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

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

After several attempts with geometry and gymnastics with number theory, what I believe I found a general proof and simple one as Fermat would wish. But anyway, I am not skilled in Matematics, nor in any subject except chemistry, I just get sometimes the creative ideas.... So I got one last night...I would really like to ask you guyz, where is the mistake?As probably there is one. 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 so if we sqare the equation it will hold true that (a^n + b^n)^2 = (c^n)^2 so -----> a^2n + b^2n + 2a^nb^n = c^2n ----------> 2a^nb^n = c^2n - b^2n - a^2n, from the number theory we know that it follows that c^2n - b^2n is devidable by a^n, so c^2n - b^2n = a^n*k where k is the element of natural numbers. so lets multiply the original Fermats equation by factor k, so ------> a^n*k + b^n*k = c^n*k, lets now substitute the term a^n*k by c^2n - b^2n so:-------> c^2n - b^2n + b^n*k = c^n*k -------->b^n*(k - b^n) = c^n*(k - c^n), since b and c are coprimes b^n = (k - c^n)*m and c^n = (k - b^n)*m where m again is the element of Natural numbers. so--------> c^n + b^n*m = b^n + c^n*m, we see that m is 1, so -----> k = c^n + b^n lets now put that into 2a^nb^n = c^2n - b^2n - a^2n -----------> lets divide now the whole eqation by a^n ----------------> 2b^n = c^n + b^n - a^2 -------------> b^n = c^n - a^2 and since a^n + b^n = c^n ---------> b^n = a^n + b^n - a^2 ------------> a^n = a^2 -------------> n = 2 if the solution of the fermats last theorem exists. Robert Kulovec Mueller, Slovenija correction: k is the element of odd numbers, not natural