Generalization of FLT

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Fermat's last theorem states that the diophantine equation

[tex] x^n + y^n = z^n [/tex]

does not have solutions if [tex] xyz \neq 0 [/tex] and n > 2. Now and Indian mathematician, Dhananjay P. Mehendal have conjectured a generalization of FLT. His (or her, I'm no good at Indian names) states that the diophantine equation

[tex] \sum_{i=1}^{n} x_i^k = z^k [/tex]

with 1 < n < k does not have a solution if [tex] \prod_{i=1}^{n} x_i \neq 0 [/tex]. This conjecture is true for n=2, becaue this case is Fermat's last theorem which Wiles proved in the mid 90ies. However do you think the conjecture is true for all n, or can you easily find a counter-exampel to the suggested conjecture?
 

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  • #2
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That, as written, can be easily disproved:

[tex] 27^5+84^5+110^5 +133^5 = 144^5 [/tex]

I believe that Euler had at one time conjectured what you mention, but in 1967 Lander and Perkins discovered the above. I think it is found in Hardy and Wright, but I found the above at: http://mathworld.wolfram.com/DiophantineEquation5thPowers.html

Other than 5th powers, I was not aware of any similar cases, but a web search finds that in 1988 Noam Elkies showed:

[tex]2682440^4 + 15365639^4 + 18796760^4 = 20615673^4 [/tex] and it was Euler who made the original conjecture: http://library.thinkquest.org/28049/Euler's conjecture.html

Article adds: Despite the success in finding counter-examples for 4th and 5th powers, solutions of higher powers are still unknown. Was Euler totally wrong, or is there still some truth lying in his conjecture?
 
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  • #3
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Intresting, I didn't think there were any counter-examples. It's intresting if Euler was _almost_ right and no other counter-exampel for the exponent could be found.
 
  • #4
mathwonk
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It seems rash to suggest euler may have been "almost right", when no valid cases of his generalization are known, and the next two possible cases are both known to be false.
 
  • #5
matt grime
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"totally wrong" to "almost right". And who says there's no such thing as grade inflation?
 
  • #6
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Well, the conjecture for 3 is correct: 3^3+4^3+5^3 = 6^3. It might be well to remember that the cases of 4 and 5 were found using a computer search. Something, obviously, not available to Euler.

Actually, a friend told me at Burkeley that on the 5th power case, they were actually looking for 5 fifth powers that was a fifth power, but got a zero in one of the terms! Nobody even suspected the case of 4 fifth powers.
 
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  • #7
mathwonk
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you seem to have a different conjecture in mind robert. by anal;ogy with the exampels above, for three it should be the n=3 case of FLT.
 
  • #8
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mathwonk: for three it should be the n=3 case of FLT.

That is how hedlund states it. Anyway, Euler's conjecture on the website http://library.thinkquest.org/28049/Euler's conjecture.html is quoted as:

"It has seemed to many Geometers that this theorem[Fermat's Last Theorem] may be generalized. Just as there do not exist two cubes whose sum or difference is a cube, it is certain that it is impossible to exhibit three biquadrates whose sum is a biquadrate, but that at least four biquadrates are needed if their sum is to be a biquadrate, although no one has been able up to the present to assign four such biquadrates. In the same manner it would seem to be impossible to exhibit four fifth powers whose sum is a fifth power, and similarly for higher powers."

Yes, you seem correct. This then ignores the case of three cubes being a cube. (I was thinking the conjecture was that it takes N Nth powers to make an Nth power.)
 
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  • #9
mathwonk
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the most common open version of a general FLT type conjecture known to me is the "ABC" conjecture. I think one keeps two terms on the left and one on the right and allows different exponents. probably something out there by andrew granville explains this.
 
  • #11
shmoe
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mathwonk said:
probably something out there by andrew granville explains this.
There was one (co-authored with Thomas J.Tucker) in Notices a few years back:

http://www.ams.org/notices/200210/fea-granville.pdf

You need an AMS-web account to access it.
 
  • #12
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Isn't it confusing to write n>k? Actually n is 133 not 4.
 

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