Find all polynomials $p(x)$ such that $(p(x))^2 = 1 + x \cdot p(x + 1)$ for all $x\in \mathbb{R}$ 
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?
 A: 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 $p(x)=x+1$. 
A: Our OP SunShine has correctly deduced that $\deg p(x) = 1$, so that
$p(x) = ax + b; \tag 1$
we compute:
$p^2(x) = a^2x^2 + 2ab x + b^2; \tag 2$
$p(x + 1) = ax + a + b; \tag 3$
$xp(x + 1) = ax^2 + ax + bx; \tag 4$
$xp(x + 1) + 1 = ax^2 + ax + bx + 1; \tag 5$
$a^2 x^2 + 2ab x + b^2 = ax^2 + ax + bx + 1; \tag 6$
$(a^2 - a)x^2 + 2abx + b^2 = ax + bx + 1; \tag 7$
$a^2 = a, \; b^2 = 1, \; 2ab = a + b; \tag 8$
$a = 0, 1; \; b = \pm 1; \tag 9$
$b = 1 \Longrightarrow 2a = a + 1 \Longrightarrow a = 1 \in \{0, 1 \}; \tag{10}$
$b = -1 \Longrightarrow -2a = a - 1 \Longrightarrow a = \dfrac{1}{3} \notin \{0, 1 \}; \tag{11}$
thus the only solution is $a = b = 1$, so that
$p(x) = x + 1. \tag{12}$
CHECK:
$p^2(x) = (x + 1)^2 = x^2 + 2x + 1$
$= x(x + 2) + 1 = x((x + 1) + 1) + 1 = xp(x + 1) + 1; \tag{13}$
it thus appears that the only solution is
$p(x) = x + 1. \tag{14}$
