Find all polynomials $P(x)$ such that $2 P\left(2 x^{2}-1\right)=P(x)^{2}-2$ Question -
(Romania, 1990)Find all polynomials
$P(x)$ such that
$$
2 P\left(2 x^{2}-1\right)=P(x)^{2}-2
$$
for all $x \in \mathbb{R}$.
Solution
Putting $x=1,$ we get a quadratic equation in $P(1)$ and hence $P(1)=1 \pm \sqrt{3} .$ If $P(1)=1+\sqrt{3},$ we may write $P(x)=(x-1) Q(x)+1+\sqrt{3} .$ This gives a relation for $Q(x)$.
$$
4(x+1) Q\left(2 x^{2}-1\right)=(x-1) Q(x)^{2}+2(1+\sqrt{3}) Q(x)
$$
Taking $x=1,$ we see that $Q(1)=0 .$ Thus $(x-1)$ divides $Q(x) .$ 
We may use the induction to prove that $(x-1)^{n}$ divides $Q(x)$ for all positive integers $n .$ Thus $Q(x) \equiv 0$ and $P(x)=1+\sqrt{3}$ for all $x .$ Similarly, $P(1)=1-\sqrt{3}$ gives $P(x)=1-\sqrt{3}$ for all $x$
now i did not understand how to prove that $(x-1)^{n}$ divides $Q(x)$ for all positive integers $n .$ using induction....
Any help will be appreciated 
thankyou
 A: You have the relation $P(x)=(x-1)Q(x)+(1+\sqrt{3})$. And you have also proved that $Q(x)=(x-1)T(x)$ with $\text{deg}~ T(x)=\text{deg}~ Q(x)-1$, thus you plug in this value of $Q(x)$ in the relation to get $P(x)=(x-1)^2T(x)+(1+\sqrt{3})$ and you can apply the same trick as before (plugging in $x=1$) to get $x-1 \mid T(x)$ as well and thus $(x-1)^2 | Q(x)$. Thus you can continue this process $\text{deg}~ P(x)-1$ (degree of $Q(x)$) times and degree of $P$ can be arbitrary so you have the conclusion.
A: Firstly, when using induction, you need the initial/base case. We conclude that $(x-1) = (x-1)^1$ divides the polynomial $Q(x)$.
Note that I will be using the following notation; $Q(x) = (x-1) \cdot Q^{(1)}(x)$, and in general \begin{equation}Q^{(n)}(x) = (x-1) \cdot Q^{(n+1)}(x)\end{equation} or \begin{equation}Q(x) = (x-1)^n \cdot Q^{(n)}(x).\end{equation}
Now for induction-step we will assume that $(x-1)^n$ divides $Q(x)$. We get the following: \begin{equation}P(x)=(x−1)^n Q^{(n)}(x) + 1 \pm \sqrt3 \end{equation}
To answer one must prove "the inductive step". Prove that if we assume the statement holds for $n$, it also holds for $n+1$.
So using the above, assumed identity, and plugging it into the original equation, we get:
\begin{equation}2 \left(((2x^2-1)−1)^n Q^{(n)}(2x^2-1) +1\pm\sqrt3\right) = \left((x−1)^n Q^{(n)}(x) +1\pm\sqrt3\right)^2 - 2\end{equation}
You should work out the brackets to yield:
\begin{equation} 4(x-1)^n(x+1)^n Q(2x^2-1) + 2(1\pm \sqrt3) =\\ (x-1)^{2n} Q^{(n)}(x)^2 + (x-1)^n Q^{(n)}(x) + 1 \pm 2\sqrt3 +3 -2\end{equation}
The constant cancels, we divide by $(x-1)^n$ and we get:
\begin{equation}4(x+1)^n Q(2x^2-1) = (x-1)^n Q^{(n)}(x)^2 +Q^{(n)}(x)  \end{equation}
Filling in $x=1$ gives:
\begin{equation} 2^{n+2} Q^{(n)}(1) = Q^{(n)}(1)\end{equation}
\begin{equation} (2^{n+2}-1) Q^{(n)}(1) = 0 \end{equation}
We conclude $Q^{(n)}(1) = 0$, thus $(x-1)$ is a divisor of $Q^{(n)}(x)$. 
