TITLE: Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$
QUESTION [8 upvotes]: Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$.

I assume that $O(p(x)=n)$   (Where O(p) denotes the order of p)
let $p(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2} \dots \dots +a_{1}x^{1}+a_{n-1}$ also $a_n \neq 0$
Then we have $a_n^2x^{2n} +\dots +a_0^2=1+a_nx^{n+1}+ \dots a_0x$
since $a_n \neq 0$ we must have $2n=n+1$ and hence $n=1$ hence the polynomial must be linear and hence there doesn't exist any solution(I proved assuming $p(x)=ax+b$)
Is it correct?

REPLY [4 votes]: After observing that the degree is at most $1$ you can simplify the rest of the argument as follows: $p(0)^{2}=1$ from the equation, so $p(0)=\pm 1$. Also $p(-1)^{2}=1-p(0)$. By considering the cases $p(0)=1$ and $p(0)=-1$ separately it is quite easy to conclude that the only solution is $p(x)=x+1$.