anyone??? 500th post

Notice first that by plugging $x=0$ in equation we get $P(0)=0$, and by plugging first $x=1$ and then $x=-1$ we get $P(-1)=-1$. Let $deg\ P = n$ and $deg\ Q = m$. Now let $P(x)=x^{n}+p_{n-1}x^{n-1}+\cdots+p_{1}x+p_{0}$ and $Q(x)=x^{m}+q_{m-1}x^{m-1}+\cdots+q_{1}x+q_{0}$. First compare degrees of both sides of given equation: On $LHS$ trivially it is $n$ and on $RHS$ we have: $Q\left(\frac{(x+1)^2}{2} \right)-Q\left(\frac{(x-1)^2}{2} \right)=\left(\frac{(x+1)^2}{2} \right)^{m}-\left(\frac{(x-1)^2}{2} \right)^{m}+q_{m-1}\left(\left(\frac{(x+1)^2}{2} \right)^{m-1}-\left(\frac{(x-1)^2}{2} \right)^{m-1}\right)+\cdots+q_{1}\left(\left(\frac{(x+1)^2}{2} \right)-\left(\frac{(x-1)^2}{2} \right)\right)\ (1)$ and we see that degrees $2m$ will cancel out leaving $2m-1$, so $n=2m-1$. Now we compare leading coefficients on $LHS$ and $RHS$, and again on $LHS$ it is trivially 2. Now we look at the $k^{th}$ term on RHS without the coefficient $q_k$: $\left(\frac{(x+1)^2}{2} \right)^{k}-\left(\frac{(x-1)^2}{2} \right)^{k}=\left(\frac{x^2+2x+1}{2}-\frac{x^2-2x+1}{2}\right)\left(\left(\frac{(x+1)^2}{2} \right)^{k-1}+\left(\frac{(x+1)^2}{2} \right)^{k-2}\left(\frac{(x-1)^2}{2} \right)+\cdots+\left(\frac{(x-1)^2}{2} \right)^{k-1}\right)=2x\left(\left(\frac{(x+1)^2}{2} \right)^{k-1}+\left(\frac{(x+1)^2}{2} \right)^{k-2}\left(\frac{(x-1)^2}{2} \right)+\cdots+\left(\frac{(x-1)^2}{2} \right)^{k-1}\right) (2)$. Now plugging $k=m$ in $(2)$ it is easily verified that coeffecient with $x^{2m-1}$ is $\frac{2m}{2^{m-1}}=\frac{m}{2^{m-2}}$ and now $2=\frac{m}{2^{m-2}}$ so $m=1$ or $m=2$. Now we check cases: $1)$ $m=1$ It follows that $n=1$ so because $P(0)=0$ it follows $P(x)=x$ and by plugging into original equation we get $Q(x)=x+c$ for some real constant $c$. $2)$ $m=2$ It follows that $n=3$. Because $P(1)=1$, $P(0)=0$, $P(-1)=-1$ we can get $P(x)=x^3$. Plugging this into original equation and after some algebraic manipulation we get $Q(x)=x^2-x+c$ for some real constant $c$.