Syracuse (or Collatz) conjecture

[Sami1]

We define the following z function:

[math]z=(2^x-2^y)/(2^x-3^y )[/math]
x and y are two positive integers.
Show that z can only be integer and positive in one case when x = 2 and y = 1.

 

[Steven G]

Nice problem.
What progress have you made in proving this? We are a help forum and as such we do not solve problems for students. Instead, we give leading hints so the student can solve the problem on their own.

Hint: z=(2^x−2^y)/(2^x−3^y) =(2^x−3^y + 3^y-2^y)/(2^x−3^y) = (2^x−3^y) /(2^x−3^y) + (3^y-2^y)/(2^x−3^y) =....

 

[Sami1]

I am a researcher and not a student and maybe I posed the problem in an inappropriate forum.
For some time now, I have been trying to solve the Syracuse (or Collatz) conjecture which is apparently very simple but has remained unsolved since 1928.
In my calculation steps, I arrived at this form of z which also appears simple. But after several attempts I realized that showing that z is not integer might be difficult than the Syracuse conjecture itself.
So I thought about posting this problem and whoever can solve it, would have been of great help in solving a problem that has not been resolved for nearly a century.

 

[BigBeachBanana]

Not sure this will lead somewhere but from inspection, real roots exist when [imath]x \neq 0 \land y=x[/imath]