Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation. 
Find all pairs (p,q) of real numbers such that whenever $\alpha$ is a root of $x^2 + px+q=0$ $\alpha^2-2$ is also root of the equation.

My Approach:
I could not find any elegant method that is why I tried applying quadratic formula: the roots will be $\frac{-p±\sqrt{p^2-4q}}{2}$
Now we have 2 cases:

*

*$$\left(\frac{-p+\sqrt{p^2-4q}}{2}\right) =\left (\frac{-p-\sqrt{p^2-4q}}{2}\right)^2 -2$$

*$$\left(\frac{-p-\sqrt{p^2-4q}}{2}\right)=\left(\frac{-p+\sqrt{p^2-4q}}{2}\right)^2-2$$
It's very tedious to solve these. I searched on Wolfram alpha and got these I II
Is there an elegant way to approach this problem?

For each $\alpha \in \mathbb{R}$ we can define a quadratic with zeros $x_1 = \alpha $ and $x_2 = \alpha^2 - 2$ by using Vieta's Theorem, which states $x^2+px+q$ with $-p = x_1 + x_2$ and $q = x_1 x_2$ has the desired zeros $x_1,x_2$. Hence for a given $\alpha$ the quadratic you are looking for is $x^2 - (\alpha^2 + \alpha - 2)x + (\alpha^3 - 2\alpha)$.
Edit: In order to find solutions s.t. it works indepently from the choice of the root, which we call $\alpha$, the both roots must satisfy $x_1^2 -2 = x_2$ and $x_2^2-2 = x_1$. How to go on from there is already shown in the answer by markvs.
