Question 1: Find all values of $a,b$ such that roots of $x^2+ax+b=0$ be of the type $(p,p^2)$.
There can be infinitely many pairs of $(a,b)$ such as $a=-(p+p^2),b=p^3$. One may also eliminate $p$ from these two equations: $b^{2/3}+b^{1/3}=-a$ cubing both sides we have $b^2+b-3ab=-a^3$. Next we can relate $b$ to $a$ as $$b=\frac{3a-1\pm\sqrt{(1-3a)^2+4a^3}}{2}$$ to get infinitely many quadratic equations, ordinarily.
The standard and the fate of this equation changes if we ask:
Question 2:Find all pairs of $a$ and $b$ so that if $p$ is a root of $x^2+ax+b=0$ then $p^2$ is also its root.
We can assume that that $p,p^2$ are the roots. When $p^2$ is a root, so $p^4$ should also be a root. Since a quadratic cannot have three roots two of $(p,p^2,p^4)$ need to be equal to determine the allowed values of $p$, then the roots will be $(p,p^2)$.
Case 1: $p=p^2\implies p=0,1$ so roots are $(0,0),(1,1)$ and then $a=0,b=0$; $a=-2,b=1.$
Case 2: $p^2=p^4 \implies p=0, p=\pm 1$ gives new roots as $(-1,1)$. then $a=0, b=-1$.
Case 3: $p=p^4 \implies p=0,1, w, w^2$ give us new roots as $(w,w^2), (w^2,w^4=w)$, giving us $a=1,b=1$. Here $w$ is cube root of unity.
In all there exist only 4 such quadratic equations: $x^2=0, x^2-1=0, x^2-2x+1=0, x^2+x+1=0.$
What could be other rephrasing/solutions of the question 2, stated above?
This question can be re-stated as:
Find all $(a,b)$ such that $x^2+ax+b=0$ is invariant if the roots are squared. $y=x^2, x=\sqrt{y}$the the quadratic changes to $$y+a\pm \sqrt{y}+b=0 \implies (y+b)^=a^2y \implies y^2+(2b-a^2)+b^2=0$$ This quadratic should be identical to the given quadratic. Comparing the coeffcients we get $$ \frac{1}{1}=\frac{(2b-a)}{a}=\frac{b^2}{b}.$$
From (i) and (ii), we get $b^2-b\implies b=0,1$
From (i) and (ii) we get $a^2+a-2b=0$
When $b=0$, we get $a=0,-1$
When $b=1$, we get $a=-2,a=1$
Combining we recover four sets of $(a,b)=(0,0),(-1,0),(-2,1),(1,1).$