<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] Momentarily setting aside the trivial solution, one option is to rearrange in the form of a finite difference equation, like so:
$$(x^3+1)(P(x+1)-2P(x)+P(x-1))-mx^2(P(x+1)-P(x-1))+2mxP(x)=0$$
Adjusting the definition of a polynomial $P(x)$ might make sense in this case, where it appears that finite differences play a significant role.  In particular, using the binomial form of polynomial terms would result in a definition of an $n$th degree polynomial as
$$P(x)=\sum_{i=0}^na_i{x\choose i}$$
and then the values of $P(x+1)-2P(x)+P(x-1)$ and $P(x+1)-P(x-1)$ are easily determined as
$$P(x+1)-2P(x)+P(x-1)=\sum_{i=0}^na_i\left({x+1\choose i}-2{x\choose i}+{x-1\choose i}\right)
=\sum_{i=2}^na_i{x-1\choose i-2}$$
$$P(x+1)-P(x-1)=\sum_{i=0}^na_i\left({x+1\choose i}-{x-1\choose i}\right)
=\sum_{i=1}^na_i\left({x\choose i-1}+{x-1\choose i-1}\right)$$
and then the full difference equation looks like
$$(x^3+1)\sum_{i=0}^{n-2}a_{i+2}{x-1\choose i}-mx^2\sum_{i=0}^{n-1}a_{i+1}\left({x\choose i}+{x-1\choose i}\right)+2mx\sum_{i=0}^na_i{x\choose i}=0$$
Reorganizing, we have extra terms $Q(x)=-a_nmx^2\left({x\choose n-1}+{x-1\choose n-1}\right)+2a_nmx{x\choose n}+2a_{n-1}mx{x\choose n-1}$ which rearranges to
$$Q(x)=a_nmx^2\left(\frac2n-1\right){x-1\choose n-1}+mx(2a_{n-1}-a_nx){x\choose n-1}$$
and main sum
$$Q(x)+\sum_{i=0}^{n-2}\left[\left(a_{i+2}(x^3+1)-a_{i+1}mx^2\right){x-1\choose i}+\left(2a_imx-a_{i+1}mx^2\right){x\choose i}\right]=0$$
Clearly, the low-exponent monomials in $x$ will have many terms to deal with, but it appears that the high-exponent monomials in $x$ will only occur a few times each, so it would be best to analyze this sum starting from the high-exponent terms first.  So starting with all terms where $x^{n+1}$ appears, we have
$$a_nx^3{x-1\choose n-2}-a_nmx^2\left({x\choose n-1}+{x-1\choose n-1}\right)+2a_nmx{x\choose n}=0$$
So either $a_n = 0$ or is irrelevant for the solution.  We can further divide through by $x$ to get
$$x^2{x-1\choose n-2}-mx\left({x\choose n-1}+{x-1\choose n-1}\right)+2m{x\choose n}=0$$
$$x^2\frac{(x-1)!}{(n-2)!(x+1-n)!}-mx\frac{x!}{(n-1)!(x+1-n)!}-mx\frac{(x-1)!}{(n-1)!(x-n)!}+2m\frac{x!}{n!(x-n)!}=0$$
$$n(n-1)x^2(x-1)!-mnx^2(x-1)!-nmx(x+1-n)(x-1)!+2mx(x+1-n)(x-1)!=0\\
n(n-1)x^2-mnx^2-nmx(x+1-n)+2mx(x+1-n)=0\\
n(n-1)x-mnx-mn(x+1-n)+2m(x+1-n)=0$$
In order for all $x$-dependent terms to cancel, $n$ must be either $0$ or $1$.  If $n=0$ then $P(x)=c$ for some constant $c$, but this fails a simple sanity check with the original equation ($0=-2mxc,m\neq 0$) unless $c=0$.  So we have either $P(x)=0$ which means that $m$ is unlimited except for not being $0$, or $P(x)=a_1x+a_0$ which means $n=1$.  In that case, our main sum is $0$ already, and since $Q(x)$ has only a single term with $a_{n-1}$ present and that term has a lower exponent of $x$ than all the others, we must have $a_0=0$, so our solution set is $P(x)=ax$.
Given this solution set, what is the limitation on $m$ when $n=1$?  All terms in $Q(x)$ cancel, so we have to go back to the original again, which works out as
$$(x^3 - mx^2 +1 )a(x+1)  + (x^3+mx^2+1)a(x-1) =2(x^3 - mx +1 )ax\\
ax^4+ax^3-amx^3-amx^2+ax+a+ax^4-ax^3+amx^3-amx^2+ax-a=2ax^4-2amx^2+2ax\\
0=0$$
From this, it is apparent that $m$ can take on any non-zero integer value.
Sources:  my own work over the past $20$+ years, as supplemented by mathworld and wikipedia.
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