\epsilon$. At this point we take the derivative with respect to $z$ and take the limit as $z \to 0$. One can compute that $y \to -x$ and $\tfrac{dy}{dz} \to -(2x^2 + 1)$, so a lot of stuff cancels. In the end, one gets \[-(2x^2 + 1)(P(-x)-xP'(-x)) + P(0) = -x^2(P(2x)+2P(-x)).\] It turns out that this solves the problem; by comparing leading coefficents one gets that $P$ is either zero or quadratic, and the rest is a finite computation. The solutions are $c(x^2 + 3)$ for real $c$, which can be shown to work through some effort.

The given can be rewritten as saying that \begin{align*} Q(x,y,z) &\overset{\text{def}}{=} xP(x) + yP(y) + zP(z) \\ &- xyz \left( P(x-y) + P(y-z) + P(z-x) \right) \end{align*}is a polynomial vanishing whenever $xyz \ne 0$ and $2xyz = x+y+z$, for real numbers $x$, $y$, $z$. Claim: This means $Q(x,y,z)$ vanishes also for any complex numbers $x$, $y$, $z$ obeying $2xyz=x+y+z$. Proof. Indeed, this means that the rational function \[ R(x,y) \overset{\text{def}}{=} Q\left( x,y,\frac{x+y}{2xy-1} \right) \]vanishes for any real numbers $x$ and $y$ such that $xy \ne \frac{1}{2}$, $x \ne 0$, $y \ne 0$, $x+y \ne 0$. This can only occur if $R$ is identically zero as a rational function with real coefficients. If we then regard $R$ as having complex coefficients, the conclusion then follows. $\blacksquare$

Now we regard $P$ and $Q$ as complex polynomials instead. First, note that substituting $(x,y,z) = (t,-t,0)$ implies $P$ is even. We then substitute \[ (x,y,z) = \left(x, \frac{i}{\sqrt2}, \frac{-i}{\sqrt2}\right) \]to get \begin{align*} &\phantom= xP(x) + \frac{i}{\sqrt2} \left( P\left( \frac{i}{\sqrt2} \right) - P\left( \frac{-i}{\sqrt2} \right) \right) \\ &= \frac{1}{2} x\left( P(x-i/\sqrt2) + P(x+i/\sqrt2) + P(\sqrt2i) \right) \end{align*}which in particular implies that \[ P\left( x + \frac{i}{\sqrt2} \right) + P\left( x - \frac{i}{\sqrt2} \right) - 2P(x) \equiv P(\sqrt 2i) \]identically in $x$. The left-hand side is a second-order finite difference in $x$ (up to scaling the argument), and the right-hand side is constant, so this implies $\deg P \le 2$. Since $P$ is even and $\deg P \le 2$, we must have $P(x) = cx^2 + d$ for some real numbers $c$ and $d$. A quick check now gives the answer $P(x) = c(x^2+3)$ which all work.

This is impossible

I plugged in $i \frac{\sqrt{2}}{2}, -i \frac{\sqrt{2}}{2}, z$, which finished the problem.

Nonzero reals?

expiLnCalc wrote: Nonzero reals? You can take the difference between the resulting polynomials and argue that it can be extended to complex since it has to vanish.

Here's a sketch of my janky solution: Define the linear transformation from polynomials $P$ to other polynomials by $f(P)$ is the result when you plug in $(x, y, z) = (2t, t, \frac{3t}{4t^2 - 1})$ into the expression $2xP(x) + 2yP(y) + 2zP(z) - (x + y + z)(P(x - y) + P(y - z) + P(z - x))$. Note that the expression is the problem statement rearranged and that triple $(x, y, z)$ satisfies 2xyz = x + y + z, so $f(P)$ should be identically 0. Expanding everything and doing big-O notation stuff so we can ignore the $z$, we get that $f(P)$, when we plug in $P = x^n$, has leading term $t^{n + 1}$, unless $n = 2$, in which case it has leading term $t$. This means that unless $P$ is degree 2, we get screwed by the leading $t^{n + 1}$ term. Bash or whatever to see that $P = k(x^2 + 3)$.

v_Enhance wrote: to get \begin{align*} &\phantom= xP(x) + \frac{i}{\sqrt2} \left( P\left( \frac{i}{\sqrt2} \right) - P\left( \frac{-i}{\sqrt2} \right) \right) \\ &= \frac{1}{2} x\left( P(x-i/\sqrt2) + P(x+i/\sqrt2) + P(\sqrt2i) \right) \end{align*}which in particular implies that \[ P\left( x + \frac{i}{\sqrt2} \right) + P\left( x - \frac{i}{\sqrt2} \right) - 2P(x) \equiv 0 \]identically in $x$. Minor error: you get $2P(x) = P(x-i/\sqrt{2}) + P(x+i/\sqrt{2}) + P(\sqrt{2}i)$. Substituting $x\to x+\sqrt{2}$ and subtracting then gives a third-order finite difference, which then yields a quadratic solution $P$.

First we claim that we can plug in any nonzero complex $x,y,z$ satisfying $2xyz=x+y+z$ into the equation. If we let $F(x,y,z)=\sum_{cyc} xP(x)-xyzP(x-y)$, this essentially boils down to showing $F=0\pmod{2xyz-x-y-z}$ as polynomials. Let $n$ be the degree of $z$ in $F$, then consider $(2xy-1)^nF$, it suffices to show this is congruent to the $0$ polynomial because $2xy-1$ is relatively prime to $2xyz-x-y-z$. But $(2xy-1)z=x+y\pmod{2xyz-x-y-z}$, so we can replace all instances of $(2xy-1)z$ with $x+y$ to get a polynomial $L(x,y)=(2xy-1)^nF\pmod{2xyz-x-y-z}$, and this polynomial is $0$ whenever $x,y,z$ are nonzero reals with $2xyz-x-y-z=0$, that is, $x,y,x+y,2xy-1\not=0$. But then this is a nonempty open set in $\mathbb{R}^2$ on which $L$ is $0$ thus $L$ must be identically $0$ as desired. Now that we have that, plug in $(\frac{i}{\sqrt{2}},\frac{-i}{\sqrt{2}},z)$ for arbitrary nonzero $z$. We get \[\frac{1}{z}\left(\frac{P(\frac{i}{\sqrt{2}})}{\frac{-i}{\sqrt{2}}}+\frac{P(\frac{-i}{\sqrt{2}})}{\frac{i}{\sqrt{2}}}\right)+2P(z)=P(i\sqrt{2})+P(-z-\frac{i}{\sqrt{2}})+P(z-\frac{i}{\sqrt{2}})\]Now everything except the $c\cdot\frac{1}{z}$ term is a polynomial, so the coefficient of $\frac{1}{z}$ must be 0, and also the right hand side is unchanged under $z\to-z$, so $P(z)=P(-z)$. Thus, $2P(z)=P(z+\frac{i}{\sqrt{2}})+P(z-\frac{i}{\sqrt{2}})+P(i\sqrt{2})$. If we let $g(z)=P(z+\frac{i}{\sqrt{2}})-P(z)$, then this implies $g(z)-g(z-\frac{i}{\sqrt{2}})$ is constant so $g$ is linear thus $P$ is at most quadratic, and since it's even it must be $ax^s+b$ and then plugging in and using the identity $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2)$ yields $b=3a$ to finish.

stupid question that had to be asked: is the solution set worth any points?

We claim that all such $P(x)$ satisfy $P(x)=c(x^2+3)$ for some real $c$. We now note that $2xyz-x-y-z$ must divide $xP(x)+yP(y)+zP(z)-\left(\frac{x+y+z}{2}\right)(P(x-y)+P(y-z)+P(z-x)$, since if we were to divide out $2xyz-x-y-z$ from the latter expression, we would get something that is equal to $0$ at infinitely many points and has degree $0$. We now note that, if we were to repeatedly substitute $xyz=\frac{x+y+z}{2}$ into the highest degree, where the degree of $ax^iy^jz^k$ is defined as $i+j+k$, then we must get $0$. In particular, this means that we must be able to factor out $xyz$ from each of the highest degree terms, since otherwise there would be a fraction which, if we were to keep substituting, would not be able to cancel out with anything else, meaning that we will not be able to get to an expression identically $0$. We now consider the maximum degree terms of $xP(x)+yP(y)+zP(z)-\left(\frac{x+y+z}{2}\right)(P(x-y)+P(y-z)+P(z-x))$. If we let the maximum degree term of $P(x)$ be $a_nx^n$, then we have the maximum degree terms of the above expression are $a_n(x^{n+1}+y^{n+1}+z^{n+1}-\left(\frac{x+y+z}{2}\right)((x-y)^n+(y-z)^n+(z-x)^n)))$. We now consider the $x^ny$ term. We may find that it is $\frac{a_n}{2}(n-(1+(-1)^n))$, which is nonzero for all $n\neq 2$. However, we recall that $xyz$ must divide it, a contradiction. Thus, $n=0,2$. We can show that P is even with $y=-x,z=0$, where we may plug in $0$ because $P$ is continuous. Thus, we get $P$ is of the form $P(x)=ax^2+b$. Plugging in $x=y=1,z=2$ gives us $b=3a$, as desired.

Let $x=N+\tfrac{1}{N},y=N,z=\tfrac{1}{N}.$ Then we get \[P(N+\tfrac{1}{N})+\frac{P(N)}{1+\tfrac{1}{N^2}}+\frac{P(\tfrac{1}{N})}{N^2+1}=P(\tfrac{1}{N})+P(N-\tfrac{1}{N})+P(-N).\]Multiplying by $N^2+1$, we get $$(N^2+1)P(N+\tfrac{1}{N})+N^2P(N)+P(\tfrac{1}{N})=(N^2+1)\left(P(\tfrac{1}{N})+P(N-\tfrac{1}{N})+P(-N)\right).$$ Comparing the leading coefficients of both sides as $N\to \infty,$ it becomes obvious that $\text{deg}(P)$ is even. Now WLOG $P$ monic and write $P(x)=x^d+ax^{d-1}+bx^{d-2}+\cdots.$ Assume $\text{deg}(P)> 2.$ Now compare the $N^{d+1},N^d$ coefficients of both sides. They must match since we can take $N\to \infty.$ We can also purge the $P(\tfrac{1}{N})$ terms since they contribute at most an $N^2$ term and we assumed $\text{deg}(P)> 2.$ Note that expanding $(N+\tfrac{1}{N})^d$ gives us a $dN^{d-2}$ with Binomial-Theorem, etc. Doing the expansion and pruning the irrelevant terms gives $$(N^2+1)(N^d+aN^{d-1}+(b+d)N^{d-2})+N^2(N^d+aN^{d-1}+bN^{d-2})=(N^2+1)(N^d+aN^{d-1}+(b-d)N^{d-2}+N^d-aN^{d-1}+bN^{d-2}).$$and matching the $N^d$ coefficient shows that $1+b+d+b=2+2b-d\implies d=\tfrac{1}{2},$ absurd. Thus $P$ must be quadratic and from here it's easy.

Here is an algebraic solution: Let $S(x,y,z) = 2xP(x) + 2yP(y) + 2zP(z) - (x+y+z)(P(x-y) + P(y-z)+P(z-x))$, then $2xyz=x+y+z\neq 0\implies S(x,y,z) = 0$. Write $(2yz-1)^d S(x,y,z) = (2xyz-x-y-z)Q(x,y,z) + R(y,z)$, for $d = \deg S$, then we get $R(y,z) = 0$ for all $yz \neq 1$, $y+z\neq 0$. In particular, this shows that $R(y,z)$ is the zero polynomial. Now, if $2xyz = x+y+z = 0$, WLOG $z = 0$, and then $(2yz-1)^d \neq 0$ so $S(x,y,z) = 0$ as well. Thus $S(x,y,z) = 0$ for all $2xyz = x+y+z$. Now in particular $z = 0$, $x+y = 0$ implies that $P$ is even. Write $P(x) = a_0 + a_1x^2 + \cdots + a_d x^{2d}$. Then we have that if $2xyz = x+y+z$ implies $\sum_{i=0}^d a_i (2x^{2i+1} + 2y^{2i+1} + 2z^{2i+1} - (x+y+z)((x-y)^{2i} + (y-z)^{2i} + (z-x)^{2i})) = 0$, which implies that \[ Z(x,y,z) = \sum_{i=0}^d a_i(2x^{2i+1} + 2y^{2i+1} + 2z^{2i+1} - (x+y+z)((x-y)^{2i} + (y-z)^{2i} + (z-x)^{2i}))(x+y+z)^i (2xyz)^{d-i} = 0 \]for $2xyz=x+y+z$. Now $Z$ is a homogenous polynomial, so its variety $V$ lies in $\mathbb R\mathbb P^2$; furthermore any $(x,y,z)$ satisfying $2xyz = x+y+z$ satisfies $[x\colon y \colon z]\in V$; by scaling appropriately we can show that for all $x,y,z > 0$ $[x\colon y \colon z] \in V$, which of course forces $V = \mathbb R\mathbb P^2$ and $Z = 0$. Now take $Z(x,y,0)$. We get $2x^{2d+1} + 2y^{2d+1} = (x+y)((x-y)^{2d} + x^{2d} + y^{2d})$, or $2(x^{2d} - x^{2d-1}y + \cdots + y^{2d}) = (x-y)^{2d}+x^{2d} + y^{2d}$; by comparing $x^{2d-1}y$ coefficients we conclude that $d = 1$. Then take $Z(x,x,x) = 0$ to get \[ a_1(6x^3)(3x) + a_0 (-3x)(2x^3) = 0, \]which forces $a_0 = 3a_1$ and we conclude. $\blacksquare$

MathStudent2002 wrote: Now $Z$ is a homogenous polynomial, so it generates a variety $V$ on $\mathbb R\mathbb P^2$; furthermore any $(x,y,z)$ satisfying $2xyz = x+y+z$ satisfies $[x\colon y \colon z]\in V$; by scaling appropriately we can show that for all $x,y,z > 0$ $[x\colon y \colon z] \in V$, which of course forces $V = \mathbb R\mathbb P^2$ and $Z = 0$. Will you marry me <3

Answer: The only possible $P(x)$ is $c(x^2+3)$. Firstly rewrite the equation as $$g(x,y,z)=2(xP(x)+yP(y)+zP(z))-(x+y+z)(P(x-y)+P(y-z)+P(z-x))=0$$For $2xyz=x+y+z$ . I wouldn't bother proving that $P(x)=Q(x^2)$ for $Q\in \mathbb{R}[x]$ but due to continuity you can simply set $(x,-x,0)$ and get that $P(x)=P(-x)$. We are going to prove that $Q$ is linear. Suppose that the highest degree term of $P$ is $x^{2k}$ for $k>1$. We use the following substitution: $$(x,y,z)=(\sqrt{\frac{a^2+b^2+c^2}{2}}\frac{a^2}{abc},\sqrt{\frac{a^2+b^2+c^2}{2}}\frac{b^2}{abc},\sqrt{\frac{a^2+b^2+c^2}{2}}\frac{c^2}{abc})$$. Then consider $f(a,b,c)=\frac{g(x,y,z)}{\sqrt{\frac{a^2+b^2+c^2}{2}}}=0$ which is a rational function that vanishes. We shall only investigate what happens with $x^{2k}$.It turns out that $x^{2k}$ adds the following expression to $f$. $$( \frac{a^2+b^2+c^2}{2})^k\frac{2(a^{4k+2}+b^{4k+2}+c^{4k+2})-(a^2+b^2+c^2)((a^2-b^2)^{2k}+(b^2-c^2)^{2k}+(c^2-a^2)^{2k})}{(abc)^{2k+1}}$$The term $a^{4k+1}$ vanishes but $a^{4k-1}$ doesn't instead it appears as $$\frac{(2k(b^2+c^2)-2(b^2+c^2))}{2^k}a^{4k-1}$$.Which is a term that will not cancel in $f$ given that $k>1$.Thus $k\le 1$and then it is straightforward to show that $P(x)=c(x^2+3)$.

Let $z \rightarrow \frac{x+y}{2xy-1},$ then the condition of the problem rewrites as $$x(2xy-1)P(x)+y(2xy-1)P(y)+(x+y)P\left(\frac{x+y}{2xy-1}\right)=xy(x+y)\left(P(x-y)+P\left(y-\frac{x+y}{2xy-1}\right)+P\left(\frac{x+y}{2xy-1}-x\right) \right)$$Let $d=\text{deg}P$ and suppose that $P$ is monic. After amplifying with $(2xy-1)^d$ we can view both sides as polynomials in $\mathbb{R}[X,Y],$ so their monomials with highest power must be equal, this implies the following $$2x^{d+2}y+ 2xy^{d+2}= xy(x+y)((x-y)^d+y^d+(-x)^d) \Longleftrightarrow 2(x^{d+1}+y^{d+1})=(x+y)((x-y)^d+y^d+(-x)^d)$$Because the expression is symmetric in $x$ and $y,$ this implies that $d$ is even, now let $x\rightarrow 3$ and $y\rightarrow 2,$ after some manipulations we obtain that $3^d-2^d=5,$ which clearly implies that $d=2,$ and $P(X)=c(x^2+3)$ easily follows.

How are you supposed to realize that the condition $2xyz = x + y + z$ is mostly irrelevant? I tried basing my approach off of the given condition (as in, assuming it's integral to the problem), and tried doing things like $$(x,y,z) \to \left( \frac{\tan A}{\sqrt{2}}, \frac{\tan B}{\sqrt{2}}, \frac{\tan C}{\sqrt{2}} \right), \quad (x,y,z) \to \left(\frac{a-b}{c\sqrt{2}}i, \frac{b-c}{a\sqrt{2}}i, \frac{c-a}{b\sqrt{2}}i \right)$$(Little did I know my second substitution gets close to the solution...) What's the motivation for discarding the condition almost completely, and instead just treating the problem as a normal FE? (i.e. substituting $\pm i/\sqrt{2}$) @below: i thought titu wanted me to do a substitution

Th3Numb3rThr33 wrote: What's the motivation for discarding the condition almost ocmpletely, and instead just treating the problem as a normal FE? (i.e. substituting $\pm i/\sqrt{2}$) Weird condition-> Titu-> Bash

Here is an analytic solution. TL;DR: Take $y=1$, $z = \frac{x+1}{2x-1}$ and let $x\to\infty$. We will carefully analyze the growth order of the equation in variable $x$ which turns out to be sufficient to determine the degree of $P$. Let $n = \deg P$ and WLOG, let $P$ be monic. Trivial substitution $(x,y,z) = (x,-x,0)$ gives $P$ is even polynomial. Thus rewrite the equation as $$ yP(y) + zP(z) - xyzP(y-z) = xyz(P(x-y) + P(x-z)) + xP(x)$$$$yP(y) + zP(z) - xyzP(y-z) = x(2yz-1)P(x) - xyz(2P(x) - P(x-y) - P(x-z)) $$$$\frac{yP(y) + zP(z)}{x} - yzP(y-z) = (2yz-1)P(x) - yz(2P(x) - P(x-y) - P(x-z))$$Now, plug in $y=1$ and $z=\frac{x+1}{2x+1}$, we get $$\frac{P(1) + \frac{x+1}{2x+1}P\left(\frac{x-1}{2x-1}\right)}{x} + O(1) = \frac{3}{2x+1}P(x) - \frac{x+1}{2x+1}\left[2P(x) - P(x-1) - P\left(x-\frac{x+1}{2x+1}\right)\right].$$Clearly $\tfrac{x-1}{2x-1}\to 0.5$ as $x\to \infty$. Thus we get that LHS is $O(1)$. Moreover, it's evident that the first term of the LHS is $\frac{3} {2}x^{n-1} + O(x^{n-2})$. This simplifies the equation to $$\frac{x+1}{2x+1}\left[2P(x) - P(x-1) - P\left(x-\frac{x+1}{2x+1}\right)\right] = \frac{3}{2}x^{n-1} + O(x^{n-2}).$$Moreover, as $\frac{x+1}{2x+1} = \frac{1}{2} + O\left(\frac{1}{x}\right)$. Thus $$2P(x) - P(x-1) - P\left(x-\frac{x+1}{2x+1}\right) = 3x^{n-1} + O(x^{n-2}). \qquad (*)$$To get further estimate, we need the following lemma. Lemma : Let $Q$ be a monic polynomial of degree $n$ with real coefficient and fix $a\in\mathbb{R}$. Then $$P(x+a) - P(x) = anx^{n-1} + O(x^{n-2})$$Proof : Just notice that $\deg(P(x+a) - P(x))= n-1$ and its leading coefficient is $na$. Back to the main problem, we easily get the following estimate. $$P(x) - P(x-1) = nx^{n-1} + O(x^{n-2}) \qquad (\spadesuit)$$The other term is slightly more tricky. Fix $\varepsilon > 0$. Take $x$ large enough such that $\frac{x+1}{2x+1} < 0.5 + \varepsilon$. By our lemma, we get $$P(x) - P\left(x-\frac{x+1}{2x+1}\right) \geqslant \left(\frac{1}{2}+\varepsilon\right)nx^{n-1} + cx^{n-2}$$for some constant $c$. Taking $\epsilon\to 0$, we see that $$P(x) - P\left(x-\frac{x+1}{2x+1}\right) = \frac{1}{2}nx^{n-1} + o(x^{n-1})$$(Note that we used small-oh notation.) Thus, combining with $(\spadesuit)$, we see $$2P(x) - P(x-1) - P\left(x-\frac{x+1}{2x+1}\right) = \frac{3n}{2}x^{n-1} + o(x^{n-1})$$Hence from $(*)$ , we get $n=2$. It's easy to verify now that only $c(x^2+3)$ is a solution.

Here is a very easy finish after noting that $P(x)$ is even: First, suppose that $\deg(P)=2n>2$. Then, by plugging in $\left(x,\frac{1}{x},x+\frac{1}{x}\right)$, we get $$\left(P\left(x+\frac{1}{x}\right)-P\left(x-\frac{1}{x}\right)\right)(x^2+1)=P(x)+x^2P\left(\frac{1}{x}\right)$$, we find that the leading coefficient on the LHS is $4na_n$, whereas the leading coefficient on the RHS is $a_n$, which is a contradiction. So, we must have $\deg(P)\le 2$, and the answer follows.

Fast solution using "substitution": Plug in $(x,-x,0)$ to get $P$ even after multiplying by $xyz$ on both sides. Set $P(x) = Q(x^2)$ and rewrite the givens using $u=\frac{1}{xy}, v = \frac{1}{yz}, w = \frac{1}{zx}$ and $Q$. This is good because it lets us rewrite the given $2xyz = x+y+z$ as $u+v+w = 2$. Now set $w = 2, u = -v$ to get $$2Q(-2z^2) = Q(-1/2 – 2z – 2z^2) + Q(-1/2 + 2z – 2z^2) + Q(-2), z = \frac 1u$$Suppose $Q$ has degree $n$ and compare coefficients of $z^{2n-2}$ on both sides to finish.

stroller wrote: Fast solution using "substitution": Plug in $(x,-x,0)$ to get $P$ even after multiplying by $xyz$ on both sides. Set $P(x) = Q(x^2)$ and rewrite the givens using $u=\frac{1}{xy}, v = \frac{1}{yz}, w = \frac{1}{zx}$ and $Q$. This is good because it lets us rewrite the given $2xyz = x+y+z$ as $u+v+w = 2$. Now set $w = 2, u = -v$ to get $$2Q(-2z^2) = Q(-1/2 – 2z – 2z^2) + Q(-1/2 + 2z – 2z^2) + Q(-2), z = \frac 1u$$Suppose $Q$ has degree $n$ and compare coefficients of $z^{2n-2}$ on both sides to finish. Can't use $0$ !

^No, but you can consider $x + \epsilon, -x-\epsilon, \epsilon$ or else define a new polynomial to satisfy $$2xP(x) + 2yP(y) + 2zP(z) = (x + y + z) \cdot (P(x - y) + P(y - z) + P(z - x))$$.

Also, to add on: In the original polynomial, you can’t plug in 0. However after multiplying both sides by $xyz$, there are no division by $0$ errors. Since $P$ is continuous, as it is a polynomial, it easily extends to $0$. Another way to think about it is the following: $z=\frac{x+y}{2xy-1}$. Then consider the equation reached after multiplying by $xyz$, except replace $z$ with the new expression. This new equation is an identity for all nonzero real $x,y$. However, the fact that we have polynomials means that this new identity we have must be true for all numbers. This includes the reals, $0$, and even complex numbers.

I didn't include this part because it's already been said by many other posters (e.g. v_Enhance). Any other issues?

Thanks now i get it.

Observe that constant multiples of $x^2+3$ work; confirm by using the $x^3+y^3+z^3-3xyz$ identity. First clear denominators so both sides are polynomials in $x,y,z$; by continuity we can now take $(x,y,z)=(x,-x,0)$ to deduce that $P$ is even. Now assume $P(t)=\displaystyle\sum_{k=0}^na_kt^k$ with $n$ even, $a_n\neq0$, and $a_{n-1}=0$. Substitute $(x,y,z)=(x,1,\frac{x+1}{2x-1})$ for $x\neq0,-1,\frac{1}{2}$ and multiply by $x(x+1)(2x-1)^n$ to get \begin{align*} &x(2x-1)^{n+1}P(x)+(2x-1)^{n+1}P(1)+(x+1)(2x-1)^nP(\frac{x+1}{2x-1})\\ =\quad&(x^2+x)(2x-1)^nP(x-1)+(x^2+x)(2x-1)^nP(\frac{x-2}{2x-1})\\&+(x^2+x)(2x-1)^nP(\frac{-2x^2+2x+1}{2x-1}) \end{align*}where each term is a polynomial. Now we compute the $x^{2n+1}$ coefficient of each term. Write \begin{align*} &x(2x-1)^{n+1}P(x)\\ =&x(2^{n+1}x^{n+1}-(n+1)2^nx^n+O(x^{n-1}))(a_nx^n+O(x^{n-2}))\\ =&2^{n+1}a_nx^{2n+2}-(n+1)2^na_nx^{2n+1}+O(x^{2n}) \end{align*}so the coefficient in the first term is $-(n+1)2^na_n$. Write \begin{align*} &(x^2+x)(2x-1)^nP(x-1)\\ =&(x^2+x)(2^nx^n-n2^{n-1}x^{n-1}+O(x^{n-2}))(a_n(x-1)^n+O(x^{n-2}))\\ =&(x^2+x)(2^nx^n-n2^{n-1}x^{n-1}+O(x^{n-2}))(a_nx^n-na_nx^{n-1}+O(x^{n-2}))\\ =&2^na_nx^{2n+2}+(2^na_n-n2^{n-1}a_n-n2^na_n)x^{2n+1}+O(x^{2n}) \end{align*}so the coefficient in this term is $2^na_n-n2^{n-1}a_n-n2^na_n$. Write \begin{align*} &(x^2+x)(2x-1)^nP(\frac{-2x^2+2x+1}{2x-1})\\ =&(x^2+x)\sum_{k=0}^na_k(-2x^2+2x+1)^k(2x-1)^{n-k}\\ =&(x^2+x)(a_n(-2x^2+2x+1)^n+O(x^{2n-2}))\\ =&(x^2+x)((-2)^na_nx^{2n}+2n(-2)^{n-1}a_nx^{2n-1}+O(x^{2n-2}))\\ =&(-2)^na_nx^{2n+2}+((-2)^na_n+2n(-2)^{n-1}a_n)x^{2n+1}+O(x^{2n}) \end{align*}so the coefficient here is $(-2)^na_n+2n(-2)^{n-1}a_n$. Observe that $(2x-1)^{n+1}P(1)$, $(x+1)(2x-1)^nP(\frac{x+1}{2x-1})=(x+1)\displaystyle\sum_{k=0}^na_k(x+1)^k(2x-1)^{n-k}$, and $(x^2+x)(2x-1)^nP(\frac{x-2}{2x-1})$ (expand the same way) are of degree less than $2n+1$, so there is no contribution here. Now, the $x^{2n+1}$ coefficient on both sides of the equation above must be the same, since it is a polynomial identity which holds for infinitely many real $x$. Thus \[-(n+1)2^na_n=2^na_n-n2^{n-1}a_n-n2^na_n+(-2)^na_n+2n(-2)^{n-1}a_n.\]Using the fact that $a_n\neq0$ and $n$ is even, this reduces to $n=2$. So if $P$ is not the zero polynomial, then $P$ has degree $2$. We can quickly confirm that $cx^2+d$ works only when $d=3c$, giving us the solution claimed at the beginning.

tastymath75025 wrote: Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$. Proposed by Titu Andreescu and Gabriel Dospinescu Solution. The answer is $P(x)=c(x^2+3)$ where $c\in\mathbb R,$ it is easy to verify that it works. Consider the following claims: Claim 1. $P(x)=P(-x).$ Proof. Solving $2xyz=x+y+z,$ we get $z=\tfrac{x+y}{2xy-1},$ therefore the polynomial equation can be written as \[\frac{P(x)(2xy-1)}{y}+\frac{P(y)(2xy-1)}{x}+\frac{(x+y)P\left(\tfrac{x+y}{2xy-1}\right)}{xy}=(x+y)\left(P(x-y)+P\left(y-\frac{x+y}{2xy-1}\right)+P\left(\frac{x+y}{2xy-1}-x\right)\right).\]The above is true for all $x,y\in\mathbb R$ such that $xy\ne \tfrac12, xy\ne 0$ and $x+y\ne 0.$ Both the LHS and RHS are bivariate rational functions, so they must be the same rational functions. Putting $y=-x,$ we get $-P(x)+P(-x)=0.$ Hence the claim. $\square$ Claim 2. $P(0)\ne 0$ if $P$ is not the zero polynomial. Proof. Assume the contrary that $P(0)=0.$ Now, putting $y=x$ and $z=\tfrac{2x}{2x^2-1}$ and using Claim 1, we get \[2\cdot\frac{P(x)(2x^2-1)}{2x^2}+\frac{P\left(\tfrac{2x}{2x^2-1}\right)}{x^2}=P(0)+2P\left(x-\frac{2x}{2x^2-1}\right)\]which can be simplified as \[P(x)+\frac{1}{2x^2-1}P\left(\frac{2x}{2x^2-1}\right)=\frac{2x^2}{2x^2-1}P\left(\frac{2x^3-2x}{2x^2-1}\right)\]Now put $P(x)=x^nQ(x)$ where $n\ge 1$ and $Q(0)\ne 0,$ \[Q(x)+\frac{2^n}{(2x^2-1)^{n+1}}Q\left(\frac{2x}{2x^2-1}\right)=\left(\frac{2x^2-2}{2x^2-1}\right)^n\left(\frac{2x^2}{2x^2-1}\right)Q\left(\frac{2x^3-2x}{2x^2-1}\right)\]Substituting $x=0,$ \[Q(0)+\frac{2^n}{(-1)^{n+1}}Q(0)=0\implies Q(0)=0\]which is a contradiction to the assumption that $Q(0)\ne 0.$ Hence our claim is true. $\square$ Now suppose $H(x)$ is a solution which is not a scalar multiple of $x^2+3,$ then consider the polynomial \[T(x)=H(x)-\frac{H(0)}3(x^2+3),\]Note that $T$ satisfies the polynomial equation and $T(0)=0.$ Hence by Claim 1, $T(x)\equiv 0,$ thus $H(x)=\tfrac{H(0)}{3}(x^2+3)$ which is a contradiction to our assumption that $H$ is not a scalar multiple of $x^2+3.$ And we are done. $\blacksquare$

Solution with awang11 and claserken: It's easy to see that the only constant solution is $P(x)=0$, which works. Note that if $P(x)$ works, then $kP(x)$ works for any real constant $k$. Thus let us assume that $P$ has degree $\geq 1$ and is monic. Multiply both sides by $xyz$, and plug $(x,\frac{x+\varepsilon}{2x \varepsilon -1}, \varepsilon)$ into this equation. Taking the limit as $\varepsilon \to 0$ gives $P(x)=P(-x)$, so $P$ is even. Plugging $(x,\tfrac{1}{x},x+\tfrac{1}{x})$ into the original functional equation, we get $$\frac{P(x)}{1+\tfrac{1}{x^2}}+\frac{P(\tfrac{1}{x})}{x^2+1}+P(x+\tfrac{1}{x}) = P(x-\tfrac{1}{x})+P(x)+P(\tfrac{1}{x})$$which rearranges as $$\boxed{(x+\tfrac{1}{x})(P(x+\tfrac{1}{x})-P(x-\tfrac{1}{x})) = \tfrac{1}{x}P(x)+xP(\tfrac{1}{x})}.$$Since $P(x)$ is even, it only has terms of even degree. We compare the coefficients of the highest degree terms of the two sides of the above boxed equation. Note that the highest degree term of $(x+\tfrac{1}{x})^d-(x-\tfrac{1}{x})^d$ is $2dx^{d-2}$; thus we can see that the coefficient of the highest degree term of the left side is $2 \text{deg} (P)$ (because we assumed $P$ was monic). If $\text{deg} (P) >2$, then the coefficient of the highest degree term of the right side is $1$. But since $2 \text{deg} (P) \neq 1$, this is a contradiction. Thus $\text{deg} (P) \leq 2$, and since we already did the case $\text{deg} (P)=0$, we can let $P(x)=x^2+c$ for some constant $c$ (becaue $P$ is even). It can be checked that $P(x) = x^2+3$ works. Assume there was some $c \neq 3$ such that $Q(x)=x^2+c$ worked; then we can easily see that $P(x)-Q(x)=3-c$ must also work. This is a contradiction since we know the only constant solution is $0$, and we're done; scaling back gives $\boxed{P(x) = k(x^2+3)}$ for any (possibly 0) real constant $k$.

v_Enhance wrote: The left-hand side is a second-order finite difference in $x$ (up to scaling the argument), and the right-hand side is constant, so this implies $\deg P \le 2$. Can someone explain this part to me.

Thanks Titu. Rewrite the given equation as \[xP(x)+yP(y)+zP(z)=xyz[P(x-y)+P(y-z)+P(z-x)].\]Note that since $x=\frac{y+z}{2yz-1}$ may be varied in $y$, the given is a polynomial identity in $y$ that is satisfied infinitely many times so we need not worry about the $y$ real or $x\ne 0$ restrictions. A similar argument implies that it suffices to solve the previous equation except for complex numbers $2xyz=x+y+z$. Taking $x=0$ implies $yP(y)$ is odd, so we can write it as $yQ(y^2)$ for some polynomial $Q$. Thus, we may write \[xQ(x^2)+yQ(y^2)+zQ(z^2)=xyz[Q((x-y)^2)+Q((y-z)^2)+Q((z-x)^2)].\]Let $y=\frac{i}{\sqrt{2}},z=-\frac{i}{\sqrt{2}}$ so the given is always true. Then \[xQ(x^2)=\frac{x}{2}[Q((x-i/\sqrt{2})^2)+Q((x+i/\sqrt{2})^2)+Q(-2)].\]Since the result upon division by $x/2$ holds for all values of $x\ne 0$, it must be a polynomial identity so \[Q(x^2)-Q((x-i/\sqrt{2})^2)=Q((x+i/\sqrt{2})^2)-Q(x^2)+Q(-2).\]But this implies the difference polynomial $R(x)=Q((x+i/\sqrt{2})^2)-Q(x^2)$ is linear, so $Q(x^2)$ is quadratic and $Q(x)=ax+b$ for some $a,b$. Note $a=0$ would imply $Q(-2)=0$ so $b=0$ and $P\equiv 0$. Otherwise, $a\ne 0$ and we can scale to set $a=1$. Thus the original given rewrites as \[x^3+bx+y^3+by+z^3+bz=3bxyz+2xyz[x^2+y^2+z^2-xy-yz-zx]=3bxyz+(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=\]\[x^3+y^3+z^3-3xyz+3bxyz.\]Thus $2bxyz=b[x+y+z]=(3b-3)xyz$. Taking $x,y,z$ all nonzero immediately implies $b=3$. Thus $P(x)=c(x^2+3)$ for some constant $c$ and the previous manipulation implies it is the only working choice of $P$.

Solution (finished basically like william122): We claim that the answer is $P(x) = c(x^2+3)$, where $c$ is a real number. Firstly, due to continuity, since $P$ is a polynomial, the given equation can be extended to $0$. Now take $(x, -x, 0)$ to get that $P$ is even. Now let $deg(p) = k$. We will show that $k \leq 2$. Taking $(x, \frac{1}{x}, x+\frac{1}{x})$ into the original equation and rearranging some terms, we have $(x^2 + 1)(P(x+\frac{1}{x})-P(x-\frac{1}{x})) - P(x) - x^2P(\frac{1}{x}) = 0$. Since $P$ is even, the terms only have even degree. So now, if $k \ge 4$, then the coefficient of $x^{k-1}$ is nonzero, which is absurd. And so we can have $P(x) = c(x^2+a)$, and so if we plug in any real numbers for $c$, we can easily get that $a = 3$, so we are done. $\square$ Remark: $\frac{\pm i}{\sqrt{2}}$ also brrrr

Rewrite as\[xP(x) + yP(y) + zP(z) = xyz[P(x-y) + P(y-z) + P(z-x)]\]for all reals $x + y + z = 2xyz$. For each specific $x, y$, by the condition, we must plug $z = \tfrac{x + y}{2xy - 1}$. Say we take $y$ to be a tiny $\epsilon$ and consider $\epsilon \to 0$. Then, by polynomial continuity, we are allowed to take limits and get\[xP(x) - xP(-x) = 0 \implies P(x) = P(-x)\]so $P$ is even. Now, consider $t, \tfrac 1t, t + \tfrac 1t$ across all $t \neq 0$. We have\[tP(t) + \tfrac 1t P(\tfrac 1t) + (t + \tfrac 1t)P(t + \tfrac 1t) = (t + \tfrac 1t)[P(t - \tfrac 1t) + P(t) + P(\tfrac 1t)]\]which actually simplifies to\[(t + \tfrac 1t)[P(t + \tfrac 1t) - P(t - \tfrac 1t)] = \tfrac 1t P(t) + tP(\tfrac 1t)\]so these two must actually be equivalent everywhere. Let $k$ be the leading coefficient of $P$. Then, the largest degree term of the RHS is $kt^{d - 1}$ where $d = \text{deg} P$. Since $P$ is even, in consideration of the LHS, the highest degree terms is $2kt^{d-1}$ from which it follows that $k = 1$ and $P$ is constant, in which case $\boxed{P = 0}$ works, or $d \leq 2$. In the latter case, we have $P$ of the form $kx^2 + m$. Analyzing $(1, 1, 2)$ yields $m = 3k$, and it is easy to check all polynomials of the form $\boxed{P(x) = k(x^2 + 3)}$ work. $\blacksquare$

jj_ca888 wrote: Rewrite as\[xP(x) + yP(y) + zP(z) = xyz[P(x-y) + P(y-z) + P(z-x)]\]for all reals $x + y + z = 2xyz$. For each specific $x, y$, by the condition, we must plug $z = \tfrac{x + y}{2xy - 1}$. Say we take $y$ to be a tiny $\epsilon$ and consider $\epsilon \to 0$. Then, by polynomial continuity, we are allowed to take limits and get\[xP(x) - xP(-x) = 0 \implies P(x) = P(-x)\]so $P$ is even. Now, consider $t, \tfrac 1t, t + \tfrac 1t$ across all $t \neq 0$. We have\[tP(t) + \tfrac 1t P(\tfrac 1t) + (t + \tfrac 1t)P(t + \tfrac 1t) = (t + \tfrac 1t)[P(t - \tfrac 1t) + P(t) + P(\tfrac 1t)]\]which actually simplifies to\[(t + \tfrac 1t)[P(t + \tfrac 1t) - P(t - \tfrac 1t)] = \tfrac 1t P(t) + tP(\tfrac 1t)\]so these two must actually be equivalent everywhere. Let $k$ be the leading coefficient of $P$. Then, the largest degree term of the RHS is $kt^{d - 1}$ where $d = \text{deg} P$. Since $P$ is even, in consideration of the LHS, the highest degree terms is $2kt^{d-1}$ from which it follows that $k = 1$ and $P$ is constant, in which case $\boxed{P = 0}$ works, or $d \leq 2$. In the latter case, we have $P$ of the form $kx^2 + m$. Analyzing $(1, 1, 2)$ yields $m = 3k$, and it is easy to check all polynomials of the form $\boxed{P(x) = k(x^2 + 3)}$ work. $\blacksquare$ I don't get the "polynomial continuity" part of the solution.. can you explain more about it?

The answer is $P(x)=0$ and $P(x)=a(x^2+3)$. The checking is obvious. Moreover, if $n$ is constant obviously $P=0$, so assume that it is non-constant We now prove that these are the only ones. The key substitution is $(x,y,z)=(x,\frac{x+\frac{3}{2x}}{2},\frac{3}{2x})$. Let $x\to\infty$, then $z\to 0$ and $y\to\frac{x}{2}$, hence $yz\to\frac{3}{4}$. Let $\deg P=n$, we have $P(y), P(x-y),P(y-z)$ all tend to $\frac{1}{2^n}P(x)$, while $P(z-x)$ tends to $P(x)$ and $\frac{P(z)}{P(x)}$ tends to $0$. Therefore, dviding the equation by $P(x)$ we have $$\frac{4}{3}+\frac{2}{3}\cdot \frac{1}{2^n}=\frac{1}{2^{n-1}}+(-1)^n$$From which we obviously have $n=2$. Substituting $f(x)=ax^2+bx+c$, noting that\begin{align*} \frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}-3 &=\frac{x^3+y^3+z^3-3xyz}{\frac{x+y+z}{2}}\\ &=2(x^2+y^2+z^2-xy-yz-zx)\\ &=(x-y)^2+(y-z)^2+(z-x)^2 \end{align*}we can obtain $$\frac{2b(x^2+y^2+z^2)}{x+y+z}=c-3a$$Replacing $(x,y,z)$ by $(-x,-y,-z)$ we have $b=0$, hence $c=3a$ as desired.

tastymath75025 wrote: Find all polynomials $P$ with real coefficients such that $$\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)$$holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$. Proposed by Titu Andreescu and Gabriel Dospinescu Firstly, \(2xyz-x-y-z\) divides the LHS minus RHS (argue by long division and rational functions, such a technique is mentioned in PFTB chapter 1 example 4) and \(P\) is even (interchange two variables). So, we can consider \(P\) as a polynomial which takes complex values. Plugging \((x,y,z)=\left(x,\frac{i}{\sqrt{2}},-\frac{i}{\sqrt{2}}\right)\) gives us that \[2P(x)=P\left(x-\frac{i}{\sqrt{2}}\right)+P\left(x-\frac{i}{\sqrt{2}}\right)+P\left(\frac{2i}{\sqrt{2}}\right)\]implying that \[P\left(x-\frac{i}{\sqrt{2}}\right)+P\left(x-\frac{i}{\sqrt{2}}\right)-2P(x)\]is constant. Replace \(x\) with \(\frac{i}{\sqrt{2}}x\) and let \(Q(x)=P\left(\frac{i}{\sqrt{2}}x\right)\). This gives us that \[Q(x+1)+Q(x-1)-2Q(x)\]is constant. From here we can deduce that \(\deg P\leq 2\) by taking second derivative on that expression. Checking gives us that \(P(x)=c(x^2+3)\) is the only solution.

Multiply both sides by $xyz$ to get \[xP(x)+yP(y)+zP(z)=xyz(P(x-y)+P(y-z)+P(z-x))\] Let $Q(x,y,z)$ denote the given assertion. We have $y=\frac{x+z}{2xz-1}$. Take the limit as $z\to 0$ since $P$ is continuous. We get $y\to -x$ and so $xP(x)-xP(-x)=0\implies P(x)=P(-x)$ so $P$ is even, which implies all of its terms have even degree. $Q(z,\frac{1}{z},z+\frac{1}{z}): zP(z)+\frac{1}{z}P(\frac{1}{z})+(z+\frac{1}{z})P(z+\frac{1}{z})=(z+\frac{1}{z})(P(z-\frac{1}{z})+P(z)+P(\frac{1}{z}))$ We simplify this, \[zP(z)+\frac{1}{z}P(\frac{1}{z})+(z+\frac{1}{z})P(z+\frac{1}{z})=(z+\frac{1}{z})P(z-\frac{1}{z})+zP(z)+\frac{1}{z}P(z)+zP(\frac{1}{z})+\frac{1}{z}P(\frac{1}{z}),\]so \[(z+\frac{1}{z})(P(z+\frac{1}{z})-P(z-\frac{1}{z}))=zP(\frac{1}{z})+\frac{1}{z}P(z).\] Let $a$ be the leading coefficient of $P$ and suppose $P$ has degree $d$, where $d$ is even. Now suppose $d>2$. Since $zP(\frac{1}{z})$ doesn't have any highest degree term, the term with the highest degree in the RHS is $az^{d-1}$. Looking at the LHS, we find that the highest degree term is $2\cdot a \cdot d\cdot z^{d-1}$. $az^{d-1}=az^{d-1}\cdot 2d\implies 2d=1$ (since $a\ne0$), a contradiction. So $d\le 2$, which implies either $P$ is constant or $d=2$. Case 1: $P$ is constant. This obviously implies $P\equiv 0$. Case 2: $\deg(P)=2$. Since none of the terms have degree $1$, let $P(z)=az^2+c$. We look at this again \[(z+\frac{1}{z})(P(z+\frac{1}{z})-P(z-\frac{1}{z}))=zP(\frac{1}{z})+\frac{1}{z}P(z).\] The highest degree term of the LHS is $4az^{d-1}=4az$. But, the highest degree term of the RHS is in fact $az^{d-1}+cz=(a+c)z\implies a+c=4a\implies c=3a$. Thus, $P(x)=\boxed{a(x^2+3)}$ for real constants $a$ and $c$. $P\equiv 0$ is also covered here. It suffices to show that this works. The LHS is \[x^3+y^3+z^3+3x+3y+3z=x^3+y^3+z^3+6xyz=x^3+y^3+z^3-3xyz+9xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)+9xyz.\]The RHS is \[\frac{x+y+z}{2}(2(x^2+y^2+z^2-xy-yz-xz)+9)=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)+9xyz,\]which implies LHS=RHS, so we are done.

Very weird? First, notice that $P(x) = c(x^2+3)$ is a solution. Next, notice that the solution is a vector space, so assume there exists another solution, then subtract a constant factor of $c(x^2+3)$ to obtain a nonzero solution for $P(x)$ such that $P(0) = 0$, so $\text{deg} P \geq 2$. Asymptotically, it is clear that $\deg P$ is even. Then consider $(x, y, z) = (w, w, \frac{2w}{2w^2-1})$ for $\lim w \to 0$ Rearranging, we have $2P(x) - \frac{1}{x^2}P(x)+\frac{P(\frac{2x}{2x^2-1})}{x^2} = P(x-\frac{2x}{2x^2-1})+P(\frac{2x}{2x^2-1}-x)$. So $\frac{P(\frac{2x}{2x^2-1})}{x^2} = \frac{P(x)}{x^2}$ in the limit, so by double hospital's rule, we know $x^2 \mid P(x)$. Returning to the original equation, notice that we can now induct to contradiction on the multiplicity of $x$ in $P(x)$ by dividing both sides by $x^k$ then using hospital, so we are done.

We claim that $P(x)=cx^2+3c$ for $c\in\mathbb R$ are the only solutions. It's easy to check that these are the only quadratic solutions, now suppose FTSOC that $\deg P=n>2$. Note that the only constant solution is $0$, so we can assume that $n\ne0$ also. Final, by scaling, WLOG $P$ is monic. The equation rewrites as: $$xP(x)+yP(y)+zP(z)=xyz(P(x-y)+P(y-z)+P(z-x)).$$Taking $z\to0$ and $y=-x$ gives $xP(x)=xP(-x)$, so $P(x)=P(-x)$ for all $x\ne0$. But since $Q(x):=P(x)-P(-x)$ is zero for infinitely many points, we must have $Q$ be identically zero (since it's polynomial), so $P$ is even. Then $n$ is even, so $n\ge4$. Now we make use of the substitution: $$y=\frac1x,z=x+\frac1x.$$The equation becomes: $$A(x):=\left(x+\frac1x\right)P\left(x-\frac1x\right)-\left(x+\frac1x\right)P\left(x+\frac1x\right)+\frac1xP(x)+xP\left(\frac1x\right)=0.$$Note that the $x^{n-1}$ term of $A(x)$ must be $0$. The highest-degree term of $P\left(\frac1x\right)$ is at most constant, so the $xP\left(\frac1x\right)$ contributes no $x^{n-1}$ terms, since the most it contributes is a degree $1$ term. The highest-degree term of $P(x)$ is $x^n$, so the $\frac1xP(x)$ term contributes only $x^{n-1}$. The two highest degree-terms of $P\left(x-\frac1x\right)$ are $x^n-nx^{n-2}$, so $\left(x+\frac1x\right)P\left(x-\frac1x\right)$ contributes only $(1-n)x^{n-1}$. The two highest degree-terms of $P\left(x+\frac1x\right)$ are $x^n+nx^{n-2}$, so $-\left(x+\frac1x\right)P\left(x+\frac1x\right)$ contributes only $(-1-n)x^{n-1}$. Adding all of these, we have that the coefficient of $x^{n-1}$ in $A(x)$ is $1-2n$, so we need to have $2n=1$, a contradiction.

Rewrite the equation as $$xP(x)+yP(y)+zP(z)=xyz( P(x-y)+P(y-z)+P(z-x))$$Since any polynomials are continous, take $x,y,z \rightarrow x,-x,0$. Then we have: $$x(P(x)-P(-x))=0$$Which implies $P(x)=P(-x)$, or our polynomial is even. Now, in order to cancel $xyz$, take $(x,y,z)=(x,\frac{1}{x},x+\frac{1}{x})$ We have $$x^2P(x)+P(\frac{1}{x})+(x^2+1)P(x+\frac{1}{x})= (x^2+1)(P(x-\frac{1}{x})+P(-x)+P(\frac{1}{x}))$$Which simplifies into $$P(x)+x^2P(\frac{1}{x})=(x^2+1)(P(x+\frac{1}{x})-P(x-\frac{1}{x}))$$Now, let $d= \deg (P(x))$, suppose $d>2$. Let $a_n$ be the leading coefficient of our polynomial. Note that leading coefficient of $RHS$ is leading coefficient of $(x+\frac{1}{x})^d-(x-\frac{1}{x})^d$, by binomial theorem, it equals to $2a_nd(x^{d-2})$, while in $LHS$ it is $a_n$, but it is contradiction since $2a_nd \neq a_n$. Hence $d=2$ . Now just plug in $P(x)=ax^2+c$ to get $P(x)=a(x^2+3)$, hence we are done

Here is a simple algebraic proof. We will just use binomial expansion and polynomial divisibility. The answer $\boxed{P(x) \equiv c(x^2 + 3)}$ for any constant $c \in \mathbb R$. It is easy to verify these solutions work. Define the polynomial

There exists polynomials $R(x,y,z)$ and $S(x,y,z)$ such that \begin{align*} R(x,y,z) &:= (x+y+z) \left( P(x-y) + P(y-z) + P(z-x) \right) - \left( xP(x) + yP(y) + zP(z) \right) \\ &= S(x,y,z) \cdot \left(2xyz - (x+y+z) \right) \end{align*}This means in $R(x,y,z)$, if we replace every instance of $xyz$ by $\frac{x+y+z}{2}$, then we must eventually obtain the all $0$ polynomial. As a corollary, this implies any term of $R(x,y,z)$ not divisible by $xyz$ must cancel out. Now put $z=0$ and $x+y = 0$ to get $P$ is even. If $\deg P \le 2$, it is easy to see the mentioned solutions are the only ones. Assume on the contrary that $\deg P \ge 4$. Let $x^{2d}$ be the largest degree term of $P$. For convenience, we assume $x^{2d}$ has coefficient $1$ (by scaling $P$ by constant). Look at the expansion of $R(x,y,z)$. Let $\lambda$ denote the coefficient of $x^{2d}y$. Note $x^{2d}y$ is largest degree term. Due to previous analysis, it must also cancel. This forces $\lambda = 0$. But on the other hand:- Focus on $$ xP(x-y), yP(x-y), yP(z-x) $$in expansion of $R(x,y,z)$. These are precisely which contribute to $\lambda$. We obtain $$ \lambda = -2d + 2 $$Since $\lambda = 0$, this forces $d=1$, which is a contradiction! This completes the proof. $\blacksquare$

As the condition is equivalent to $x = \frac{y+z}{2yz - 1}$, this implies $2xyz-x-y-z$ divides $P$. Hence we can actually extend the problem to the complexes: now set $y_0 = \frac i{\sqrt 2}$ and $z_0 = -\frac i{\sqrt 2}$, such that $(x, y_0, z_0)$ satisfies the condition for any choice of $x$. This implies $$P\left(x+\frac i{\sqrt 2}\right) + P\left(x-\frac i{\sqrt 2}\right) - 2P(x)$$is constant. But this implies the second-order finite difference of $P$ is constant, thus $\deg P \leq 2$. It is simple to verify $P \equiv c(x^2+3)$ are the only solutions in this case.