Find the number of positive integer pairs (a,b) such that ${2a-1 \over b} ,{2b-1 \over a}$ are integers
(A)1 (B)2 (C)3 (D) more than 3 
Find the number of positive integer pairs (a,b) such that $${2a-1 \over b} ,{2b-1 \over a}$$ are integers
(A)1
(B)2
(C)3
(D) more than 3

Now I don't really know what the general approach to these type of questions are. All I could deduce is that a,b are odd as the numerator of both expressions in numerator are odd. Also by checking (1,1) (3,5) and (5,3) are a few cases which appear to satisfy it, but how to prove how many are their ??
 A: There are several ways to determine how many solutions exists, with one method being to find what all of the solutions are. This is what I will show here for your particular problem.
Since $a,b$ are positive integers, then $2a - 1$ and $2b - 1$ are odd, positive integers. For $\frac{2a - 1}{b}$ and $\frac{2b - 1}{a}$ to both be integers requires that $a$ and $b$ be odd, with the resulting integers from those fractions also being positive & odd. In particular, you have for some odd, positive integers $i$ and $j$ that
$$\frac{2a - 1}{b} = i \implies 2a - 1 = ib \implies 2a = ib + 1 \tag{1}\label{eq1A}$$
$$\frac{2b - 1}{a} = j \implies 2b - 1 = ja \implies 4b - 2 = j(2a) \tag{2}\label{eq2A}$$
Substituting \eqref{eq1A} into \eqref{eq2A}, rearranging and factoring gives
$$\begin{equation}\begin{aligned}
4b - 2 & = j(ib + 1) \\
4b - jib & = j + 2 \\
(4 - ji)b & = j + 2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $b$ and $j + 2$ are both positive, $4 - ji$ must also be positive. As $i,j$ are odd, positive integers, the only $3$ possibilities for them are $i = j = 1$, $i = 1, j = 3$ and $i = 3, j = 1$. Substituting these into \eqref{eq3A} to get $b$ and then substituting the values into \eqref{eq2A} to get $a$ gives $(a,b) = (1,1)$, $(a,b) = (3, 5)$ and $(a,b) = (5,3)$, respectively. This matches what you already determined to be $3$ solutions, but it also shows there are no others.
A: So $a=\dfrac{\alpha b+1}{2}$  for some integer $\alpha$, putting in $\dfrac{2b-1}{a}=\dfrac{4b-2}{\alpha b+1}=k \in \mathbb{N}$, if $b>0$, and $a>0$ $\alpha\in [1, 3]$ since for $|\alpha|\ge 4$, $\left|\dfrac{4b-2}{\alpha b+1}\right|<1$.  $k$ can be $1$ when $\alpha=3 $; $\alpha=2$ is not possible,  $\alpha=1$ you can get $k=1, 3$.  There are $3$ positives solutions.  The one you found.
