I have been working through an exercise and I have found out that $a^2(x) = b^2(x) (1 - x^2)$, where $a(x), b(x) \in \mathbb{R}[x]$. Is this enough to deduce $a = b = 0$?
I think so, because from the first equality you can deduce that $a(x)$ is divisible by $1 - x$ and $1 + x$, from that you can deduce that $b(x)$ is divisible by $1 - x$ and $1 + x$, and so on. Is this proof correct? Can anyone write it down in a more formal way?
This is just the usual proof that the square root of some element is irrational, suitably generalized. One version is the following: Let $R$ be a factorial domain and let $a \in R$ be a non-zero non-unit which contains a prime element $p$ with multiplicity $1$. Then $T^2-a \in R[T]$ is irreducible by Eisenstein's criterion, which implies that $\sqrt{a} \notin Q(R)$ (but you can also show this directly as usual). Now apply this to $R=\mathbb{R}[x]$ and $a=1-x^2$, $p = 1-x$.