Polynimials $P(x)$ and $Q(x)$ satisfying $P(x^2+1)=(Q(x))^2+2x, Q(x^2+1)=(P(x))^2$

Question

I am looking for all polynomials $P(x)$ and $Q(x)$ satisfying the simultaneous functional equations given in the title, i.e.

$P(x^2+1)=(Q(x))^2+2x$
$Q(x^2+1)=(P(x))^2$

I've already found one solution: $P(x)=x$ and $Q(x)=x-1$ but I'm not sure if this is the only one solution and if it's not, then, how to show it that there is no other possible solutions. I've tried to google something but I couldn't find anything. Please, help me.

Answer 1

Hint: In the first equation, replace $x$ by $-x$. You get $Q(x)^2-Q(-x)^2=-4x$. From $$(Q(x)-Q(-x))(Q(x)+Q(-x))=-4x$$ you can easily deduce that $Q(x)=ax+b$, and then it is easy to conclude.