Find the number of the pairs $(p,q)$ such that the quadratic equation with roots $(\alpha)^2,(\beta)^2$ is still $x^2 - px + q = 0$ 
$\alpha,\beta$ are the real roots of the equation $x^2 - px + q = 0$ . Find the number of the pairs $(p,q)$ such that the quadratic equation with roots $\alpha^2,\beta^2$ is still $x^2 - px + q = 0$ .

I want to verify if my working for this solution is correct or not .
My Attempt :- From Vieta's Formula we get that :-
$$\alpha + \beta = \alpha^2 + \beta^2$$
$$ \alpha\beta = (\alpha\beta)^2$$
So from the $2$nd equation we get $\alpha\beta$ $= 1$ , but I don't know how to use it here .
My Approach :- We get in $1$st equation :- from $(\alpha\beta = 1)$,
$$\alpha + \beta + 2 = (\alpha + \beta)^2$$
$$2 = (\alpha + \beta)^2 - (\alpha + \beta)$$
$$2 = (\alpha + \beta)(\alpha + \beta - 1)$$
From here I can see that both the numbers are consecutive . But I can't say that rigorously as $\alpha,\beta$ are real numbers .
Can anyone help after this step? Also I only tried using $\alpha$ and $\beta$ , where the question asks for different $(p,q)$ , which I have no idea how to find them .
 A: If $p$ and $q$ are the same for the two sets of roots, then the sets of roots must be the same. So we need $\alpha=\alpha^2$ and $\beta=\beta^2$ or $\alpha=\beta^2$ and $\beta=\alpha^2$. For the first two equations we have $\alpha,\beta\in \{0,1\}$, and we can get various $p$ and $q$ that way. For the second, we have $\alpha=\alpha^4$ and $\beta=\beta^4$, which have the same real roots.
Thus, for $\alpha=0$ and $\beta=0$ we have $p=q=0$, for $\alpha=1$ and $\beta=0$ we have $p=-1$, $q=0$, and for $\alpha=1$ and $\beta=1$ we have $p=-2$ and $q=1$. The number of solutions is $3$.
A: This is a strange question.  As $x^2 -px +q=0$ will have at must two solutions isn't this a matter that $\{\alpha^2, \beta^2\} =\{\alpha, \beta\}$ which means either
1: $\alpha^2 = \alpha$ and $\beta^2 = \beta$ or
2: $\alpha^2 = \beta$ and $\beta^2 = \alpha$
?
If 1: we have three options
a) $\alpha = \beta = 0$ and $x^2 - px+q = (x-0)(x-0)=x^2$ and $p = q=0$.
b) $\alpha = \beta = 1$ and $x^2 - px + q = (x-1)(x-1) = x^2 + 2x +1$ and $p=-2$ and $p = 1$.
c) $\alpha|\beta = 0$ and $\beta |\alpha = 1$ and $x^2 - px + q = (x-0)(x-1)=x^2 -x$ and $p =-1$ and $q=0$.
If 2: we have $\beta= \beta^4=\alpha^2; \alpha = \alpha^4=\beta^2$ so $\alpha=\beta =0$ or $\alpha = \beta = 1$ so a) or b) above.
so it that the question? Or do I not understand.
A: Here is a solution without calculating $\alpha$ and $\beta$ explicitly:
You have $q=\alpha\beta=\alpha^2\beta^2$. That means $$\alpha\beta(\alpha\beta-1)=0$$
The solutions are $$q=\alpha\beta\in\{0,1\}$$
The other equation is $p=\alpha+\beta=\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$. We can rewrite this as
$$(\alpha+\beta)^2-(\alpha+\beta)-2\alpha\beta=0$$
Using $p$ and $q$:$$p^2-p-2q=0$$
For $q=0$ you get $p=0$ or $p=1$. For $q=1$ you get $$p=\frac{1\pm\sqrt{1+8}}{2}=\frac{1\pm3}{2}$$
So $p=-1$ or $p=2$.
Now let's check:

*

*$q=0, p=0$ yields $x^2=0$, with $x_{1,2}=0$. $0^2=0$

*$q=0, p=1$ yields $x^2-x=0$, with $x=0$ and $x=1$. $0^2=0$ and $1^2=1$

*$q=1, p=2$ yields $x^2-2x+1=0$, with $x_{1,2}=1$. $1^2=1$

*$q=1, p=-1$ yields $x^2+x+1=0$. These are the complex solutions for $x^3-1=0$. With $$x_{1,2}=-\frac 12\pm i\frac{\sqrt 3}2,$$you can easily verify that $x_1^2=x_2$ and $x_2^2=x_1$.

In conclusion, the number of solutions is 3, since the last option does not give real solutions.
