Elegant proof of fermats last theorem 2. try

  • Thread starter robert80
  • Start date
  • #1
64
0

Main Question or Discussion Point

Sorry guys about yesterday. I did a fundamental mistake in proof and I was so embaressed, that I have decided, I will give it a try for the last time. Here is the proof

Proof:

Let us suppose that a,b,c are coprimes, so if we construct the from a,b,c the smallest triangle for solution of the Fermats Last Theorem.

so lets suppose that the sollution exist, a^n + b^n = c^n lets suppose a,b,c are coprimes

Let us check the problem for odd powers of n. We can write now the equation (c^n + b^n)*(c^n - b^n) = (c^2n - b^2n) so that holds everytime, not specifically for the problem. ---------------> now the Fermats Last theorem is included: if c^n - b^n = a^n than we can easily see that the factors on the left are coprimes. (c^n + b^n) equals 2b^n + a^n and that shure is coprime to a, since a and b are coprimes. So lets rearange the equation for odd powers of n. (c^n + b^n)*(c - b)* (c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c^2 - b^2)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-----------> (c^n + b^n)*(c - b)* (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) = (c - b)*(c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-----------> so we can exclude (c - b) factor from the eqation: (c^n + b^n)*((c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-------------------------------------->so since we can rearange the expresion: c^(n-1) + c^(n-2)*b..............+ b^(n-1) into c^(n-1) + c^(n-2)*b = c^(n-2)*(c+b) first two are devidable by (c+ b) we carry on doing that and we see if we take first two members and than second two and third two, we see that all are devidable by (b + c) BUT since n is odd we have the odd numbers of those members so we come till the last one: and we get somehow: (b + c)*z + b^(n-1) so (c + b) and (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) are coprimes. So the only possibility is now, that c^n + b^n is devidable by (c + b) if the solution exists. Ok lets suppose that c^n + b^n = (c + b)*l where l is the member of natural numbers ---------------->c*l - c^n = b^n -l*b ------------------------>c*(l - c^(n-1) = b*(b^(n-1) -l) -----------------> (l+c^(n-1)) = b*m and (b^(n-1)-l) = c*m since b and c are coprimes.now sum both equations: (l+c^(n-1)) + (b^(n-1)-l) = b*m + c*m, l goes out so: c^(n-1) + (b^(n-1) = m* (c + b) where m is the member of natural numbers. So if we state that c^n + b^n is devidable by (b + c) it follows out of that that c^(n-1) + b^(n-1) is devidable by (b + c). Since (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) has odd number of members when n is odd, we saw that (c+ b ) does not devide that expression, now if we start from the begining and sum 2 by 2 till the last one is out: (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) = (b + c) *z + b^(n-1), now lets turn this procedure around: lets sum all the members till the last one 2 by 2 by the backside so: c*b^(n-2) + b^(n-1) = b^(n-2)*(c + b)..........so we carry on till we come to the c^(n-1) -------------> c^(n-1) + (c + b)*t where t is again number of natural numbers. BUT if we multiply (c^(n-1) + c^(n-2)*b..............+ b^(n-1) by 2 we can sum the equations: c^(n-1) + (c + b)*t + (b + c) *z + b^(n-1) since (b + c) devides c^(n-1) + b^(n-1), we got that if we multiply the expression by 2 (c+ b ) devides 2*(c^(n-1) + c^(n-2)*b..............+ b^(n-1)) since (c + b) is odd -------------------------------> (c+b) devides (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) too. Thats a contradiction (c + b) is coprime to (c^(n-1) + c^(n-2)*b..............+ b^(n-1), so (c^n + b^n) and(c^n - b^n) are coprimes when a,b,c are coprimes -------------------> so (c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1)) is the member of rational numbers. This is again a contradiction, since we know that this expression is the member of natural numbers. So we can not find solutions in whole number sistem of a,b,c for n is odd and > 2.

I want to make better impression, so I got back to the problem for a few hours today.

Sincerelly

Robert Kulovec Mueller, Slovenija
 

Answers and Replies

  • #2
22,097
3,282
It would be a HUGE help for us if you would
1) put some things in tex, just put [tex] and [ /tex] (without space) around your formula...
2) leave some spacing. The wall of text you wrote is very uninviting to read...
 
  • #3
64
0
Hello, I will do that in few days, since these days I have planty of work for the University...

Thanks for all the assistance.

Robert
 
  • #4
64
0
There could be a mistake, or something that is not prooven and it should be, but I believe that general idea about the proof is correct.
 
  • #5
64
0
The following part is not a mistake, it was just writen wrong :

****so we can exclude (c - b) factor from the eqation: (c^n + b^n)*((c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))***

it should be so so we can exclude (c - b) factor from the FORMER equation and we get-------->: (c^n + b^n)*((c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))

Just the funny (--------->) is missing :) Anyway, I will post it in a better form soon.
 
  • #6
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
3,750
99
So if we state that c^n + b^n is devidable by (b + c) it follows out of that that c^(n-1) + b^(n-1) is devidable by (b + c)
If b=2 and c=3, and n=3

2+3=5
23+33 = 35 is divisible by 5
22+32=13 is not divisible by 5
 
  • #7
2,967
5
I took the liberty of performing some formatting to your text. I hope I did not change any of its meaning.

Sorry guys about yesterday. I did a fundamental mistake in proof and I was so embarrassed, that I have decided, I will give it a try for the last time. Here is the proof

Proof:

Let us suppose that [itex]a, b, c[/itex] are coprimes, so if we construct the from [itex]a, b, c[/itex] the smallest triangle for solution of the Fermat's Last Theorem.

So let us suppose that the solution exist, [itex]a^{n} + b^{n} = c^{n}[/itex]; let's suppose [itex]a, b, c[/itex] are coprimes.

Let us check the problem for odd powers of n. We can write now the equation:

[tex]
(c^{n} + b^{n})*(c^n - b^n) = (c^{2n} - b^{2n}) \ (1)
[/tex]

so that holds every time, not specifically for the problem.

---------------> Now the Fermat's Last theorem is included:

If [itex]c^{n} - b^{n} = a^{n}[/itex], then we can easily see that the factors on the left are coprimes. [itex](c^{n} + b^{n})[/itex] equals [itex]2 b^{n} + a^{n}[/itex] and that sure is coprime to a, since a and b are coprimes.

So lets rearrange the equation for odd powers of n.

[tex]
(c^{n} + b^{n})*(c - b)* (c^{n-1} + c^{n-2}*b + \ldots+ b^{n-1}) = (c^{2} - b^{2})*(c^{2(n-1)} + c^{2(n-2)}*b^{2} + \ldots + b^{2(n-1)}) \ (2)
[/tex]

----------->

[tex]
(c^{n} + b^{n})*(c - b)* (c^{n-1} + c^{n-2}*b + \ldots + b^{n-1}) = (c - b)*(c + b)*(c^{2(n-1)} + c^{2(n-2)}*b^{2} + \ldots + b^{2(n-1)}) \ (3)
[/tex]

----------->

So we can exclude a [itex](c - b)[/itex] factor from the equation:

[tex]
(c^{n} + b^{n})*(c^{n-1} + c^{n-2}*b + \ldots + b^{n-1}) = (c + b)*(c^{2(n-1)} + c^{2(n-2)}*b^{2} + \ldots + b^{2(n-1)}) \ (4)
[/tex]

-------------------------------------->

So, since we can rearrange the expression:

[tex]
c^{n-1} + c^{n-2}*b + \ldots+ b^{n-1} \ (5)
[/tex]

into

[tex]
c^{n-1} + c^{n-2}*b = c^{n-2}*(c+b) \ (6)
[/tex]

first two are divisible by [itex](c+ b)[/itex]; we carry on doing that and we see if we take first two members and than the second two and the third two, we see that all are divisibla by [itex](b + c)[/itex], BUT since n is odd we have the odd numbers of those members so we come till the last one: and we get somehow:

[tex]
(b + c)*z + b^(n-1) \ (7)
[/tex]

so [itex](c + b)[/itex] and [itex](c^{n-1} + c^{n-2}*b + \ldots + b^{n-1})[/itex] are coprimes. So the only possibility now is that [itex]c^{n} + b^{n}[/itex] is divisible by [itex](c + b)[/itex] if the solution exists.

Ok lets suppose that:

[tex]
c^{n} + b^{n} = (c + b)*l \ (8)
[/tex]

where l is the member of natural numbers

---------------->

[tex]
c*l - c^n = b^n -l*b \ (9)
[/tex]

------------------------>

[tex]
c*(l - c^{n-1}) = b*(b^{n-1} -l) \ (10)
[/tex]

----------------->

[tex]
(l+c^(n-1)) = b*m \ (11)
[/tex]

and

[tex]
(b^(n-1)-l) = c*m \ (12)
[/tex]

since b and c are coprimes.

Now, sum both equations:

[tex]
(l+c^(n-1)) + (b^(n-1)-l) = b*m + c*m \ (13),
[/tex]

l goes out so:

[tex]
c^(n-1) + (b^(n-1) = m* (c + b) \ (14),
[/tex]

where m is the member of natural numbers. So if we state that [itex]c^{n} + b^{n}[/itex] is divisible by [itex](b + c)[/itex], it follows out of that that [itex]c^{n-1} + b^{n-1}[/itex] is divisible by [itex](b + c)[/itex].

Since [itex](c^{n-1} + c^{n-2}*b + \ldots + b^(n-1)}[/itex] has odd number of members when n is odd, we saw that [itex](c+ b)[/itex] does not divide that expression, now if we start from the beginning and sum 2 by 2 till the last one is out:

[tex]
(c^{n-1} + c^{n-2}*b + \ldots+ b^{n-1}) = (b + c)*z + b^{n-1} \ (15),
[/tex]

now lets turn this procedure around: lets sum all the members till the last one 2 by 2 by the backside so:

[tex]
c*b^{n-2} + b^{n-1} = b^{n-2}*(c + b) + \ldots \ (16)
[/tex]

so we carry on till we come to the [itex]c^{n-1}[/itex]

------------->

[tex]
c^{n-1} + (c + b)*t \ (17),
[/tex]

where t is again number of natural numbers. BUT if we multiply [itex](c^{n-1} + c^{n-2}*b + \ldots + b^{n-1})[/itex] by 2 we can sum the equations:

[tex]
c^{n-1} + (c + b)*t + (b + c)*z + b^{n-1} \ (18)
[/tex]

since [itex](b + c)[/itex] divides [itex]c^{n-1} + b^{n-1}[/itex], we got that if we multiply the expression by [itex]2 (c+ b)[/itex] divides [itex]2*(c^{n-1} + c^{n-2}*b + \ldots + b^{n-1})[/itex] since [itex](c + b)[/itex] is odd

------------------------------->

[itex](c+b)[/itex] divides [itex](c^{n-1} + c^{n-2}*b + \ldots + b^{n-1})[/itex] too.

That's a contradiction:

[itex](c + b)[/itex] is coprime to (c^{n-1} + c^{n-2}*b + \ldots + b^{n-1})[/itex], so [itex](c^{n} + b^{n})[/itex] and [itex](c^{n} - b^{n})[/itex] are coprimes when [itex]a, b, c[/itex] are coprimes

------------------->

so [itex](c^{2(n-1)} + c^{2(n-2)}*b^{2} + \ldots + b^{2(n-1)})[/itex] is the member of rational numbers. This is again a contradiction, since we know that this expression is the member of natural numbers. So we can not find solutions in whole number system of [itex]a, b, c[/itex] for n is odd and [itex]n > 2[/itex]. Q.E.D.

I want to make better impression, so I got back to the problem for a few hours today.

Sincerelly

Robert Kulovec Mueller, Slovenija
 
  • #8
64
0
If b=2 and c=3, and n=3

2+3=5
23+33 = 35 is divisible by 5
22+32=13 is not divisible by 5
Thanks for better readable form :) I owe you a drink.

@Office_Shredder this is just the assumption and b,c, are solution of the fermats eqation. It is seen later that this assumption is wrong and we get a contradiction...so if (c+b) is coprime with other two multiplied factors and they are coprimes to each other, the last part with square parts is rational. That is a contradiction and so the first assumption that the a,b,c solution exists for odd powers of n is wrong by itself. because if the solution exists, they do not sattisfy the first writen equation which holds true in any case for the b,c being members of natural numbers. When we make the conditions more stricts with including coprimes which follows from fermat eqations, we come to the contradicition. And the first eqation is not true eny more...
 
  • #9
64
0
@Dickfore in 7, 11,12,13, 14 and in text under 14 there are powers of (n-1) I believe thats not clear... Thank you so much.
 
  • #10
64
0
In the last lines where is writen thats a contradiction. It should be better to state, so (c+b) does not divide c^n + b^n if (c+b) does not divide c^n + b^n and (c+b) is coprime to the (c^(n-1) + c^(n-2)*b..............+ b^(n-1) it follows that (c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1)) is the element of rational numbers... Thats better I believe.
 
  • #11
64
0
Î think that my assuption in the original proof is wrong, its not necessary to proof that c^n + b^n is coprime to (b + c) its enough to state that factors on the left in first equations are coprimes ++ (c +b) does not divide c^n + b ^n and (c+b) is coprime to (c^(n-1) + (c^(n-2)......b^(n-1))(c+b) doesnt have to be coprime to c^n + b^n. its enough that this expression is not dividable by (c+b). (maybe its even not coprime:) so the rationality of the sqare expression on the right comes from this 3 assuptions.
 
  • #12
64
0
Î think that my assuption in the original proof is wrong, its not necessary to proof that c^n + b^n is coprime to (b + c) its enough to state that factors on the left in first equations are coprimes ++ (c +b) does not divide c^n + b ^n and (c+b) is coprime to (c^(n-1) + (c^(n-2)......b^(n-1))(c+b) doesnt have to be coprime to c^n + b^n. its enough that this expression is not dividable by (c+b). (maybe its even not coprime:) so the rationality of the sqare expression on the right comes from this 3 assuptions.
Ok I didint state this in the proof, I am just a bit tired from my uni learning, sorry about that. Coprimety I stated in the reply to Office_Shredder so I wanted to make clear, that is not necessary to proove that c^n + b^n and (b+c) are coprimes. Dividability, or better undividability would be more than enough.

Thank you guys to be so patient with me. :)
 
  • #13
2
0
in lines 11 and 12 don't you assume that both (l-c^(n-1)) and (b^(n-1)-l) = m? It looks like you accidentally picked the same variable to represent two different quantities, but didn't notice this and then started doing calculations with expressions involving this variable.

This of course can't be true unless l is exactly halfway between c^(n-1) and b^(n-1), so your proof loses generality.
 
  • #14
2
0
Just to say, I think its cool that you are working on this, but it you think you've found a proof that simple of Fermat's Last Theorem, its overwhelmingly likely a mistake - if you think not, then you are basically making the assumption that thousands of amazing mathematicians who have capabilities far beyond that of this style of proof have just been foolishly missing it for years. Seeing the sorts of mathematics that many mathematicians are capable of, especially of this type, I have extreme doubts that a proof of this style is the missing simple proof of Fermat's Last Theorem.

I know its possible of course, but there are so many proofs like this that always turn out to have really silly errors in them. Its so much more likely that you made a common freshman algebra/logic error than that you solved one of the greatest puzzles in mathematical history that you should really double and triple check your work and be a little more self-critical before releasing your proofs.
 
  • #15
64
0
In line 11 is a fundamental mistake, minus changed into plus :) Unfortunatelly the proof is wrong. So the topic is closed. Anyway its somehow funny to work with equation with 4 variables :) I mean everytime you get a new idea how to prove it you use the fermats equation twice :) I am just trying to solve it from 2 points of view.

1.The theorem is already solved.
and second assumption is that fermat couldnt be wrong in his statement he has an elegant proof. Anyway I will stop at this point for some time.

So the post from the member was correct, my assuption of division by (c+b) was wrong because one part of the equation 11 should be minus and not plus.
 
  • #16
64
0
Just to say, I think its cool that you are working on this, but it you think you've found a proof that simple of Fermat's Last Theorem, its overwhelmingly likely a mistake - if you think not, then you are basically making the assumption that thousands of amazing mathematicians who have capabilities far beyond that of this style of proof have just been foolishly missing it for years. Seeing the sorts of mathematics that many mathematicians are capable of, especially of this type, I have extreme doubts that a proof of this style is the missing simple proof of Fermat's Last Theorem.

I know its possible of course, but there are so many proofs like this that always turn out to have really silly errors in them. Its so much more likely that you made a common freshman algebra/logic error than that you solved one of the greatest puzzles in mathematical history that you should really double and triple check your work and be a little more self-critical before releasing your proofs.
Thanks for that kind msg. Anyway I will post the topic about how to expand the fermats last theorem...
 
  • #17
HallsofIvy
Science Advisor
Homework Helper
41,833
955
You were coming perilously close to becoming a "crank" but with
In line 11 is a fundamental mistake, minus changed into plus :) Unfortunatelly the proof is wrong. So the topic is closed.
you turned away in time! Congratulations.

Often when it is pointed out to people that it is unlikely that a person with little or no education in mathematics could solve a problem that stopped professional mathematicians for many years, they will respond that the professionals may be looking for deep, difficulty ways and overlook something simple. That's simply not true. One of the first things a mathematician learns is to look at the simplest possiblities and the simplest cases first. Even if the simple possiblities don't work, you learn about the problem that way.
 
  • #18
64
0
@HalsofIvy, thank you for your response :) I just like doing it, because I believe you could have some creative thoughts about it. I am pretty sure I will not prove it :) Thousands have tried and they have all failed. I believe that the more simple proof exists that this from the 90ies. But I strongly believe its not reducable to one page. Perhaps 100. :)

Take care,

Robert
 
  • #19
64
0
@HalsofIvy and By the way I like your profound English :)
 
  • #20
mathwonk
Science Advisor
Homework Helper
10,940
1,096
I once gave something resembling the following question in a logic class:

Give fermat's last theorem holds as follows:

Assume:
1. If FLT were false, there would exist a non modular elliptic curve.
2. All elliptic curves however are modular.

Give your conclusion with explanation of the reasoning:
 
  • #21
124
0
negation of 1: if every elliptic curve is modular then FLT is true

2 holds therefore FLT is true

[I hope this was easy for your students]
 
  • #22
124
0
Deleted
 
  • #23
64
0
Assume:
1. If FLT were false, there would exist a non modular elliptic curve.
2. All elliptic curves however are modular.

Give your conclusion with explanation of the reasoning:

Thats not a tautonomy, is just an implication I guess...
 
  • #24
mathwonk
Science Advisor
Homework Helper
10,940
1,096
contrapositive, not negation. Good! see how easy math is with the right hypotheses?
 

Related Threads on Elegant proof of fermats last theorem 2. try

Replies
1
Views
2K
Replies
9
Views
6K
  • Last Post
Replies
1
Views
3K
Replies
20
Views
35K
Replies
150
Views
25K
Replies
26
Views
5K
Replies
5
Views
4K
Replies
3
Views
4K
  • Last Post
3
Replies
52
Views
10K
  • Last Post
Replies
2
Views
2K
Top