# 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 ??

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.

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.