In my opinion, if the statement is true (I feel that it probably is), then it must be quite hard to prove.
It would follow directly from Beal's conjecture but unfortunately it is not proven.
Hi,
I wonder whether somebody can advise me how to solve in natural numbers (or integers) the following equation:
I have spent a couple of weeks thinking about it without any considerable progress. The obvious guess is that the only solution is (x,y,z)=(2,2,2). I've looked at the equation modulo different numbers 3,4,5,9,16,25 ... etc, but so far that wasn't very conclusive in proving that (2,2,2) is the only solution. I've tried a lot of other different things with the equation with the same negative result. Honestly I've run out of ideas of how to solve the equation.
Any help is appreciated.
In my opinion, if the statement is true (I feel that it probably is), then it must be quite hard to prove.
It would follow directly from Beal's conjecture but unfortunately it is not proven.
I done it in Haskell: http://en.wikipedia.org/wiki/Haskell...mming_language)
Dear segasai,
I have attached a solution that I have achieved for this problem. Please go through it and reply me soon about your views regarding the solution. If you have any problem regarding the attachment please don't hesitate to reply me. I value your reply as I enjoyed solving the problem. I have corrected this attachments they are in the order page 2, page 3, page 1, page 4
Hope this helps.
Hi everyone,
Sorry if I confused you. (0,0,0) is not a solution. Shame on me. And thank you very much Shanks for telling me that. But could anyone tell me whether my idea is correct or not neglecting this error. I have corrected the attachments.
Thanks again Shanks.
Given that we know that .
Then and . Therefore and we can write .
We see that both factors can't be a power of 3 since there sum isn't divisible by 3. Therefore we've the following cases:
1) and
2) and
For the first case, using mod3 on the equality implies that must be even and must be odd. This is true since .
For we see that (using that a is odd) but clearly so there is no solution for
We know that is even so we need to consider and .
For y=0 we obtain which clearly has no solutions (use mod4).
For y=2 we obtain and this implies that which is and a solution is therefore
For the second case, clearly can't be true so there are no new solutions.
Therefore is the only solution, as expected.
If we add the possibilty for x,y,z to be 0 we've to consider
This gives us
It's easy to see that is a solution.
We have to consider
z must be odd since and
But using mod8 we see that and so clearly z must be even.
We've reached a contradiction so there are no more solutions than (2,2,2) and (0,1,1) for
Hi ... I've been disturbed by the proof by Sudharaka, because it somehow excludes the solution (0,1,1).
So, I've been looking into it, and I spotted what I think is a problem.
, , and are integers, so when you let , and , then , , and can be negative (but not less than -2).
This means that , , and are not necessarily integers, they may be fractions ... but what you have done is limited them to integer solutions ... I guess this means that your contradiction does not mean there are no other solutions to the given problem, but that can't be an integer ....
Defunkt, I think you're thinking about it the wrong way ... indeed, what he has shown is that the solutions for cannot be of that form, hence no other solutions exist