a recursion in roots of the polynomial let $P(x)$ is a polynomial which satisfies property $\psi$ where property $\psi$ is given by

whenever r is a root of $P(x) = 0$ then $r^2 - 4$ is also a root of the given equation.

i) if $P(x)$ is a quadratic polynomial of the form $x^2 + ax + b$ then find all the possible equations which satisfy $\psi$ and has distinct real roots.
ii) if $P(x)$ is a cubic polynomial of the form $x^3 + ax^2+bx+c$ then find all the cubic equations which do not share any root with the equations calculated in part (i)

My attempt
for part (i)
let $r$ is a root of $P(x)$  then $r^2 - 4$ should also be a root of the equation
and as $r^2 -4$ is a root then is $(r^2-4)^2-4$ so atleast two of them should be equal otherwise the $P(x)$ would be the zero polynomial which is a contradiction so
either  $$ r = r^2 - 4 \ or \ r = r = (r^2-4)^2-4 $$ as the third equation leads to complex roots
and again the first is false as the roots should be distinct so $r$ satisfies $(r^2-4)^2-4 = r$
but I am not able to solve further and also not sure that what i have done is correct
I just need the idea to solve the cubic part

by calculator i found four values of r for first part but in the question it was mentioned that there are only 2 polynomials
 A: Part i):
Let $r, s$ be the two roots of $P$.
By assumption, $r^2 - 4$ and $s^2 - 4$ all belong to the set $\{r, s\}$. Thus there are several possibilities:

*

*$r^2 - 4 = r$, $s^2 - 4 = s$.
In this case, $r$ and $s$ are the two roots of $x^2 - x - 4$, and we have $P(x) = x^2 - x - 4$.


*$r^2 - 4 = r$, $s^2 - 4 = r$.
In this case, $r$ is one of the two roots of $x^2 - x - 4$, which are $\frac {1 \pm \sqrt{17}}2$, and since $s^2 = 4 + r = r^2$, we have $s = - r$.
Therefore $P(x) = (x - r)(x + r) = x^2 - r^2 = x^2 - \frac{9 \pm \sqrt{17}}2$.


*$r^2 - 4 = s$, $s^2 - 4 = r$.
In this case, we have $(r^2 - 4)^2 - 4 = r$, which gives $(r^2 - r - 4)(r^2 + r - 3) = 0$.
Since $r\neq s$, we have $r^2 - r - 4 \neq 0$, so $r$ is a root of the polynomial $x^2 + x - 3$.
The same argument shows that $s$ is also a root of $x^2 + x - 3$. Therefore $P(x) = x^2 + x - 3$.

Part ii):
We now have three roots $r, s, t$ (which are a priori not necessarily real numbers).
The number $r^2 - 4$ cannot be equal to $r$, so it is one of $s$ and $t$, let's say $s$.
And $s^2 - 4$ cannot be equal to neither $s$ nor $r$ (otherwise $s$ satisfies $(s^2 - 4)^2 - 4 = s$ and coincides with one of the previous roots), so it must be $t$.
Finally, $t^2 - 4$ cannot be equal to neither $t$ nor $s$, for the same reason as above, so it must be $r$.
Therefore we have $((r^2 - 4)^2 - 4)^2 - 4 = r$. After factorization, we get: $$(r^2 - r - 4) (r^3 - r^2 - 6r + 7)  (r^3 + 2r^2 - 3r - 5) = 0.$$ The first factor contains only previous roots, so we see that $r^3 - r^2 - 6r + 7 = 0$ or $r^3 + 2r^2 - 3r - 5 = 0$.
If $r^3 - r^2 - 6r + 7 = 0$, then $s = r^2 - 4$ also satisfies $s^3 - s^2 - 6s + 7 = 0$, as can be verifies as follows: $$(r^2 - 4)^3 - (r^2 - 4)^2 - 6(r^2 - 4) + 7 = (r^3 - r^2 - 6r + 7) (r^3 + r^2 - 6r - 7) = 0.$$
For the same reason, $t$ also satisfies $t^3 - t^2 - 6t + 7 = 0$. Therefore $r, s, t$ are exactly the three different roots of the polynomial $x^3 - x^2 - 6x + 7$, so that $P(x) = x^3 - x^2 - 6x + 7$.
Similarly, if $r^3 + 2r^2 - 3r - 5 = 0$, then we conclude that $P(x) = x^3 + 2x^2 - 3x - 5$.
