Puzzle: The number of quadratic equations which are unchanged by squaring their roots is My friend asked me this puzzle:

The number of quadratic equations which are unchanged by squaring their roots is

My answer is: 3
$x^2-(\alpha+\beta)x +\alpha\beta = 0$
where $\alpha$ and $\beta$ be the roots.
case 1: 
$\alpha$ = 0
$\beta$ = 0
case 2: $\alpha$ = 1
$\beta$ =1
case 3:$\alpha$ = 0
$\beta$ = 1
But my friend says answer is 4. How come?
 A: Your approach is good; you've just overlooked a solution. As you implied, we must have
$$\alpha^2+\beta^2=\alpha+\beta$$
and
$$\alpha^2\beta^2=\alpha\beta\;.$$
The second equation is fulfilled if at least one root is $0$; those are your cases $1$ and $3$. It's also fulfilled if $\alpha\beta=1$. Substituting that into the first equation yields
$$\alpha^2+\frac1{\alpha^2}=\alpha+\frac1\alpha\;,\\
\alpha^4-\alpha^3-\alpha+1=0\;.$$
One solution is $\alpha=1$; that corresponds to your case $2$. Dividing through by $\alpha-1$ yields
$$\alpha^3-1=0\;.$$
This again has $\alpha=1$ as a solution, but also $\alpha=\exp(2\pi\mathrm i/3)$ and $\alpha=\exp(2\pi\mathrm i/3)^2$. These two together yield the fourth solution you were missing.
A: How about just writing: $(x-\alpha)(x-\beta)=(x-\alpha^2)(x-\beta^2)$, so either $\alpha = \alpha^2$ and $\beta=\beta^2$ or $\alpha=\beta^2$ and $\beta=\alpha^2$.
If $\alpha=\alpha^2$, then $\alpha=0\text{ or }1$.  Similarly, $\beta=0\text{ or } 1$.  So there are three such equations (because $(\alpha,\beta)=(0,1)$ and $(\alpha,\beta)=(1,0)$ yield the same quadratic.)
On the other hand, if $\alpha = \beta^2$ and $\beta=\alpha^2$, then $\alpha^4=\alpha$. If $\alpha=0\text{ or } 1$, then $\beta=\alpha$ and we already covered those quadratics above.  So assume $\alpha\neq 0,1$.  Then $0=\alpha^2+\alpha +1 = \frac{\alpha^4-\alpha}{\alpha^2-\alpha}$.  But then, $\beta=\alpha^2$ is also a root of $x^2+x+1$, so that's one last quadratic.  ($\beta = \alpha^2 = -(\alpha+1)$ so $\beta^2 + \beta + 1 = \alpha^2+2\alpha+1 - (\alpha+1) + 1 = \alpha^2 + \alpha +1 = 0$.)
So you get four quadratics, $x^2$, $x^2-x$, $x^2-2x+1$, and $x^2+x+1$.
So the total is $4$.
