1. ## Solve the equation

Solve the equation for integers:
$2^{\frac {x-y}y}-\frac 32y=1$.

2. by guessing.....................lol'
x=4,y=2
want to get an integer, so y should be multiple of 2, so i like 2.....

3. Hello, james_bond!

Solve the equation for integers: . $2^{\frac {x-y}y}-\frac 32y\;=\;1$
We have: . $2^{\frac{x-y}{y}} \;=\;\frac{3}{2}y + 1\;\;{\color{blue}[1]}$

Since the right side is rational, the exponent must be an integer.
. . Hence, the right side is a power-of-2.

Then: . $\frac{3}{2}y + 1 \:=\:2^k$ for some positive integer $k.$
. . And: . $y \:=\:\frac{2}{3}\left(2^k-1\right)$

So $2^k-1$ must be divisible by 3.
. . This happens when $k$ is even: . $k \,=\,2n$

. . Hence, we have: . $\boxed{y \;=\;\frac{2}{3}\left(2^{2n}-1\right)}\;\;{\color{blue}[2]}$

Then [1] becomes: . $2^{\frac{x-y}{y}} \;=\;2^{2n} \quad\Rightarrow\quad \frac{x-y}{y} \:=\:2n$

. . and we have: . $x \:=\:y(2n+1)\;\;{\color{blue}[3]}$

Substitute [2] into [3]: . $\boxed{x \;=\;\frac{2}{3}\left(2^{2n}-1\right)(2n+1)}$

Solution: . $\begin{Bmatrix}x &=&\frac{2}{3}\left(2^{2n}-1\right)(2n+1) \\ \\[-3mm] y &=& \frac{2}{3}\left(2^{2n}-1\right) \end{Bmatrix}\quad\text{ for }n \in I^+$

The first few solutions are: . $(6,2),\;(50,10),\;(294,42),\;(1530,170),\;\hdots$