$\alpha,\beta$ 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 .
Best Answer
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 $3$.