The answer is $P(x) = bx^2 + a(x^4+6)$ which clearly works. Clearly we can extend this as a poly identity by multiplying $x^{\deg P}$ so we can plug any $x\in \mathbb{C}$. Let $Q(x)$ denote the assertion. $Q(x), Q(1/x)$ imply $P$ even. When $n\ge 6$ let $cx^n$ be the leading coefficient of $P$. Then there is a $dx^{n-4}$ term that doesn't cancel in $RHS-LHS$, so $\deg P \le 4$. Since $P$ is a polynomial in $x^2$, we can see that $P(x)$ must be of the form $bx^2 + a(x^4+6)$

We are given that $$P(x)+P\left(\frac1x\right) =\frac{P\left(x+\frac1x\right) +P\left(x-\frac1x\right)}2.$$We claim that the only solution is $P(x)=bx^2$ where $b\in\mathbb{R}$ is a constant. Plugging $\frac1x$ instead of $x$ gives $$P\left(\frac1x\right)+P(x)=\frac{P\left(x+\frac1x\right)+P\left(\frac1x-x\right)}2\Longrightarrow P\left(\frac{1}{x}-x\right)=P\left(x-\frac{1}{x}\right)$$Since $x-\frac{1}{x}$ assume values for all reals, we conclude that the polynomial is even. So, $P(x)=Q\left(x^2\right)$ for some polynomial $Q\in\mathbb{R}[x]$. We have $$2Q\left(x^2\right)+2Q\left(\frac{1}{x^2}\right)=Q\left(x^2+\frac{1}{x^2}+2\right)+Q\left(x^2+\frac{1}{x^2}-2\right)$$So, for every $x>0$ we have $$2Q(x)+2Q\left(\frac{1}{x}\right)=Q\left(x+\frac{1}{x}+2\right)+Q\left(x+\frac{1}{x}-2\right)$$Suppose that $Q(x)=a_0+a_1x+\cdots+a_nx^n$, where $a_0,a_1,\ldots,a_n\in\mathbb{R}$ and $a_n\ne0$. We claim that $Q(x)$ is a linear polynomial. Suppose $n>1$. Then, coefficient of $x^{n-2}$ in LHS is $2a_{n-2}$ while that in RHS is $2a_{n-2}+8a_n$, which gives $a_n=0$, impossible. So, $n\leq1$. Note that if $Q(x)=a+bx$, then $$Q\left(x+\frac{1}{x}+2\right)+Q\left(x+\frac{1}{x}-2\right)=a+b\left(x+\frac{1}{x}+2\right)+a+b\left(x+\frac{1}{x}-2\right)$$$$=2\left(Q(x)+Q\left(\frac{1}{x}\right)\right)=2\left(a+bx+a+b\left(\frac{1}{x}\right)\right)\Longrightarrow a=0\Longrightarrow P(x)=bx^2,$$as claimed. $\square$

Both solutions are wrong; indeed, $P(x)=x^4+6$ is a solution. Further, the set of solutions is an $\mathbb{R}$ vector space. It seems to be that (since $P$ is even) we can find a working polynomial with degree $2k$ for every $k$, and then generate the solution set by taking linear combinations. I cannot seem to find a nice closed form of the above though...

Indeed Here is a short python (Sympy) script that solves 12-dim polynomials: from sympy import symbols, Function, collect, simplify, solve s = " ".join(f"a{i}" for i in range(100)) a = symbols(s) x = symbols('x') def f(x, degree = 12): return sum(a[i] * x**i for i in range(degree + 1)) def expression(f, x): return 2 * f(x) + 2 * f(1/x) - f(x+1/x) - f(x-1/x) print(expression(Function('P'), symbols('x'))) d = solve(collect(simplify(expression(f, x)), x), a) d = {int(key.name[1:]) : value for key, value in d.items()} for k,v in sorted(d.items()): print(f"a{k}","=",v) 2*P(1/x) + 2*P(x) - P(x - 1/x) - P(x + 1/x) a0 = 6*a4 a1 = 0 a3 = 0 a5 = 0 a6 = 0 a7 = 0 a8 = 0 a9 = 0 a10 = 0 a11 = 0 a12 = 0 It looks like the general solution will be $\boxed{P(x) = a x^4 + bx^2 + 6a}$

Replace x by 1/x we have P(x)=P(-x) for all x, so P(x)=Q(x^2) We have Q(x)+Q(1/x)=(Q(x+1/x+2)+Q(1/x+x-2))/2. Let k=degQ. if k>=3 then multiple both side by x^k and compare coefficient of x^(2k-2) we have conditraction, so k<= 2. By computing we have all forms of Q is Q(x)=ax^2+bx+6a for some const a,b real. So all P(x) satisfy is ax^4+bx^2+6a.

I hope this is ok By swapping $x,\frac {1}{x}$ we'll get that $P$ is even.Let $P(x)=\sum b_ix^i$ and $d=degP$ Now , consider the terms in $(x+\frac{1}{x})^k + (x-\frac{1}{x})^k)$ for even $k$.It's $2x^k+a_{k-1}x^{k-1}+a_{k-2}x^{k-2}+a_{k-3}x^{k-3}+a_{k-4}x^{k-4}+sth$ where the rest of degrees are between $k-5,-k$. Clearly $a_{k-1},a_{k-3}$ are $0$ since they don't even exist at all , and the terms with degree $d-2$ will cancel each other in that equation.So the largest nonzero component is $x^{d-4}$. So if $b_{d-4}$ is nonzero , $RHS-LHS = O(x^{d-4})$ which is impossible if $d \geq 4$ . Similarly we'll get that all $b_{even}$'s are zero(for those with index smaller or equal than $d-4$) and since $P$ was even , $b_{odd}$'s are also zero. So $P(x)=ax^d+bx^{d-2} + c$. The rest is bashing and we'll get the general solution$ P(x) = a x^4 + bx^2 + 6a$

We claim the only answer is $P(x)=ax^4+bx^2+6a$ for some real constants $a,b$, which works. We now prove no other polynomial works. Note that by plugging $x \rightarrow 1/x$ in the given equation we obtain $P(x-1/x)=P(1/x-x),$ and since $x-1/x$ is surjective, we obtain that $P$ is even. Now, if $\deg P=2$, then $P(x)=cx^2+s,$ which works for any $c,s$. Moreover, if $\deg P=4,$ then $P(x)=ax^4+bx^2+cx$ for some constants $a,b,c$. After plugging back in the initial equation we obtain that $c=6a$. Henceforth assume that $\deg P=2n \geq 6$. Let $P(x)=a_{2n}x^{2n}+a_{2n-2}x^{2n-2}+\ldots+a_2x^2+a_0$ Note that the given identity $2P(x)+2P(1/x)=P(x+1/x)+P(x-1/x)$ can be interpreted as a corresponding polynomial equality, by multiplying both sides with $x^m$ for some sufficiently large $m$. Hence, we perform the following two steps: Step 1: Comparing the coefficient of $x^{2n-2}$ in both sides. From this comparison it's easy to obtain that $(n-1)a_{2n-2}=0,$ and so $a_{2n-2}=0$. Step 2: Comparing the coefficient of $x^{2n-4}$ in both sides. From this comparison it's easy to obtain that $2n(2n-1)a_{2n}+2a_{2n-4}=2a_{2n-4},$ and so $a_{2n}=0,$ absurd. Hence, we reach a contradiction if $\deg P \geq 6,$ as desired. Remark: Step 1 is basically unnecessary, but is nonetheless included for more clarity. Edit: corrected the solution.

What are all the solutions to this equation? I might have made a mistake in my solution as I have not included $x^4+6$, as pointed out by @tadpoleloop, although I can't find my mistake

Replacing $x$ with $\frac1x$ implies $P$ is even. Notice that $$x^{2n}+\frac1{x^{2n}}-\frac12\left(x+\frac1x\right)^{2n}-\frac12\left(x-\frac1x\right)^{2n}$$is a polynomial in $x$ and $\frac1x$ with degree exactly $2n-4$ for $n\geq2$ and it has degree at most $0$ for $n\leq1$. This means that if $P(x)=a_{2n}x^{2n}+a_{2n-2}x^{2n-2}+\ldots+a_2x^2+a_0$, then we must have $a_{2n}=a_{2n-2}=\ldots=a_6=0$, and since $x^4+\frac1{x^4}-\frac{\left(x+\frac1x\right)^4+\left(x-\frac1x\right)^4}2=-6$, we must have that $P(x)=ax^4+bx^2+6a$, which works.

vsamc wrote:

Thank you

My problem! Original problem statement wrote: Find all polynomials $P(x) \in \mathbb{R}[x]$ satisfying \[P(x) + P\!\left(\frac{1}{x}\right) = \frac{P\!\left(x + \frac{1}{x}\right) + P\!\left(x - \frac{1}{x}\right)}{2}\]for all nonzero $x \in \mathbb{R}$.

we replace $x \mapsto \frac{1}{x}$ to get: $P\left(\frac{1}{x}\right)+P(x)=\frac{P\left(\frac{1}{x}+x\right)+P\left(\frac{1}{x}-x\right)}{2}$ so we have $P\left(x-\frac{1}{x}\right)=P\left(\frac{1}{x}-x\right)$, since $z=x-\frac{1}{x}$ takes all real values for $x \in \mathbb{R}$ we have: $P(z)=P(-z)$ $\forall$ $z \in \mathbb{R}$ , which implies $P(x)=Q(x^2)$ for some $Q(x) \in \mathbb{R}[x]$ we set $\deg(P(x))=2n$, where $n \in \mathbb{Z}^{+}$ Claim:- $n \leqslant 2$ Pf:- FTSOC we assume $n>2$ set $P(x)=a_{2n}x^{2n}+a_{2n-2}x^{2n-2}+a_{2n-4}x^{2n-4}+\cdots+a_{0}$ , where $a_{2i} \in \mathbb{R}, i \in \{0,1,\cdots, n-1\}$ and $a_{2n} \in \mathbb{R} \setminus \{0\}$ we compare the degree of $x^{2n-4}$ in the original given reation to get: $2a_{2n-4}=2\left (\binom{2n-2}{2n}a_{2n}+\binom{2n-3}{2n-2}a_{2n-2}+\binom{2n-4}{2n-4}a_{2n-4}\right) \implies 2n(2n-1)a_{2n}+(2n-2)a_{2n-2}=0$ but if we compare degree of $2n-2$ in the given relation we get $a_{2n-2}$ is also $0$ , which gives $a_{2n}=0$ for $n>2$ which is absurd, hence we get a contradiction so we have $n \leqslant 2$ , claim follows. $\square$ Also we see that if $P(x)=ax^4+bx^2+c$, where $a,b,c \in \mathbb{R}$ then plugging this again in the given relation we get $c=6a$ which gives $\boxed{\{a , b \in \mathbb{R}:ax^4+bx^2+6a\}}$ to be the set of polynomials that satisfy the given relation. $\blacksquare$

We have $P(x-\frac{1}{x})=P(\frac{1}{x}-x)$ which implies the function is even. Thus, we could conclude the degree of the polynomial is even only. Note $(x+\frac{1}{x})^n+(x-\frac{1}{x})^n$ has terms $\binom{n}{2}x^{n-4}$ which doesn't exist in $x^n+\frac{1}{x}^n$, thus $n-4\leq 0\implies n_{max}=4$. So we limit the degrees of the polynomial to $0,2,4$ now. Note $\frac{1}{2}\cdot (x+\frac{1}{x})^4+(x-\frac{1}{x})^4=x^4+6+\frac{1}{x}^4\implies P(x)=a(x^4+6)$ When deg=2, $(x+\frac{1}{x})^2+(x-\frac{1}{x})^2=x^2+\frac{1}{x}^2\implies P(x)=bx^2$ Combine two equations together, the polynomial is $a(x^4+6)+bx^2+c=P(x)$

Main thing was to observe that polynomial is even after that coefficient matching and straightforward