<s>[INST] How find this P(x) if $ (x^3 - mx^2 +1 ) P(x+1) + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $ Let $m \neq 0 $ be an integer. Find all polynomials $P(x) $ with real coefficients such that 
$$ (x^3 - mx^2 +1 ) P(x+1)  + (x^3+mx^2+1) P(x-1) =2(x^3 - mx +1 ) P(x) $$
This problem is IMO Shortlist 2013 
let $$P(x)=\sum_{i=0}^{n}a_{i}x^i,a_{i}\in R$$
then
$$\sum_{i=0}^{n}[a_{i}x^3(x+1)^i-mx^2(x+1)^i+(x+1)^i]+\sum_{i=0}^{n}[x^3(x-1)^i+mx^2(x-1)^i+(x-1)^i]=2\sum_{i=0}^{n}[x^{3+i}-mx^{i+1}+x^i]$$
then I can't
 [/INST] EDIT (by abiessu):  the original is wrong, but as it has been accepted, it seems that it would be good to make the accepted answer correct.  The original approach is generally used but corrected where in error.

(original) I suspect that there is a typo in your OP, because as it is stated now it is rather uninterestingly easy and there are only trivial (linear) solutions.
Definitely not IMO Shortlist stuff.
We can assume that $n$ is the degree of $P$, so that $a_n\neq 0$. Then,
if $n \neq 0$ the coefficient before $x^{n+2}$ in 
$D=(x^3 - mx^2 +1 ) P(x+1)  + (x^3+mx^2+1) P(x-1) -2(x^3 - mx +1 ) P(x)$
is given by considering
$$x^3\sum_{j=0}^n a_j\left((x+1)^j+(x-1)^j-2x^j\right)\tag 1$$
$$-mx^2\sum_{j=0}^na_j(x+1)^j+mx^2\sum_{j=0}^na_j(x-1)^j\tag 2$$
$$2mx\sum_{j=0}^na_jx^j\tag 3=2m\sum_{j=0}^na_jx^{j+1}$$
We see that $j=n$ results in $0=0$ from $(1)$, and this is the only $x^{n+3}$ source.  The same thing happens when we try to find $x^{n+2}$ terms, and so non-zero $x^{n+1}$ terms must be considered, and these are
$$a_{n}n(n-1)x^{n+1}-2ma_{n}nx^{n+1}+2ma_nx^{n+1}=0$$
The result of this is that $a_{n}(n(n-1) - 2m(n-1))=0$.  This is not fully conclusive, but it does mean that we can claim one of three scenarios:


*

*$a_n = 0$

*$n = 2m$

*$n-1 = 0$


Of these, $a_n = 0$ contradicts our assumption, so we have either $n = 2m$ or $n = 1$.  Considering the first possibility, note that $m\ne 0$ is an integer, and therefore $n=2m \to m \ge 1$ and $n=2m \to n \ge 2$.
Now considering terms having $x^n$, we find terms
$$a_{n-1}n(n-1)x^n-2mna_{n-1}x^n+2ma_{n-1}x^n=0$$
This exactly repeats our previous scenario, but now $a_{n-1}=0$ is an acceptable possibility, and $n=2m$ or $n=1$ are simply confirmed as alternate options.  This pattern will almost certainly repeat itself ad infinitum, except that considering terms with $x^{n-1}$ will bring in the following extra terms:
$$\frac 1{12}a_nn(n-1)(n-2)(n-3)x^{n-1}-\frac 13ma_nn(n-1)(n-2)x^{n-1}=0$$
which brings us to $\frac 14n(n-1)(n-2)(n-3)-mn(n-1)(n-2)=0$.  This narrows our possibilities to either $n=1$ or $n=2m$ with $n=2$ or $n-3=4m$.
Now we can see that $n=1,2$ are the only possibilities when $a_n$ is non-zero, since $n=4m+3$ is incompatible with $n=2m$ for positive $n$.
If we assume that $n=2$ and use $P(x) = ax^2 + bx + c$, we get
$$ax^2(x^3-mx^2+1)+(2a+b)x(x^3-mx^2+1)+(a^2+b+c)(x^3-mx^2+1)\\
+ax^2(x^3+mx^2+1)+(-2a+b)x(x^3+mx^2+1)+(a^2-b+c)(x^3+mx^2+1)\\
=2ax^2(x^3-mx+1)+2bx(x^3-mx+1)+2c(x^3-mx+1)$$
resulting in
$$-2amx^3+2a^2x^3+2a^2=-2cmx$$
The only way this can be true for all $x$ is if $a = 0$ and $c=0$, thus we have that $n=1$ is the only source of polynomial solutions for our original equation, and every such solution is of the form $P(x)=ax$.
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