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I am trying to show $f=0, g=0$ are the only solutions to $(f(x))^2=(g(x))^2(1-x^2)$ for any $f(x),g(x)\in \Bbb R[X]$, which is from the answer of the question Ring of trigonometric functions with real coefficients.

I tried to expand it and plug values, but I found no way to prove it.

xyz
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  • Does $g^2(1-x^2)$ mean $[g(x)]^2 (1-x^2)$ or $[g(1-x^2)]^2$ ? – ypercubeᵀᴹ Dec 06 '20 at 09:59
  • sorry for the confusion, I will edit it. @ypercubeᵀᴹ – xyz Dec 06 '20 at 10:00
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    Working over $\mathbb{C}$: what can you say about the multiplicities of the roots of $f^2$? Of the right hand-side? – D. Thomine Dec 06 '20 at 10:10
  • @D.Thomine sorry, I cannot see it. I know 1 and -1 are roots for $f$ – xyz Dec 06 '20 at 12:58
  • @xyz In other words, what can you say about the parity of the highest power of $(1-x)$ which can divide LHS? What about the same for the RHS? – Macavity Dec 06 '20 at 13:10
  • I got it!!!! Thank you very much for your help!@Macavity. $f(x)$ has $(1-x)$ as a factor, let $n$ be the max power of $1-x$ in $f(x)$, therefore $f(x)^2=h(x)^2(x-1)^{2n}=g(x)^2(x-1)(x+1)$. So $g(x)^2(x+1)=h(x)^2(x-1)^{2n-1}$, thus $g(x)$ has a $1-x$ as a factor, so the parity of the power $1-x$ on the left-hand side is odd, while is even in the right-hand side. I am wondering if I got it right. – xyz Dec 06 '20 at 13:32

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