# An algebraic proof of Fermat's conjecture

1. Jan 30, 2012

### suchness

I'm sure there are mistakes here so criticism is welcome. In the unlikely event someone would like to sponsor my article for ArXiv, please let me know.

http://www.non-ducor.com/fj/Algebraic_proof_of_Fermats_conjecture_001.png [Broken]
http://www.non-ducor.com/fj/Algebraic_proof_of_Fermats_conjecture_002.png [Broken]

Last edited by a moderator: May 5, 2017
2. Jan 31, 2012

### dodo

Hi, suchness,
you have made a clean exposition of your argument, that's nice. I do have a question.

The heart of the matter seems to be that, when n>2, it is presumed that$${n \choose k} \frac {y^k + (-1)^k c^k} {b^k}$$cannot be an integer. I'm not sure I follow your reason here. Even if you assume y,c,b to be pairwise coprime, $n \choose k$ has no obligation, I think, to be coprime to b. And, for a large n and small k, $n \choose k$ could be larger than b^k. Actually, even if all fractions in the sum are proper fractions, why can't they end up adding to an integer? I am probably missing something, though.

3. Jan 31, 2012

### suchness

I'm not a mathematician so I probably won't understand your question, but I'll give it my best shot.

The remaining series together with the value of n would have to resolve to a value of either 0 for when n is an odd number and -2 when n is an even number.

When n is odd and equal to 3 or greater, the series should resolve to a positive number and therefore prevent -n plus the series being equal to 0.

When n is an even number greater than 2, the series will also resolve to a positive number which together with -n with be greater than -2.

I think this can be shown by substituting c + (-b) for y and showing that even if c^k/b^k is negative it will be canceled out by the first term of the binomial expansion of (c + (-b))^k. Then the remaining terms of (-b)^k plus the remainders of that binomial expansion multiplied by the value of the factorial terms would end in the results noted above.

It occurred to me to explicitly proof this but I wasn't sure it was necessary. If you are suggesting that it is, I think you may be right and I will work on that first thing this morning once I get a bit more rest.

Thank you.

4. Jan 31, 2012

### dodo

Given that the argument has been crystal clear up to the end, I think it would help to justify the last statement. It would lessen the burden on the reader and move it to the writer, where it belongs. By all means, go ahead!

5. Jan 31, 2012

### micromass

Staff Emeritus
"The only value of n that satisfies the equation is 2"

6. Jan 31, 2012

### suchness

I had thought that it would be unnecessary, but Dodo had already convinced me of my error. I will be working on this and will post when I either have something useful to add or it becomes obvious to me that I was premature.

7. Jan 31, 2012

### micromass

Staff Emeritus
Furthermore, when you multiply with $b^{-n}$. I believe that you multiplied the second sum incorrectly. It should be

$\sum_{k=1}^{n-1} \binom{n}{k}(-1)^{n-k}b^{-k}c^k$

8. Jan 31, 2012

### Char. Limit

This is nice. :)

I hope it works out!

9. Feb 1, 2012

### morphism

It should be pointed out that all you did, up until the last step, is rearrange the equation. That alone should have made you skeptical about your proof, because certainly any proof will have to offer something more than just a rearrangement of terms.

10. Feb 1, 2012

### suchness

Since I'm not sure at this point if the approach I had attempted will be successful, I'll just say what I had intended that approach to be. The thing that makes the n=2 version of the functions in the generalized version of the Pythagorean theorem unique, is the fact that the second derivative is a constant.

I had hoped to be able to show that was a requirement for the validity of Fermat's conjecture by using binomial expansion to express the conjecture in terms of the difference of differences.

11. Feb 2, 2012

### dodo

That was one part I didn't understand (nor I do yet) from your exposition. A difference of differences would look something like$$((c+2)^n - (c+1)^n) - ((c+1)^n - c^n)$$Maybe I'm being obtuse, but I don't quite see that in your rearrangement of $a^n+b^n = c^n$. Could you elaborate a bit?

P.S. Arrangements like $(c^n - b^n) - (b^n - a^n)$ would not necessarily represent a "difference of differences" in the sense you need, as a,b,c are not necessarily consecutive integers.

Last edited: Feb 2, 2012
12. Feb 2, 2012

### suchness

I don't mean to be terse or impolite, but that was one reason I used the word "attempted" in my previous post. It well may be that, a) the approach I attempted is not suitable to the task; and/or b) I have not implemented the approach properly.

However, with that being said, I think the fact that I have been able to reduce the conjecture to the point of requiring it to produce one of 2 constants is promising, and for that reason I continue to work on this.

13. Feb 2, 2012

### dodo

No offense taken, and I hope I haven't been too harsh. I'm just being curious about a set of ideas that look interesting. I look forward to more!