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

Note that, $P(x)=cx$ is a trivial solution, we will prove that it is the only one. 
Suppose that degree of this polynomial is $>1$. 
From your work, comparing coefficients of $x^{n+1}$ gives , $n=2m$, and so $m>0$. 
Now,after doing some manipulations(multiply by $x$ and then regroup) , we get, 
$$ (x^3-mx^2+1)F(x)= (x^3+mx^2+1)F(x-1) \cdots (1)$$ 
Where, $F(x) = xP(x+1)-(x+1)P(x)$ . 
Comparing, degree of $F(x)$ with $P(x)$ we get that they both have same degree($n=2m$). 
Consider $G(x)= x^3+mx^2+1$ Since it is cubic it has either $1$ or $3$ real roots. Now, $G'(x) = 3x^2+2mx$ . $G'(x)=0  \implies x = -2m/3, x=0$ So local extrema at $x=0$ or $x=-2m/3$ . $G(0)=1>0, G(-2m/3) = 1+\frac{4m^3}{27} $
So three real roots if $ m \le -2 $ or, one real root if $m \ge -1$ 
Let $H(x) = x^3-mx^2+1 $. Suppose that , $G(x)$ and $H(x)$ have unequal number of roots. This is impossible since, in $(1)$ $LHS/RHS$ will have one root more than $RHS/LHS$ which is absurd. 
So $G(x)$ and $H(x)$ have same number of real roots. $m \le -2, m \ge 2 \implies ABSURD $. $m \ge -1, m \le 1 \implies m=1 \implies n=2 $(since m is $>0$)
Now, it is not hard to check that $n=2$ has no solutions.(Since you are attempting $IMOSL$ , i don't think it will be hard to prove this part)
