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Topic: --- --- --- FLT example
Replies: 18   Last Post: Feb 6, 2014 11:53 PM

 Messages: [ Previous | Next ]
 gnasher729 Posts: 419 Registered: 10/7/06
Re: --- --- --- FLT example
Posted: Feb 6, 2014 6:20 PM

Just wondering...

Has anyone seriously looked at the question:

For n >= 3, how close can a^n + b^n get to c^n, if a <= b < c ?

With n = 3, there are quite a few combinations where a^3 + b^3 = c^3 +/- 1. There is a parametric solution for a^3 + b^3 = c^3 - 2, with the first example 5^6 + 6^6 = 7^6 - 2. So there are a huge number of near misses.

For n = 4, it took my computer a very short time to find some values where a^4 + b^4 = c^4 when calculated with double precision floating point arithmetic, but the correctly calculated values where about a million apart.

Is there anything known for n >= 4? For example the obvious next question would be: Is it possible that a^n + b^n = c^n +/- 1 if n >= 4 and 1 <= a <= b < c?

Date Subject Author
1/30/14 Deep Deb
1/30/14 quasi
1/30/14 Clark Smith
1/30/14 quasi
1/31/14 Brian Q. Hutchings
2/3/14 Port563
2/3/14 quasi
2/4/14 Port563
2/4/14 quasi
2/4/14 Port563
2/4/14 quasi
2/6/14 gnasher729
2/6/14 gnasher729
2/5/14 quasi
2/5/14 Port563
2/1/14 Dr J R Stockton
2/6/14 gnasher729
2/6/14 Brian Q. Hutchings
2/6/14 THE COLONEL