Solutions:

Case – 1 \( \alpha = \beta \)
\( a = \alpha^{2} - 2 \)
\( \alpha^{2} - \alpha - 2 = 0 \)
\( \alpha = 2 \) or \( -1 \)
Now \( \alpha + \alpha = -a \)
\( 2\alpha = -a \)
\( a = -2\alpha \)
\( a = -4 \) or \( a = 2 \)
\( \alpha \cdot \alpha = b \)
\( \alpha^{2} = b \)
\( b = 4 \) or \( b = 1 \)
\( (a,b) = (-4,4) \) or \( (2,1) \)
Case – 2 \( \alpha \ne \beta \)
(i) \( \alpha = \alpha^{2} - 2 \) and \( \beta = \beta^{2} - 2 \)
\( \alpha, \beta = 2, -1 \)
\( \alpha + \beta = -a \) and \( \alpha \beta = b \)
\( a = -1 \quad b = -2 \)
\( (a,b) = (-1,-2) \)
(ii) \( \alpha = \beta^{2} - 2 \) and \( \beta = \alpha^{2} - 2 \)
\( \alpha - \beta = (\beta - \alpha)(\beta + \alpha) \)
\( \Rightarrow \ \beta + \alpha = -1 \)
\( -a = -1 \)
\( a = 1 \)
\( \alpha + \beta = (a + \beta)^{2} - 2\alpha \beta - 4 \)
\( -1 = -3 - 2\alpha \beta \)
\( \Rightarrow \ \alpha \beta = -1 = b \)
\( b = -1 \)
\( (a,b) = (1,-1) \)
(iii) \( \alpha = \alpha^{2} - 2 = \beta^{2} - 2 \)
\( \alpha = -\beta \)
and \( \alpha = 2 \) or \( -1 \)
\( \beta = -2 \) or \( 1 \)
\( \alpha + \beta = -a \Rightarrow a = 0 \)
\( \alpha \beta = b = -4 \) or \( -1 \)
\( (a,b) = (0,-4) \) or \( (0,-1) \)
\( \therefore \) \( 6 \) pairs of \( (a,b) \)

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