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Q.
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
KVPYKVPY 2019
Solution:
For positive integers ‘$a$’ and ‘$b$’ the numbers $\frac{2a-1}{b}$ and $\frac{2b-1}{a}$ are integers if ‘$a$’ and ‘$b$’ and odd integers because ‘$2a − 1$’ and ‘$2b − 1$’ are odd integers
Now, let $\frac{2a-1}{b}=\alpha$ and $\frac{2b-1}{a}=\beta$,
where $\alpha, \beta$ are integers
then $2a-1=ab$ and $2b-1=\beta\,a$
so $4a-2=\alpha(\beta\,\alpha+1)$
$\Rightarrow a=\frac{\alpha+2}{4-\alpha\, \beta}$
$\because$ a is an integer, then $0 <\, \alpha\, \beta<\, 4$
$\therefore $ Possible value of $\alpha=1, 2$ or $3$ and $\beta=1, 2$ or $3$
such that $0 <\, \alpha\, \beta<\,4$
Now, when $(\alpha, \beta)=(1, 1)$, then $(a, b)=(1, 1)$
When $(\alpha, \beta)=(1, 2)$, then $(a, b)$ have no value
When $(\alpha, \beta)=(1,3)$, then $(a, b)=(3,5)$
and similarly when $(\alpha, \beta)=(3, 1)$, then $(a, b)=(5, 3)$
So, number of ordered pairs $(a, b)$ is $3$