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

The correct option is C $3$

$\frac{2a-1}{b}$
Considering $a$ to be a even integer then
$2a-1$ is odd
$\therefore b$ should be odd
$\frac{2b-1}{a}$
Now if $b$ is odd
$2b-1$ is also odd
But our asumption is that $a$ is even
$\therefore \frac{2b-1}{a}=\frac{\text{odd}}{\text{even}}$ which will not give integer.
So $a$ and $b$ should be odd integer then only the given conditions are satisfied.
Let
$\frac{2a-1}{b}=m\Rightarrow 2a-1=mb\cdots \left(i\right)$
And $\frac{2b-1}{a}=n\Rightarrow b=\frac{na+1}{2}\cdots \left(ii\right)$
Using (ii) in (i),
$2a-1=m\left(\frac{na+1}{2}\right)\Rightarrow (4-mn)a=m+2$
Where $m$ and $n$ both are odd integers.
The minimum values of $a$ and $b$ will
$a=b=1\Rightarrow m=n=1$
So $1\le mn\le 3$
Possible values
$n=1;m=1\Rightarrow a=1=bn=1;m=3\Rightarrow a=5;b=3n=3;m=1\Rightarrow a=3;b=5$
Hence three possible order pairs are possible.