MHB What Polynomial Pairs Satisfy These Complex Functional Equations?

[Mathick]

Find all pairs of polynomials $$p(x)$$ and $$ q(x)$$ with real coefficients for which both equations are satisfied: $$p(x^2+1)=q(x)^2+2x$$ and $$q(x^2+1)=p(x)^2$$. These equations are set for all real $$x$$.

I tried to substitute $$x$$ for $$-x$$ and others numbers like $$-1,1$$ etc. but nothing happened... I need your help

 

[Opalg]

Find all pairs of polynomials $$p(x)$$ and $$ q(x)$$ with real coefficients for which both equations are satisfied: $$p(x^2+1)=q(x)^2+2x$$ and $$q(x^2+1)=p(x)^2$$. These equations are set for all real $$x$$.

I tried to substitute $$x$$ for $$-x$$ and others numbers like $$-1,1$$ etc. but nothing happened... I need your help

Not sure if this is going to lead anywhere or not.

In the first equation, substitute $x^2+1$ for $x$ to get $p\bigl((x^2+1)^2 + 1\bigr) = \bigl(q(x^2+1)\bigr)^2 + 2(x^2+1).$

Square both sides of the second equation to get $\bigl(q(x^2+1)\bigr)^2 = \bigl(p(x)\bigr)^4.$

Combine those two equations to get $p(x^4 + 2x^2+2) = \bigl(p(x)\bigr)^4 + 2x^2 + 2$.

Now what? Suppose that there is a solution in which $p(x)$ is a linear polynomial, say $p(x) = ax+b.$ Then $$a(x^4 + 2x^2+2) + b = (ax+b)^4 + 2x^2 + 2.$$ Compare coefficients of $x^4$ on both sides to see that $a=1$. Then compare coefficients of $x^3$ to get $b=0.$ That leads to a solution $\boxed{p(x) = x,\ q(x) = x-1}$.

Are there any other solutions? I don't know.