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Re: how to solve the integer equation Abs[3^x-2^y]=1

• To: mathgroup at smc.vnet.net
• Subject: [mg103008] Re: [mg102988] how to solve the integer equation Abs[3^x-2^y]=1
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Thu, 3 Sep 2009 19:56:02 -0400 (EDT)
• References: <200909031110.HAA24198@smc.vnet.net>

On 3 Sep 2009, at 13:10, a boy wrote:

> Does the equation |3^x-2^y|=1 give only 4 groups of solution?
> (x,y)= (0,1),
>          (1,1),
>         (1,2),
>          (2,3)
>
> can anyone give any else solution?
> when the two integers x and y become bigger and bigger, is there a
> pair integer (x,y) to give a small value for  |3^x-2^y|? Or else,how
> to prove the equation |3^x-2^y|=1having only 4 groups of integer
> solution?
>


Here is the solution to one half of your problem, showing that the  
only integer solutions of the equation 2^y-3^x == 1 are (0,1) and  
(1,2). The other half of the problem is to show that the only  
solutions of 3^x-2^y==1 are (1,1) and (2,3). The proof should be  
similar "in spirit", but it seems harder so I will leave it to you.

So, consider the equation 2^y-3^x == 1. For x==0, we must have y==1.  
Clearly, we can't have y==0. Suppose both x and y >= 1. Since 2^y ==  
(3-1)^y == (-1)^y mod 3 and 3^x + 1 == 1 mod 3, y must be even. Let y  
= 2a. Then 2^(2a)-1 == 3^x, hence (2^a-1)(2^a+1)==3^x.  This is only  
possible if both factors are powers of 3, i.e. 2a-1==3^u and 2a+1==3^v  
(where u,v>=0). Hence 3^v-3^u == 2. If both u and v >=1 then the left  
hand side is divisible by 3, a contradiction. Therefore v==1 and u==0.  
Since u+v == x, x must be 1, a==1, so y =2. So the only solutions are  
(0,1) and (1,2).

Andrzej Kozlowski

• References:
  • how to solve the integer equation Abs[3^x-2^y]=1
    • From: a boy <a.dozy.boy@gmail.com>