# Why are these two monic complex polynomials coprime?

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 $$s$$.

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.
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 $$\,\color{#c00}{P,Q}$$
• $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\ \$ – Bill Dubuque Jul 11 at 14:51