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.
 A: Is this what you want? If an equation $x^2+ax+b=0$ has roots $v,v^2-2$, then by Vieta's formulas, we have
$$v+(v^2-2)=-\frac{a}{1}=-a, v(v^2-2)=\frac{b}{1}=b.$$
Then, we have $(a,b)=(-v^2-v+2,v^3-2v)$.
So, you can get $(a,b)$ if you know the value of $v$.
Edit : OK. The answers for the "new" question is the followings :
Since $f(v)=0\iff v^2=-av-b,$
$$f(v^2-2)=(v^2-2)^2+a(v^2-2)+b=(-av-b-2)^2+a(-av-b-2)+b$$$$=\cdots =v(-a^3+2ab+4a-a^2)+(-a^2b+5b-ab-2a+b^2+4).$$
This has to be zero independently to $v$, so we have
$$-a^3+2ab+4a-a^2=0\ \text{and}\ -a^2b+5b-ab-2a+b^2+4=0$$
$$\iff\ a(a^2+a-2b-4)=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0$$
$$\iff\ "a=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0"\ \text{or}\ "a^2+a-2b-4=0\ \text{and}\ ba^2+(b+2)a-(b+1)(b+4)=0"$$
1) The former leads $(a,b)=(0,-1),(0,-4)$.
2) The latter leads $(a,b)=(-4,4),(-1,-2),(1,-1),(2,1)$. (you can set $b=(a^2+a-4)/2$ in the other equation. Then, you'll get an equation of $a$, so you can get $a$, then set them in $b=(a^2+a-4)/2$.)
Hence, we know that the followings are the necessary conditions :
$$(a,b)=(0,-1),(0,-4),(-4,4),(-1,-2),(1,-1),(2,1).$$
On the other hand, 
1) The $(0,-1)$ case has $v=1$, so this case is sufficient.
2) The $(0,-4)$ case has $v=-2$, so this case is sufficient.
3) The $(-4,4)$ case has $v=2$, so this case is sufficient.
4) The $(-1,-2)$ case does not have any $v$, so this case is not sufficient.
5) The $(1,-1)$ case has $v=\frac{-1\pm \sqrt 5}{2}$, so this case is sufficient.
6) The $(2,1)$ case has $v=-1$, so this case is sufficient.
Hence, we now reach that the answer is either of the followings :
$(a,b,v)=(0,-1,1),(0,-4,-2),(-4,4,2),(1,-1,\frac{-1\pm\sqrt 5}{2}),(2,1,-1)$.
Note that each of the $(-4,4,2),(2,1,-1)$ cases has an relation of $v=v^2-2$.
A: From the comments I understand that you are seeking some kind of relationship between $a$ and $b$ not involving $v$.
The polynomial $x^2+ax+b$ has two roots (at most) (real or complex) of the form 
 $x_{1}=\dfrac{-a- \sqrt{a^2-4b}}{2}$ and $x_{2}=\dfrac{-a+\sqrt{a^2-4b}}{2}$. And we are told that, either
$x_{1}=x_{2}^2-2$ or $x_{2}=x_{1}^2-2$.
I believe both equations lead to some relation like 
$$a^3-2a^2-2a-3ab+b^2+5b+4=0.$$
A: When $v=x+iy$ then $$a=2-v-v^2=2-x-x^2+y^2-iy(1+2x)$$
is real only if (i) $\ x=-{1\over2}$ or (ii) $\ y=0$, i.e. $v$ is in fact real. In case (i) we have
$$b=v(v^2-2)=\left(-{1\over2}+iy\right)\left(-{7\over4}-iy -y^2\right)$$
and therefore
$${\rm Im}(b)=y\left(-{5\over4}-y^2\right)\ .$$
Therefore in  case (i) the coefficient $b$ is real only if $y=0$, which refers us to case (ii).
We know now that the possible values for $a$ and $b$ are
$$a=2-v-v^2,\quad b=v(v^2-2)\qquad(v\in{\mathbb R})\ .\tag{1}$$
We may consider $(1)$ as parametric representation of the admissible pairs $(a,b)$. The following figure shows these pairs for $-2.5\leq v\leq 1.5$.

A: 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.


*

*$S=\{r\}$ 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.)
A: 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.
A: Given two numbers $r_1$ and $r_2$, to find a monic quadratic polynomial with $r_1$ and $r_2$ as roots, simply use
$$(x-r_1)(x-r_2)$$
which expands to
$$x^2-(r_1+r_2)x+r_1r_2$$
so that what the problem calls $a$ is $-(r_1+r_2)$, while $b$ is $r_1r_2$.
You just plug in $r_1 = v$ and $r_2=v^2-2$.

Edit: I interpreted the question so that $v$ was given, and all you needed to do was find $a$ and $b$ from the given $v$. There is another interpretation where it is $a$ and $b$ that are given and you need to find a criterion on $(a,b)$ necessary and sufficient to conclude that (at least one) $c$ exists. That is more complicated, see other good answers.
