
3. Find all pairs $(a, b)$ of real numbers such that whenever $\alpha$ is a root of $x^{2}+a x+b=0$, $\alpha^{2}-2$ is also a root of the equation.


## Solution:

Consider the equation $x^{2}+a x+b=0$. It has two roots(not necessarily real), say $\alpha$ and $\beta$. Either $\alpha=\beta$ or $\alpha \neq \beta$.

## Case 1:

Suppose $\alpha=\beta$, so that $\alpha$ is a double root. Since $\alpha^{2}-2$ is also a root, the only possibility is $\alpha=\alpha^{2}-2$. This reduces to $(\alpha+1)(\alpha-2)=0$. Hence $\alpha=-1$ or $\alpha=2$. Observe that $a=-2 \alpha$ and $b=\alpha^{2}$. Thus $(a, b)=(2,1)$ or $(-4,4)$.

## Case 2:

Suppose $\alpha \neq \beta$. There are four possibilities; (I) $\alpha=\alpha^{2}-2$ and $\beta=\beta^{2}-2$; (II) $\alpha=\beta^{2}-2$ and $\beta=\alpha^{2}-2$; (III) $\alpha=\alpha^{2}-2=\beta^{2}-2$ and $\alpha \neq \beta$; or (IV) $\beta=\alpha^{2}-2=\beta^{2}-2$ and $\alpha \neq \beta$

(I) Here $(\alpha, \beta)=(2,-1)$ or $(-1,2)$. Hence $(a, b)=(-(\alpha+\beta), \alpha \beta)=(-1,-2)$.

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

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

Since $\alpha \neq \beta$, we get $\beta+\alpha=-1$. However, we also have

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

Thus $-1=1-2 \alpha \beta-4$, which implies that $\alpha \beta=-1$. Therefore $(a, b)=(-(\alpha+$ $\beta), \alpha \beta)=(1,-1)$.

(III) If $\alpha=\alpha^{2}-2=\beta^{2}-2$ and $\alpha \neq \beta$, then $\alpha=-\beta$. Thus $\alpha=2, \beta=-2$ or $\alpha=-1, \beta=1$. In this case $(a, b)=(0,-4)$ and $(0,-1)$.

(IV) Note that $\beta=\alpha^{2}-2=\beta^{2}-2$ and $\alpha \neq \beta$ is identical to (III), so that we get exactly same pairs $(a, b)$.

Thus we get 6 pairs; $(a, b)=(-4,4),(2,1),(-1,-2),(1,-1),(0,-4),(0,-1)$.
