Equations

Let's try to make things simple. Let $\alpha, \beta$ be the roots of the equation $y=ax^{2}+bx+c$ . The problem requires that $\alpha^2, \beta^2$ are also the roots of the same equation. This is possible if we have the following:

$\alpha+\beta=\alpha^2+\beta^2$

$\implies\left(\alpha+\beta\right)^2-2\alpha\beta=\alpha+\beta\;\;-\left(i\right)$ and

$\alpha\beta=\alpha^2\beta^2\;\;-\left(ii\right)$

From $\left(ii\right)$ we have, $\alpha\beta\left(\alpha\beta-1\right)=0$ . This gives rise to the following cases:

$\underline{\mathbf{CASE\,I}}:$

$\alpha=0\implies\beta=0,1$ . So the equations are:

$y=\begin{cases} x^2-x+0\;\;\;\;& \quad \text{when}\;\alpha=\beta=0\\ x^2+0\cdot\,x+0\;\;& \quad \text{when}\;\alpha=0,\,\beta=1\\ \end{cases}$

$\underline{\mathbf{CASE\,II}}:$

$\beta=0\implies\alpha=0,1$ . This yields same equations as CASE I .

$\underline{\mathbf{CASE\,III}}:$

$\alpha\beta=1$ . Substituting this in $\left(i\right)$ we get

$\left(\alpha+\beta\right)^2-\left(\alpha+\beta\right)-2=0$

$\implies\;\alpha+\beta=-1,2$

So the equations are:

$y=\begin{cases} x^2+x+1\;\;\;& \quad \text{when}\;\alpha+\beta=-1,\,\alpha\beta=1\\ x^2-2x+1\;\;\;& \quad \text{when}\;\alpha+\beta=2,\,\alpha\beta=1\\ \end{cases}$

So we have a total of $\boxed{4}$ different quadratic equations of the form $y=ax^2+bx+c$ with the given conditions.