Re: how to solve the integer equation Abs[3^x-2^y]=1
- To: mathgroup at smc.vnet.net
- Subject: [mg103034] Re: [mg102988] how to solve the integer equation Abs[3^x-2^y]=1
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Fri, 4 Sep 2009 07:01:28 -0400 (EDT)
- References: <200909031110.HAA24198@smc.vnet.net> <5D1FF5A2-64C3-4153-9C10-3E40549F96C8@mimuw.edu.pl> <c724ed860909032049w210c9b08ted1619c955c41581@mail.gmail.com> <c724ed860909032109k3a1a62e6w5821ce9773a51314@mail.gmail.com>
Unfortunately your argument does not work since you are working modulo
4, which is not a prime. As Bill Rowe pointed out FindInstance shows
that the result is not true for 3x^2-2^y==1 (but it is true for
2^y-3x^2 == 1 as shown by my proof).
I am not even sure if if it is true that 3x^2-2^y==1 has only a
finite number of integer solutions. Gelfond has shown that a^x
+b^y==c^z has only a finite number of solutions, when a,b,c are all
non zero and none of them is a power of 2. However, since you do has a
power of 2 in your equation this result does not apply. It would be
interesting to find a proof one way or another.
A
On 4 Sep 2009, at 06:09, a boy wrote:
>
>
> On Fri, Sep 4, 2009 at 11:49 AM, a boy <a.dozy.boy at gmail.com> wrote:
> As your prove is so superb and clear , now, I can easly prove that
> the only solutions of 3^x-2^y==1 are (1,1) and (2,3).
> for 3^x==(4-1)^x==1+2^y,
> (-1)^x==1+2^y ( mod 4),
> when x=2a+1, it must be y<2; when x=2a, y>1 and (3^a+1)
> (3^a-1)=2^y,so only x=2.
> Proved!
>
> And more,I wonder that
> does it exist two infinite and increasing integer sequence {Xi} and
> {Yi} to
> satisfy {|3^Xi-2^Yi|} progressively decreasing?
> Could you give me Yes or NO? and why?
> Oh, the answer to this is to be No, for 1 is the last element of {|
> 3^Xi-2^Yi|} .
> Now my question is changed to :
> To construct an increasing integer pair sequence {(Xi,Yi)} satisfy
> that
> 1) {Xi} is progressively increasing
> 2) {|3^Xi-2^Yi|} progressively decreasing
> What is L=the maximum length of constructed sequence?
>
> I think it's hard to me,can you give me a good solution?
>
>
> On Fri, Sep 4, 2009 at 2:01 AM, Andrzej Kozlowski
> <akoz at mimuw.edu.pl> wrote:
>
> 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>
- how to solve the integer equation Abs[3^x-2^y]=1