# Fermat's Last theorem

by ron_jay
Tags: fermat, theorem
HW Helper
P: 3,682
 Quote by ron_jay ha...Alrighty. Agreed that the program will not run for an infinite time but if there was a solution to it, the program would have given it in a relatively short period of time.Wouldn't it? So you could conjecture that the equation is unlikely to have a solution.
Sure, you can conjecture that -- and you did, and you were right. There are lots of examples of problems with large smallest counterexamples, though, so matt took issue with your guess. Eh.
P: 83
 If you're generalization to 'all integers' is based merely on some searhc not terminating, rather than examining the reason why it does not terminate quickly, then you've clearly done something rash.
No. It's not rash at all.The search did not terminate for a reason, did it? And it is because according to the theorem there is no solution to an integral power above 2. So, the reason was examined and that is how I came across this ingenious theorem.

One way to go about solving it would be that a power of 4 for example can be expressed in terms of 2*2 or algebraically, 'p=mn'

(a^m)^n + (b^m)^n = (c^n)^n

Perhaps we could somehow use this identity(though I don't know how).
 Sci Advisor HW Helper P: 8,953 Yes, as i said you don't have to check all infinite 'n' just the infinite number of prime 'n'. Since any non-prime 'n' can be expressed as the sum of 2 prime, another tricky one to prove by the way!
HW Helper
P: 9,396
 Quote by ron_jay No. It's not rash at all.

yes it is, since you had no explanation of why it did not terminate,

 The search did not terminate for a reason, did it?
what reason would that be? That you knew something about certain modular forms? Or because you didn't have a powerful enough computer, or didn't let your program run for long enough?

 And it is because according to the theorem there is no solution to an integral power above 2. So, the reason was examined and that is how I came across this ingenious theorem.

that is complete BS. Example? OK, find an integer solution of

42323245673565835789467946805234545765831346256725764x^6+ 1356468356835783234623456724689514123412354513487654794679745x^5+264653 24643257654876925634145232534165436457546879856986986735784698659623453 4x^2+x+1

P: 994
 Quote by ron_jay ha...Alrighty. Agreed that the program will not run for an infinite time but if there was a solution to it, the program would have given it in a relatively short period of time.Wouldn't it? So you could conjecture that the equation is unlikely to have a solution.
Let's say your program stops at 9,223,372,036,854,775,807 (if you're using a signed long variable in java), for n = 3. How do you know the solution isn't at 9,223,372,036,854,775,808?

On the other hand, if you can prove that your program will not terminate, then you can conclude that there are no solutions.
HW Helper
P: 8,953
 Quote by Dragonfall On the other hand, if you can prove that your program will not terminate, then you can conclude that there are no solutions.
Unfortunately the halting problem is even less solved than fermat's last theorem!
 Sci Advisor HW Helper P: 9,396 http://primes.utm.edu/glossary/page.php?sort=LawOfSmall contains a couple of examples of the misleading behaviour of small numbers. I particularly like Skew's number. I remember another one about sums of powers as well, (perhaps the sum of 4 4th powers) that has only exceptionally large counter examples.
 P: 1,059 There was something called the Euler Conjeture: It takes n nth powers to make an nth power. This was largely accepted until it was shown--by virtue of a zero for the fifth left term, I have been told, that 27^5 + 84^5 +110^5 + 133^5 = 144^5. A larger case discovered in 1967 as well is: 85282^5 + 28969^5 +3183^5 + 55^5 = 85359^5. (It is interesting to note that 85359^5 = 4.53..x10^24, large enough to suite me.) Also I am sure people know just as 3^2+4^2= 5^2, we have 3^3+4^3+5^3 = 6^3. Well it is sort of interesting that 4^5 + 5^5 +6^5 +7^5 +11^5 = 12^5.
P: 994
 Quote by mgb_phys Unfortunately the halting problem is even less solved than fermat's last theorem!
He isn't asked to prove that his program will tell whether any other program will terminate. I'm sure there are proofs that uses the fact that an algorithm does not terminate. I can't think of anything non-trivial for the moment, but I remember seeing something like that in many proofs in graph theory.
P: 83
 Quote by matt grime yes it is, since you had no explanation of why it did not terminate,what reason would that be? That you knew something about certain modular forms? Or because you didn't have a powerful enough computer, or didn't let your program run for long enough? that is complete BS. Example? OK, find an integer solution of 42323245673565835789467946805234545765831346256725764x^6+ 1356468356835783234623456724689514123412354513487654794679745x^5+264653 24643257654876925634145232534165436457546879856986986735784698659623453 4x^2+x+1 or show none exists. Please, start your computer program now.....
What you have said is perfectly correct, but the main purpose of my question has been defeated and it is not BS.I did not positively conclude that the anomaly of the program not terminating is caused due to Fermat's Last Theorem at first thought. It was only after learning about the theorem and then ruminating that perhaps, yes, the evidence that the equation does not have a solution as said by Fermat and then proved by Andrew Wiles may have an answer to to it not terminating, even if it is for a short period of time.Yes, I had beforehand knowledge that there may not be a solution to the equation and hence for any great value(9,223,372,036,854,775,807), the program will not return an answer. Unfortunately, lot of misunderstanding has crept in.The program does not prove or validate Fermat's Last theorem(that's were I think we are going wrong),but is in fact a consequence of it.
HW Helper
P: 9,396
From post 1 by ron_jay:

 I recently ran a program to find this solution, but couldn't find for n=3,4,5... This certainly validates the theorem but how do we prove it mathematically?
from post 28 by ron_jay:

 the program does not prove or validate Fermat's Last theorem(that's were I think we are going wrong),but is in fact a consequence of it.
That's where we are going wrong???
P: 83
 Quote by Dragonfall Let's say your program stops at 9,223,372,036,854,775,807 (if you're using a signed long variable in java), for n = 3. How do you know the solution isn't at 9,223,372,036,854,775,808? On the other hand, if you can prove that your program will not terminate, then you can conclude that there are no solutions.
Correct, that if you can prove that the program does not terminate, the theorem will be proved just like the Taniyama-Shimura Conjecture was a precedent to the proof of Fermat's Last theorem.Right, we don't know whether the last possible long digit available to the program is the solution or not and it could be, but all I am doing is laying down a conjecture that if the program had not terminated and we knew that it hadn't in some way, we would arrive at a conclusion and you have rightly voiced my own opinion in every way.

Let's say I name it the "NON-Terminating Program Conjecture".
P: 83
 Quote by matt grime That's where we are going wrong???
Yes, on that post I hadn't elucidated the exact details(how misleading words can be!) and hence we went wrong in not understanding what I meant(my mistake for not expanding the first post) and went on to talk about the program instead of how we could actually prove the theorem mathematically.
 P: 994 I think by "validates" he meant "correlates", in the first post.
 HW Helper P: 4,124 ron_jay was just saying that he conjectured... he made a guess... Then he looked up Fermat's last theorem and his guess turned out right. It was just supposed to be about how he came across FLT... not that he had justified it or validated it. Whether or not the guess is unjustified, the fact is that the guess turned out to be true... Do all guesses need to have some sort of proper justification? It's just a guess after all...
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P: 3,348
 Quote by matt grime find an integer solution of 42323245673565835789467946805234545765831346256725764x^6+ 1356468356835783234623456724689514123412354513487654794679745x^5+264653 24643257654876925634145232534165436457546879856986986735784698659623453 4x^2+x+1 or show none exists.
It seems that is not an equation :(

Sometimes too much mathematical rigor destroys mathematical intuition, which in my opinion is more important. As robert_ihnot pointed out, even Euler has been guilty of the mistake: Not seeing any counter examples, so believes in the theorem for all examples. It is true, these days with the ever more complex maths, large counter examples are becoming more common, and matt grime, being a modern day mathematician, has good reason to be careful of them. He is merely trying to teach others to be the same.
 Sci Advisor HW Helper P: 9,396 Ok, root, then.
 HW Helper P: 3,348 This is getting away from the original point, but x=0 is a solution >.<

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