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The following problem comes from Vol. 1 in Goursat's Cours d'analyse mathematique. The reader is asked to show that, given two integer polynomials $P,Q$ $$ \sqrt{1-P^2}=Q\,\sqrt{1-x^2} \implies \frac{\mathrm{d}P}{\sqrt{1-P^2}}=\frac{n\,\mathrm{d}x}{\sqrt{1-x^2}},\quad n\in\mathbb{N}. $$ As a hint, Goursat notes that $1-P^2=Q^2(1-x^2)$ implies that $Q$ is prime to $P$. I tried to prove this by contradiction: assume there is a nonconstant factor $f$ such that $Qf=P$. Then $$ \begin{align} 1-Q^2f^2&=Q^2-Q^2x^2\\ 1&=Q^2(1-x^2+f^2). \end{align} $$ In order that the right-hand side be independent of $x$, and for $Q$ to be an integer polynomial, one must choose $f=x$, from which it follows that $Q=1$ and $P=x$. But $\mbox{gcd}(1,x)=1$, contrary to assumption. Hence $P$ and $Q$ must be relatively prime.

Is this proof on the right track?

CW279
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  • I would rather start with $f\mid P$ and $f\mid Q$ for a non-constant $f$, to have $\gcd(P,Q)\neq 1$. Your condition $Qf=P$ does not imply this, because for $Q=1$ we have $\gcd(P,Q)=1$ for every $P$. So your $\gcd(1,x)=1$ is not a contradiction. – Dietrich Burde Mar 30 '25 at 11:03
  • @DietrichBurde Thank you for your advice. Could you comment on whether or not the following proof holds? Given $1-P^2=Q^2(1-x^2)$, we have $1=pP+qQ$, where $p=P$ and $q=Q(1-x^2)$ are also integer polynomials. By Bezout's identity, it follows that $P,Q$ are coprime. – CW279 Mar 30 '25 at 11:45
  • Actually, by the equivalence given in Bezout, see here. – Dietrich Burde Mar 30 '25 at 11:53
  • @DietrichBurde So the proof is valid by the $(\impliedby)$ part of the link attached? – CW279 Mar 30 '25 at 14:31

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