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Let $P(x), Q(x) \in \mathbb{C}[x]$ two monic complex polynomials. It is given that $P(x)$ divides $Q(x)^2+1$ and $Q(x)$ divides $P(x)^2+1$.

Why does it follow from these conditions that $P(x)$ and $Q(x)$ are coprime?

My attempt

I tried defining $P(x)=s(x)*u(x)$ and $Q(x)=s(x)*v(x)$, where $s, u, v \in \mathbb{C}[x]$ monic complex polynomials and tried showing that $s(x)=1$ is the only solution that satisfies the conditions above, but I cannot seem to get this restriction on $s$.

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If $R(x)$ divides both $P(x)$ and $Q(x)$ then, since $P(x)\mid Q(x)^2+1$, $R(x)\mid 1$. So, $R(x)$ is constant.

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  • $\begingroup$ That's a simple proof! Thank you. $\endgroup$ – kdnooij Jul 11 at 14:42
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Hint $\ P\mid Q^2\!+1\!\iff\! RP = Q^2\!+1\!\iff\! R\,\color{#c00}P-Q\,\color{#c00}Q = 1,\, $ a Bezout equation for $\,\color{#c00}{P,Q}$

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  • $\begingroup$ $ A\mid B\ $ means $\,A\,$ divides $\,B\,$ (standard notation in number theory) Note that the proof works more generally if we replace $\, Q^2+1\,$ by $\,AQ + c\,$ for any constant $\,c\neq 0\ \ $ $\endgroup$ – Bill Dubuque Jul 11 at 14:51

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