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Victor
guest
Oct-12-05, 03:55 PM (EST)

 It is common knowledge that if the integers A and B have no common factor and A + B do not divided by simple n then the efficient (A + B) and (A^n + B^n)/(A + B) of the number A^n + B^n have no common factor.Therefore, after cancellation of common factors in the equation A^n + B^n = C^n at least two (for example A and B) from three numbers A, B, C are not divisible by n. Then each number from A^n and B^n consists of pair of the factors which have no common factor (and therefore they are n-th powers!): in particular, C - B = a^n and C - A = b^n. Therefore, if the Fermat's equation exists, then there exist the Fermat's equationsa^n  b^n = A - B = 1^n, 2^n, 3^n, etc.Absurdity is obvious. The proof of the FLT is done. Victor

Subject     Author     Message Date     ID
RE: News aboutFLT mr_homm Oct-12-05 1
RE: News aboutFLT mr_homm Oct-13-05 2
RE: News aboutFLT Victor Sorokine Oct-14-05 5
RE: News aboutFLT - More clear Victor Sorokine Oct-15-05 6
RE: - Important insertion vsorokine Oct-15-05 7
RE: News aboutFLT ramsey2879 Oct-13-05 3
RE: News aboutFLT Moshe Klein Oct-14-05 4
News about FLT neat_maths Nov-16-05 9
RE: News about FLT & coprimes tony123 Jan-26-06 11

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mr_homm
Member since May-22-05
Oct-12-05, 04:45 PM (EST)

In response to message #0

 Let me fill in the details of your argument to make sure I understand it:>It is common knowledge that if the integers A and B have no >common factor and A + B do not divided by >simple n then the efficient (A + B) and >(A^n + B^n)/(A + B) of the number >A^n + B^n have no common factor. I do not remember this fact, but I will assume it is true. Of course it is also true when you use A^n-B^n and A-B. There is a little trouble here, because for A^n-B^n can always be divided by A-B, but A^n+B^n can only be divided by A-B when n is an even number.>Therefore, after cancellation of common factors in the >equation A^n + B^n = C^n at least two (for >example A and B) from three numbers A, B, C are not >divisible by n. This is clear.>Then each number from A^n and B^n consists >of pair of the factors which have no common factor (and >therefore they are n-th powers!): in particular, C - B >= a^n and C - A = b^n. If I understand you correctly, this is true because A^n = C^n-B^n and B-C is not divisible by n. There is a trouble with this: it is possible that all of A, B, and C are not divisible by n. In that case, C-B or C-A or both may be accidentally divisible by n. In that case, there may be only one pair of numbers whose difference is not divisible by n, and you need two pairs for your argument. However, suppose that you do have C-B and C-A not divisible by n. Then the common knowledge tells you that A^n factors into two factors a_1 and a_2, which share no common factors. Therefore each of them is an n-th power. One of these factors is C-B, so you get C-B = a^n. In the same way you get C-A = b^n.Another trouble is that the numbers not divisible by n may not be C-B and C-A. One of them may instead be A+B, and in that case your common knowledge only works when n is even.>>Therefore, if the Fermat's equation exists, then there exist >the Fermat's equations >>a^n - b^n = A - B = 1^n, 2^n, 3^n, etc. >I do not follow this last part at all. How do you get 1^n, 2^n, 3^n, etc.?>Absurdity is obvious. The proof of the FLT is done. I think there are still some troubles you need to fix.--Stuart Anderson

mr_homm
Member since May-22-05
Oct-13-05, 08:11 PM (EST)

In response to message #1

 Posting to correct an error:>Of course it is also true when you use A^n-B^n >and A-B. There is a little trouble here, >because for A^n-B^n can always be divided by >A-B, but A^n+B^n can only be >divided by A-B when n is an even number. The last line should read "divided by A+B when n is an even number."--Stuart Anderson

Victor Sorokine
guest
Oct-14-05, 03:49 PM (EST)

In response to message #1

 >>"Therefore, if the Fermat's equation exists, then there exist >>the Fermat's equations >>>>a^n - b^n = A - B = 1^n, 2^n, 3^n, etc." >>>>I do not follow this last part at all. How do you get 1^n, >2^n, 3^n, etc.? >>--Stuart Anderson Answer:Let (A, B, C) is a solution of the Fermat's equation (1°), where (2°) A + B  C = U, C  B = a^n, C  A = b^n.Let's show that (A*, B*, C*) is a solution of the equation (1°), where (3°) A* + B*  C* = U, C*  B* = a*^n = a^n, C*  A* = a^n  1.(4°) 2U = 2A + 2B  2C = (A + B)  (C  B)  (C  A) = 2A* + 2B*  2C* = (A* + B*)  (C*  B*)  (C*  A*).(5°) So, C*  B* <= a*^n = a^n> is whole (cf. 3°);(6°) therefore C*  A* <= a^n  1> is whole (cf. 3°);(7°) therefore (C*  B*) + (C*  A*) is whole (cf. 5° and 6°);(8°) therefore A* + B* <= 2U  (C*  B*) + (C*  A*)> is whole;(9°) therefore C* <= U  B*  A*> is whole (cf. 2° and 4°);(10°) therefore B* <= C*  a^n> is whole (cf. 3°);(11°) therefore A* <= U  B* + C*> is whole (cf. 2° and 4°).(12°) Therefore A*, B*, C* are whole and therefore:C*  B* = a*^n, C*  A* = b*^n = a^n  1 = a*^n  1 AND b*^n  a*^n = 1 = 1^n, that is impossible!!! +++++ Lemma:If the integers A and B have no common factor and A+B do not divided by prime n then the factors (A+B) and (A^n + B^n)/(A+B) of the number A^n + B^n have no common factor.Proof:(A^n + B^n)/(A+B) = R == A^(n-1) + A^(n-2) . B + + A^(n-2)/2 . B^(n-2)/2 + A . B^(n-2) + B^(n-1) = = + + + A^(n-2)/2 . B^(n-2) == ++ AB ++  A^(n-2)/2 . B^(n-2)/2 == (A  B)^2 . P + nA^(n-2)/2 . B^(n-2)/2.The numbers A  B and nA^(n-2)/2 . B^(n-2)/2 have no common factor. The numbers A  B and (A  B)^2 . P + nA^(n-2)/2 . B^(n-2)/2 have no common factor. The Lemma is proven.Corollary:In the equation A^n + B^n = C^n (where A, B, C are integers and prime n > 2), minimum two equation from C  B = a^n, C  A = b^n, A + B = c^n are right.(Published: Journal of TRIZ, 1992, n° 3.4.92. V.M.Sorokin. Proof of the Great Fermat Theorem found with help of TRIZ-like methods. Russia.)Victor Sorokine

Victor Sorokine
guest
Oct-15-05, 07:58 AM (EST)

6. "RE: News aboutFLT - More clear"
In response to message #5

 More clear:Let (A, B, C), where: A, B, C have no common factors, (2°) A + B - C = U, A and B and C - B - 1 (or C - B + 1) are not divided by n,is an integer-valued solution of the Fermat's equation A^n + B^n - C^n (1°).Let's show that (A*, B*, C*), where(3°) A* + B* - C* = U, C* - B* = C - B <= a^n> and C* - A* = C - B - 1,is also an integer-valued solution of the equation (1°).(4°) From (2°) we have: 2U = 2A + 2B - 2C = (A + B) - (C - B) - (C - A) = = 2A* + 2B* - 2C* = (A* + B*) - (C* - B*) - (C* - A*).(5°) So, C* - B* <= a*^n = a^n> is integer-valued (cf. 3°); etc(12°) Therefore A*, B*, C* are integer, where A*, B*, C*, C* - B*, C* - A* are not divided by n and they have no common factors;therefore C* - B* = a*^n, C* - A* = b*^n = a^n - 1 = a*^n - 1 AND b*^n - a*^n = 1 = 1^n, that is impossible!!!

vsorokine
guest
Oct-15-05, 05:45 PM (EST)

7. "RE: - Important insertion"
In response to message #6

 Important insertion and (I hope) the finale"Let (A, B, C), where: A, B, C have no common factors, (2°) A + B  C = U, A and B and C  B  1 (or C  B + 1) are not divided by n,is an integer-valued solution of the Fermat's equation A^n + B^n  C^n (1°)."insertion:Let' find a real solution (A*, B*, C*) of the equation (1°) under such conditions:(3°) A* + B*  C* = U, C*  B* = C  B <= a^n> and C*  A* = C  B  1.Since the conditions are not contradictory (the determinant of the equation system (3°) =/ 0), then a real solution (A*, B*, C*) exists.Now let's show that the real numbers A*, B*, C* are integers.Farther as before

ramsey2879
guest
Oct-13-05, 10:44 PM (EST)

In response to message #0

 >It is common knowledge that if the integers A and B have no >common factor and A + B do not divided by >simple n then the efficient (A + B) and >(A^n + B^n)/(A + B) of the number >A^n + B^n have no common factor. >I believed Your proof is flawed. At best you can only show that there is no common algebraic factor. However, it is well known that two equations having no common algebraic factor are not necessarily coprime. For instance (a+b) and (a^3 + b^4) have no common algebraic factor. However, for a =3 and b=4, a+b divides a^3 + b^4!!

Moshe Klein
guest
Oct-14-05, 06:00 AM (EST)

In response to message #3

 Hi victor,Thank you for the link. How is new methode relate to the one you have before on looking on the problem on base n ? Thank you Moshe

neat_maths
Member since Aug-22-03
Nov-14-05, 08:40 AM (EST)

In response to message #3

 In this case the proposition that the two factorsF1 = (A plus B) and F2 = (A^n plus B^n)/(A plus B) are mutually prime is correct and fairly easy to prove.This assumes that n is prime and that A, B, n and (A plus B) are mutually prime.If U = A plus B and V = A*BF2 can be expressed as U*(terms in U and V) plus or - n * V^((n-1)/2)Therefore F2 = sum of two numbers (1st divisible by U) and (2nd not divisible by U)Therefore F2 is not divisible by U It is also fairly easy to prove that F2 and n are mutually prime (and we have already the prerequisite that n is mutually prime to F1)F2 can be expressed as U^(n-1) - n * V * (terms in U and V)Therefore F2 = sum of two numbers (1st not divisible by n) and(2nd divisible by n)Therefore F2 is not divisible by nI hope that is acceptable take care

neat_maths
Member since Aug-22-03
Jan-02-06, 06:45 PM (EST)

In response to message #3

 "I believed Your proof is flawed. At best you can only show that there is no common algebraic factor. However, it is well known that two equations having no common algebraic factor are not necessarily coprime. For instance (a+b) and (a^3 + b^4) have no common algebraic factor. However, for a =3 and b=4, a+b divides a^3 + b^4!!"a^3 = 27, b^4 = 256a^3 plus b^4 = 283 which is not divisible by 7I don't agree with the point you are making.take care

neat_maths
Member since Aug-22-03
Nov-16-05, 07:04 PM (EST)

In response to message #0

 How absurd. Silly that the greatest mathematicians in the world over the last 400 years didn't notice this.LOL

tony123
Member since Jan-26-06
Jan-26-06, 07:31 AM (EST)

11. "RE: News about FLT & coprimes"
In response to message #9

 >It is common knowledge that if the integers A and B have no >common factor and A + B do not divided by >simple n then the efficient (A + B) and >(A^n + B^n)/(A + B) of the number >A^n + B^n have no common factor. >I believed Your proof is flawed. At best you can only show that there is no common algebraic factor. However, it is well known that two equations having no common algebraic factor are not necessarily coprime. For instance (a+b) and (a^3 + b^3) have no common algebraic factor. However, for a =3 and b=4, a+b divides a^3 + b^3!!a^3 = 27b^3 = 64a + b = 7 (27 + 64)/7 = 91/7 = 13 !!RegardsTony123