2022/072) find positive integers x and y such that $\frac{1}{y} - \frac{1}{y+2} = \frac{1}{3 * 2^x}$

 We have multipying by $x(y+2)$ 

$\frac{2}{y(y+2)}= \frac{1}{3 * 2^x}$

or $ 3 * 2^{x+1} = y(y+2)$

let z = x+1 so

 $ 3 * 2^{x+1} = y(y+2)$

LHS is even so y is even because if y is odd then y(y+2) is odd

3 is a factor of y or y + 2

let us atke the 2 cases one by one

case 1 :3 is a factor of y

so $y = 3 * 2^a$ and $y+2 = 2^b$ for some inetger a and b 

$3 *2^a + 2 = 2^b$

or $2(3 * 2^{a-1} + 1) = 2^{b+1}$

as RHS is even so $3*2^{a-1}$ is odd or  a = 2

so y = $3 *2^a = 6$ and $y+2 =  8$ from this $3 * 2^{x +1} = 48 = 3 * 2^4$ ot x = 3

case 2 :3 is a factor of y +2

so $y+ 1 = 3 * 2^a$ and $y-2 = 2^b$ for some inetger a and b 

$3 *2^a - 2 = 2^b$

or $2(3 * 2^{a-1}-1 = 2^{b+1}$

as RHS is even so $3*2^{a-1}$ is odd or  a = 2

so y = $3 *2^a = 6$ and $y -2 =  4$ from this $3 * 2^{x +1} = 24 = 3 * 2^3$ ot x = 2  

So 2 solutions are (3,6) and (2,4)