Question: > 
> Find the number of quadratic equations of the form $x^2 + ax + b = 0,$ such that whenever $c$ is a root of the equation, $c^2 - 2$ is also a root of the equation.
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Hello, today I came across this problem on a worksheet, and I wasn't able to solve with the tools that I have learned so far on the subject.

**First attempt:** I let $(x-c)(x-c^2+2)=0$, then I tried to expand, but got something really messy and gave up with this.

**Second attempt:** I used Vieta's Formulas to get $-a=c^2+c-2 \implies a=-c^2-c+2$ and $b=c(c^2-2)=c^3-2c$, but from here I wasn't sure how to make progress after substituting to get $$x^2+(-c^2-c+2)x+c^3-2c=0.$$ Can somebody give me a solution to this problem, as I don't know what to do now? Thanks!

Answer: 
Suppose $c$ is a root. Then $c^2 - 2$ is a root. But then $(c^2 - 2)^2 - 2 = c^4 - 4c^2 + 2$ is a root as well (because $c^2 - 2$ is a root).

But there are only two roots! So either $c = c^2 - 2$ or $c = c^4 - 4c^2 + 2$ or $c^2 - 2 = c^4 - 4c^2 + 2$. If you solve all of these equations, you find that either $c = \pm 1$, $c = \pm 2$, $c = -\frac{1+\sqrt{5}}{2}$, $c = -\frac{1-\sqrt{5}}{2}$.

Note that under the map $c \mapsto c^2 - 2$, the solution $-1$ goes to itself, the solution $1$ goes to $-1$, the solution $2$ goes to itself, the solution $-2$ goes to $2$, and the solution $-\frac{1 - \sqrt{5}}{2}$ gets swapped with $-\frac{1+\sqrt{5}}{2}$.

So the possibilities are: $(x+1)^2$, $(x-2)^2$, $(x+1)(x-2)$, $(x-1)(x+1)$, $(x-2)(x+2)$, and $(x+\frac{1+\sqrt{5}}{2})(x+\frac{1-\sqrt{5}}{2})$. So the answer is six.
If $p, q$ are roots, then we must have $p^2-2\in \{p, q\}$:

Now $p^2-2=p \implies (p-2)(p+1)=0 \implies p \in \{2, -1\}$. This raises three distinct possibilities:

Case 1:  

$p^2-2=p, q^2-2=q \implies p, q \in \{2, -1\} \implies (x-2)^2, (x-2)(x+1), (x+1)^2$ are the three possible quadratics in this case.

Case 2: $p \neq q$ and $p^2-2=q, q^2-2=q \in \{2, -1\} \implies p^2-2 \in \{2, -1\} \implies (p, q) \in \{(\pm2, 2), (\pm1, -1)\} \implies (x-1)(x+1), (x+2)(x-2)$ are the only new ones belonging to this case.

Case 3: $p \neq q$ and $p^2-2=q, q^2-2=p \implies p^2-q^2=-(p-q) \implies p+q=-1 \implies p^2-2=-p-1 \implies p = \phi, \phi-1$.

So $6$ quadratics in all.