<s>[INST] 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.
 [/INST] Continuing toufik's solution, one can state the following :
Theorem Let $P=X^2+aX+b$ where $a$ and $b$ are real. Then $P$
factorizes as $(X-v)(X-(v^2-2))$ for some $v\in{\mathbb C}$ if and only if
$9-4a \geq 0$, and $b=\frac{3a-5\pm(a-1)\sqrt{9-4a}}{2}$.
Proof of theorem. In one direction, suppose that $P$
factorizes as $(X-v)(X-(v^2-2))$ for some $v\in{\mathbb C}$. Then
$X^2+aX+b=(X-v)(X-(v^2-2))$, so 
$$
a=-v^2-v+2, \ b=v^3-2v \tag{1}
$$
We deduce $(v-1)a=-v^3+3v-2$ and hence $b+(v-1)a=v-2$.
So if $a=1$, we must have $b=(-1)$. Otherwise we can write
$$
v=\frac{a-(b+2)}{a-1} \tag{2}
$$
Reinjecting this into (1) we see that
$$
a=-\big(\frac{a-(b+2)}{a-1}\big)^2+\big(\frac{a-(b+2)}{a-1}\big)+2 \tag{3}
$$
Expanding and clearing out denominators in (3), we obtain
$$
b^2 + (-3a + 5)b + (a^3 - 2a^2 - 2a + 4)=0 \tag{4}
$$
The discriminant is 
$$(-3a+5)^2-4(a^3 - 2a^2 - 2a + 4)=-4a^3 + 17a^2 - 22a + 9
=(9-4a)((a-1)^2)
$$
So that
$$
b=\frac{3a-5\pm(a-1)\sqrt{9-4a}}{2} \tag{5}
$$
Note that (5) still holds when $(a,b)=(1,-1)$.
Conversely, suppose that (5) holds, so that
$b=\frac{3a-5+\varepsilon(a-1)\sqrt{9-4a}}{2}$ for some
$\varepsilon=\pm 1$. Inspired by (2), let us put
$$
v=\frac{-1-\varepsilon\sqrt{9-4a}}{2} \tag{6}
$$
Then $v^2=\frac{5-2a-\varepsilon\sqrt{9-4a}}{2}$, so
$$
v^2-2=\frac{1-2a+\varepsilon\sqrt{9-4a}}{2} \tag{7}
$$
Multiplying (6) and (7), we see that
$$
v(v^2-2)=\frac{2a-1-(9-4a)+\varepsilon\sqrt{9-4a}(2a-1-1)}{4}=
\frac{6a-10+\varepsilon\sqrt{9-4a}(2a-2)}{4}=
b \tag{8}
$$
and adding (6) and (7), we see that
$$
v+(v^2-2)=-a \tag{9}
$$
Then (7) and (8) together imply that $X^2+aX+b=(X-v)(X-(v^2-2))$.
This concludes the proof.
</s>