The numbers of pairs

Solution:

Consider the equation $x^{2}+a x+b=0$

If has two roots (not necessarily real $\alpha \& \beta$ )

Either $\alpha=\beta$ or $\alpha \neq \beta$

Case (1) If $\alpha=\beta$, then it is repeated root. Given that $\alpha^{2}-2$ is also a root

So, $\alpha=\alpha^{2}-2 \Rightarrow(\alpha+1)(\alpha-2)=0$

$\Rightarrow \alpha=-1$ or $\alpha=2$

When $\alpha=-1$ then $(a, b)=(2,1)$

$\alpha=2$ then $(a, b)=(-4,4)$

Case $(2)$ If $\alpha \neq \beta$ Then

(I) $\alpha=\alpha^{2}-2$ and $\beta=\beta^{2}-2$

Here $(\alpha, \beta)=(2,-1)$ or $(-1,2)$

Hence $(a, b)=(-(\alpha+\beta), \alpha \beta)$

$=(-1,-2)$

(II) $\alpha=\beta^{2}-2$ and $\beta=\alpha^{2}-2$

Then $\alpha-\beta=\beta^{2}-\alpha^{2}=(\beta-\alpha)(\beta+\alpha)$

Since $\alpha \neq \beta$ we get $\alpha+\beta=\beta^{2}+\alpha^{2}-4$

$\alpha+\beta=(\alpha+\beta)^{2}-2 \alpha \beta-4$

Thus $-1=1-2 \alpha \beta-4$ which implies

$\alpha \beta=-1$ Therefore $(a, b)=(-(\alpha+\beta), \alpha \beta)$

$=(1,-1)$

(III) $\alpha=\alpha^{2}-2=\beta^{2}-2$ and $\alpha \neq \beta$

$\Rightarrow \alpha=-\beta$

Thus $\alpha=2, \beta=-2$

$\alpha=-1, \beta=1$

Therefore $(a, b)=(0,-4) \&(0,-1)$

(IV) $\beta=\alpha^{2}-2=\beta^{2}-2$ and $\alpha \neq \beta$ is same as (III)

Therefore we get 6 pairs of $(a, b)$

Which are $(2,1),(-4,4),(-1,-2),(1,-1)(0,-4)$

Option (1)