Functional equation; Find $p(x)$

Question

$p(x)$ 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?

Answer 1

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 $\,p(x)=\ldots\,$

Answer 2

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,

$$p(x)=x+2$$