Question

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

• A. 1
• B. 2
• C. 3
• D. more than $3$

Answer 1

Answer: (C)

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 $3$