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(C) Let $\frac{2a-1}{b} = \alpha$ and $\frac{2b-1}{a} = \beta$,where $\alpha, \beta \in \mathbb{Z}^+$.Since $a, b \ge 1$,we have $2a-1 \ge 1$ and $2b-

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$3$

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(C) Let $\frac{2a-1}{b} = \alpha$ and $\frac{2b-1}{a} = \beta$,where $\alpha, \beta \in \mathbb{Z}^+$.Since $a, b \ge 1$,we have $2a-1 \ge 1$ and $2b-1 \ge 1$,so $\alpha, \beta \ge 1$.From the equations,$2a-1 = \alpha b$ and $2b-1 = \beta a$.Substituting $b = \frac{2a-1}{\alpha}$ into the second equation: $2(\frac{2a-1}{\alpha}) - 1 = \beta a$.$4a - 2 - \alpha = \alpha \beta a$,which implies $a(4 - \alpha \beta) = \alpha + 2$.Since $a > 0$ and $\alpha + 2 > 0$,we must have $4 - \alpha \beta > 0$,so $\alpha \beta Possible pairs $(\alpha, \beta)$ are $(1, 1), (1, 2), (1, 3), (2, 1), (3, 1)$.Case $1$: $(\alpha, \beta) = (1, 1) \implies a(4-1) = 1+2 \implies 3a = 3 \implies a = 1$. Then $b = \frac{2(1)-1}{1} = 1$. Pair: $(1, 1)$.Case $2$: $(\alpha, \beta) = (1, 2) \implies a(4-2) = 1+2 \implies 2a = 3$,no integer solution.Case $3$: $(\alpha, \beta) = (1, 3) \implies a(4-3) = 1+2 \implies a = 3$. Then $b = \frac{2(3)-1}{1} = 5$. Pair: $(3, 5)$.Case $4$: $(\alpha, \beta) = (2, 1) \implies a(4-2) = 2+2 \implies 2a = 4 \implies a = 2$. Then $b = \frac{2(2)-1}{2} = 1.5$,not an integer.Case $5$: $(\alpha, \beta) = (3, 1) \implies a(4-3) = 3+2 \implies a = 5$. Then $b = \frac{2(5)-1}{3} = 3$. Pair: $(5, 3)$.The ordered pairs are $(1, 1), (3, 5), (5, 3)$. Total $3$ pairs.

The number of ordered pairs $(a, b)$ of positive integers such that $\frac{2a-1}{b}$ and $\frac{2b-1}{a}$ are both integers is

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