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. 
 A: For any polynomial with degree $n$ and for each constant $k\neq0$, $P(x+k)-P(x)$ is a polynomial whose degree is $n-1$; to see why, just expand $P(x+k)$.
Since, in your case, $P(x+2)-P(x)$ has degree $2$, $P(x)$ is a cubic polynomial.
Note that this assumes that $P(x)$ is a polynomial function. Otherwise, the statement is false. A counter-example is this:$$P(x)=\begin{cases}\displaystyle\frac{x^3}{6}-\frac{x^2}{2}+\frac{4 x}{3}&\text{ if $x$ is algebraic}\\[3mm]\displaystyle\frac{x^3}{6}-\frac{x^2}{2}+\frac{4 x}{3}+1&\text{ otherwise.}\end{cases}$$
A: Assuming $P$ is $C^3$, you have $P'''(x) = P'''(x+2)$ for all $x$.  Since $P'''$ is continuous, we conclude that it's constant.  When we integrate 3 times, we get a cubic polynomial.
A: The statement, as written, is false.
Take any function on $[0,2]$ with $p(2)=p(0)+2$. The linear function $p(x) = x$ between 0 and 2 will do. Again, it linear on this domain. Now use the relations $p(x+2) = p(x)+x^2+2$ to determine the function of $[2,4]$ then on $[4,6]$, etc. and $p(x-2)=p(x)-x^2-2$ on $[-2,0]$, $[-4,-2]$, etc.
For example, between $[2,4]$, 
$$p(x)=p(x-2)+x^2-4x+6=(x-2)+x^2-4x+6$$
So you have a piece-wise continuous function that satisfies the criterion but is linear on $[0,2]$ and second degree on other intervals.
Obviously, you can take any nasty function you want on $[0,2]$ that satisfies $p(2)=p(0)+2$ and get a non-polynomial solution of the recurrence relation. 
Now, if one assumes that $p(x)$ is a polynomial, I don't think you can do it in any way other than simply assuming
$$p(x)=a_nx^n+\cdots+a_1x+a_0.$$
Then use the binomial theorem to expand $p(x+2)$ then subtract $p(x)$ and set it equal to $x^2+2$. You will get $a_n =a_{n-1} =\cdots = a_4 = 0$ and $6a_3=1$, $12a_3+4a_2=0$ and $8a_3+4a_2+2a_1=2$ and $a_0$ arbitrary.
Final answer: $p(x)=\frac16x^3-\frac12x^2+\frac{13}3x+C$
A: 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:
$$2n(n-1)a_0+2(n-1)a_1=0 \Rightarrow \color{blue}{a_1}=\frac{-2}{4}=-\frac12.$$
