Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root Find $(a,b)$ such that in $x^2+ax+b$, whenever  $v$ is a root, then $v^2 - 2$ is also a root
$a,b$ are real numbers. Roots may or may not be real.
In this question, the aim is to find values of and b ,eg. 2,4.

In the question as now formulated ("whenever $v$ is a root, then $v^2−2$ is also a root"), the quadratic polynomial serves only to define its set $S$ of roots (one or two of them), so one might as well reason for$~S$. It must be closed under the map $x\mapsto x^2-2$; this can be achieved in several ways.


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*$S=\{r\}$ where $r^2-2=r$. Then either $r=-1$ or $r=2$, and since $x+ax+b=(x-r)^2$ in this case, one gets $(a,b)=(2,1)$ respectively $(a,b)=(-4,4)$.

*$S=\{r,s\}$ where $r^2-2=r=s^2-2$ and $r\neq s$. Now $r$ has the same two possibilities as above, and $s=-r$. Since $x+ax+b=(x-r)(x-s)=x^2-r^2$ in this case, one gets $(a,b)=(0,1)$ respectively $(a,b)=(0,4)$.

*$S=\{r,s\}$ where $r^2-2=s\neq r=s^2-2$. Now $r$ has to satisfy $r=(r^2-2)^2-2$, which gives the $4$th degree equation $r^4-4r^2-r+1=0$. This looks harder, but we know that the two solutions from 1. will give (unwanted) solutions to this equation, so we can divide $x^4-4x^2-x+1$ without remainder by the polynomial $x^2-x-2$ from 1., giving as quotient the polynomial $x^2+x-1$ for which $r$ has to be a root. Since the same argument holds for $s$, it is clear that $s$ must be the other root, and since $x^2+ax+b=(x-r)(x-s)=x^2+x-1$ in this case one finds a single solution $(a,b)=(1,-1)$ here, even without solving the quadratic equation.
Thus one finds exactly $5$ possible pairs for $(a,b)$, namely $\{(2,1),(-4,4),(0,1),(0,4),(1,-1)\}$.
(Solving the final quadratic in 3., while unnecessary, is not hard: $r=-\frac12-\frac{\sqrt5}2$ and $s=-\frac12+\frac{\sqrt5}2$ is one solution, the negative golden ratio and its conjugate; one can also find the two interchanged.)
