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.

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 $v=v^2-2$.
