The last try for elegant solution of fermats last theorem

[Robert80]

The reason I am writing to you is because yesterday I have sent the really amateur and wrong solution, sorry about that... I was really embaressed so I said I will try for the last time to crack the problem...

 

 

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.

 

Sorry about yesterday, I did a fundamental mistake. I want to make better impression, so I got back to the problem for a few hours today.But I am preety sure its wrong again...

 

Sincerelly

 

Robert Kulovec Mueller, Slovenija

[Bignose]

Why not write this out to that at the very least, there is one equation per line? Even better would be to use this forum's LaTeX capabilities to make it even easier to read.

 

As it is posted above, I have zero interest in even scanning the post because it is very, very, very difficult to read it.

 

Edited for grammar/spelling

Edited November 16, 2010 by Bignose

[Cap'n Refsmmat]

(There's a LaTeX tutorial available. For your purposes it should be rather easy to use.)

[Shadow]

Here is a quick LaTeX tutorial that should get you well on the way to writing readable equations:

[Robert80]

Hi guys, really thanks for being so helpful, I believe I will rearange it with the application you suggested, but right now, I have to study for my physics exam, so the readable form will have to wait for few more days. Evenif if there is a mistake in proof I wrote, I believe that the fundamental idea how to proove its the right one.

 

sincerely

 

Robert

[FLT For Fun]

Hi Robert,

 

Your statement " (c^n + b^n) equals 2b^n + a^n and that shure is coprime to a, since a and b are coprimes." needs to be proved that they are coprime. Your approach is very interesting but your current form of proof would never pass the eyes of mathematicians. I tried to prove FLT on the side as brain exercise, a la, Sudoku. But I'd also like to see how other doing their proofs.

 

I believe that I have a proof. Let me summarize what I know about FLT. FLT looks so innocent but it is as false statement as 2=1. If you want to prove it via geometry, then no Euclidian geometry will not prove it. The only geometry used to prove FLT is elliptic curve. The proof was over hundred pages long. Since I am not a mathematician but I am sure it is non-Euclidian geometry. As for algebraic proof, it can be round robin. But if you can stop it you prove it.

 

flt4fun