Functional equation; Find $p(x)$ $p(x)$ is a polynomial with real coefficients, such that:
$$2p(x^2)=p(x^2+1)+(x^2+1)$$
Thus, $$2p(x)=p(x+1)+x+1$$ and $$2p(x-1)=p(x)+x$$
From that, what I did was write p(x) as $$a_n x^n+a_{n-1}x^{n-1}+...a_1x+a_0$$, p(x+1) as $$a_n (x+1)^n+a_{n-1}(x+1)^{n-1}+...a_1(x+1)+a_0$$ 
 and use the relations above to try to get things to cancel out, which led me into a bunch of binomial expansions because of the $(x+1)^n$ and $(x-1)^n$.
After still being stuck, I don't know what to do.
What am I missing here?
 A: Alt. hint:   rewrite it as $\;2 \big(p(x) - x-2\big)=p(x+1)-(x+1)-2\,$. Telescoping for $\,n \in \mathbb{N}\,$:
$$
p(n)-n-2 = 2 \big(p(n-1) - (n-1)-2\big) = \ldots = 2^n \big(p(0) - 2\big)
$$
For $\,p\,$ to be a polynomial it is necessary that the $\,2^n\,$ term vanishes, so $\,p(0)=2\,$, then $\,p(x)=\ldots\,$
A: Got another apparent solution:
if    $2p(x)=p(x+1)+(x+1)$, both sides being polynomials:
$$2a_nx^n+2a_{n−1}x^{n−1}+...+2a_1x+2a_0=a_n(x+1)^n+a_{n−1}(x+1)^{n−1}+...+a_1(x+1)+a_0+(x+1)$$
That should mean their coefficients are also equal, so for those of $x^n$:
On the left, that coefficient is $2a_n$. On the right:
$a_n(x+1)^n=a_n(x^n+nx^{n-1}+...+nx+1)$, which leads to
$a_nx^n+a_{n-1}x^{n-1}...+a_n$
Therefore, the coefficient of $x^n$ on the right polynomial is just $a_n$. Since coefficients from both should be equal:
$$2a_n=a_n$$
$$a_n=0$$
That goes all the way until $a_2$, so all that's left is
$$2a_1x+2a_0=a_1(x+1)+a_0+x+1$$
$$2a_1x+2a_0=a_1x+a_1+a_0+x+1$$
$$2a_1x+2a_0=x(a_1+1)+a_1+a_0+1$$
On one side, the coefficient for $x^1$ is $2a_1$, on the other it's ($a_1+1)$
$$2a_1=a_1+1$$
$$a_1=1$$
Bringing the whole equation down to
$$2x+2a_0=2x+(2+a_0)$$
The coefficients for $x^0$ are $2a_0$ and $(2+2a_0)$
$$2a_0=2+a_0$$
$$a_0=2$$
Therefore,
$$p(x)=x+2$$
