5. Does there exist a pair of real numbers $(a, b)$ such that "if $\alpha, \beta$ are the two real roots of the quadratic equation $a x^{2}+b x+b-2=0$, then $\alpha^{2}, \beta^{2}$ are also the two real roots of $a x^{2}+b x+b-2=0$"? If it exists, how many such pairs are there? The correct conclusion is ( ).
(A) Such $(a, b)$ does not exist
(B) Such $(a, b)$ exists and there is only one pair
(C) Such $(a, b)$ exists and there are exactly two pairs
(D) Such $(a, b)$ exists and there are exactly three pairs
5.C. $\left\{\begin{array}{l}\alpha=0, \\ \beta=0\end{array}\right.$ or $\left\{\begin{array}{l}\alpha=0, \\ \beta=1\end{array}\right.$ or $\left\{\begin{array}{l}\alpha=1, \\ \beta=1\end{array}\right.$ (assuming $\left.\alpha \leqslant \beta\right)$.
(1) When $\alpha=0, \beta=0$, substituting $\alpha=0$ into the equation gives $b=$ 2, the equation becomes $a x^{2}+2 x=0(a \neq 0), \beta=-\frac{2}{a} \neq 0$, which is a contradiction.
(2) When $\alpha=0, \beta=1$, substituting $\alpha=0$ into the equation gives $a x^{2}+2 x=0, \beta=-\frac{2}{a}=1$, then $a=-2, b=2$, which meets the problem's requirements.
(3) When $\alpha=\beta=1$, substituting $\alpha=1$ into the equation gives $a=$ $2-2 b$, the equation becomes $(2-2 b) x^{2}+b x+b-2=0$. Let $\Delta=$ 0, we get $b=\frac{4}{3}$, then $a=-\frac{2}{3}$, which meets the problem's requirements.