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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?

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2 Answers 2

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.

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One can also arrive at the fourth pair in a less systematic fashion by noting that any solution to $\alpha^2 = \beta$ and $\beta^2 = \alpha$ would also solve the equation. Joriki's answer, however, also tells you that there isn't more than 4 quadratic polynomials with this property. –  Willie Wong Feb 26 '12 at 17:26
    
Because they are roots of unity and Euler's formula is nice for those. Alternatively, some like me may appreciate the puzzle better when the answer is stated equivalently as $\alpha = i\sqrt{3}/2-1/2$ and $\alpha = -i\sqrt{3}/2-1/2$. –  minopret Feb 26 '12 at 17:26

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 $4$.

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