Showing that $P(x)$, where $P(x) = P(x+2)-x^2-2$ for all $x$, is a third-degree polynomial I want to show that $P(x)$ is a third degree polymomial:
$P(x) = P(x+2) -x^2-2$  for every real $x$
It is not so difficult, but I am not seeing the proof straight away. 

Let:
$$\begin{align}P(x+2)&=a_0(x+2)^n+a_1(x+2)^{n-1}+a_2(x+2)^{n-2}+\cdots+a_n\\
P(x)&=\color{red}{a_0}x^n+\color{blue}{a_1}x^{n-1}+a_2x^{n-2}+\cdots+a_n \\
x^2+2=P(x+2)-P(x)&=2na_0x^{n-1}+(2n(n-1)a_0+2(n-1)a_1)x^{n-2}+\cdots \Rightarrow \\
x^2&=2na_0x^{n-1} \Rightarrow \\
2&=n-1 \quad \text{and} \quad 2na_0=1\Rightarrow \\
n&=3 \quad \text{and} \quad \color{red}{a_0}=\frac16. \end{align}$$
Thus $n=3$. You can continue to find the next coefficient:
$$2n(n-1)a_0+2(n-1)a_1=0 \Rightarrow \color{blue}{a_1}=\frac{-2}{4}=-\frac12.$$
