Question 64

Two monic quadratic polynomials $(x - \alpha)(x - \beta)$ and $(x - \alpha^2)(x - \beta^2)$ are identical if and only if their root multisets are equal: $\{\alpha, \beta\} = \{\alpha^2, \beta^2\}$. There are two cases:

1. Case 1 ($\alpha = \alpha^2$ and $\beta = \beta^2$):
   Since $z = z^2 \iff z \in \{0, 1\}$, both $\alpha$ and $\beta$ belong to $\{0, 1\}$. This yields $3$ distinct polynomials:
   • $\alpha = \beta = 0 \implies x^2$ (equal roots)
   • $\alpha = \beta = 1 \implies (x - 1)^2$ (equal roots)
   • $\{\alpha, \beta\} = \{0, 1\} \implies x(x - 1)$ (distinct roots)
2. Case 2 ($\alpha = \beta^2$ and $\beta = \alpha^2$ with $\alpha \neq \alpha^2$):
   Substituting $\beta = \alpha^2$ into $\alpha = \beta^2$ gives $\alpha^4 = \alpha \implies \alpha(\alpha - 1)(\alpha^2 + \alpha + 1) = 0$. Since $\alpha \notin \{0, 1\}$, $\alpha$ and $\beta = \alpha^2$ must be the two distinct complex cube roots of unity $\omega, \omega^2$, yielding $1$ additional polynomial:
   • $(x - \omega)(x - \omega^2) = x^2 + x + 1$ (distinct roots)

In total, there are $3 + 1 = 4$ square-invariant quadratic polynomials, of which $2$ ($x^2$ and $(x-1)^2$) have equal roots. Thus the probability is $\frac{2}{4} = 0.5$.