The polynomial $Q(x)=x^3-21x+35$ has three distinct real roots $r,s,t$.
Find reals $a,b$ so that $P(x)=x^2+ax+b$ satisfies $P(r)=s,P(s)=t,P(t)=r$ or $P(r)=t,P(t)=s,P(s)=r$.
I tried using cardano to find the actual roots and then doing lagrange interpolation, to find both polynomials and then checking which one is monic. But Cardano gets really messy since we get into complex numbers. This is from a math contest so I suppose there is a simpler solution.
I guess we can try using Vietta with the polynomial $Q(P(x))$, noticing it must have $6$ distinct real roots, which contain $r,s,t$ and are symmetric about the vertex of $P(x)$.
By Vieta's formulas, $$r+s+t=0,\quad rs+st+tr=-21,\quad rst=-35\tag1$$
Now, we have $$r+s+t=(r^2+s^2+t^2)+a(r+s+t)+3b\implies b=-14\tag2$$
By the way, as you noticed, noting that $y=P(x)$ is a parabola whose axis of symmetry is $x=-a/2$, we know that $$\small\begin{align}&Q(P(x))\\&=(x^2+ax+b)^3-21(x^2+ax+b)+35\\&=x^6+3ax^5+(3a^2+3b)x^4+(a^3+6ab)x^3+(3a^2b+3b^2-21)x^2+(3ab^2-21a)x+b^3-21b+35\end{align}$$ has six roots $r,s,t, -a-r,-a-s,-a-t$, and so by Vieta's formulas, we have $$-(3ab^2-21a)=-a^3(rs+st+tr)-a^2(r^2s+rs^2+st^2+s^2t+tr^2+t^2r)-a(r^2s^2+s^2t^2+t^2r^2)$$$$\implies 21a-3a(-14)^2=21a^3-a^2\times 3\times 35-a(-21)^2\implies a=0,2,3$$ Finally, checking if each of $(a,b)=(0,-14),(2,-14),(3,-14)$ is sufficient gives $$\color{red}{(a,b)=(2,-14)}$$