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Fermat’s Last Theorem

Fermat’s Last Theorem. Samina Saleem Math5400. Introduction The Problem. The seventeenth century mathematician Pierre de Fermat created the Last Theorem which became one of the most challenging problems in mathematics .The theorem states that:

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Fermat’s Last Theorem

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  1. Fermat’s Last Theorem Samina Saleem Math5400

  2. Introduction The Problem The seventeenth century mathematician Pierre de Fermat created the Last Theorem which became one of the most challenging problems in mathematics .The theorem states that: “The equation x n + y n = z n has no non zero integer solutions for x, y and z when where n is an integer.” In the margin of a mathematics book that Fermat read, he wrote down: “I have a truly marvelous demonstration of this proposition which this margin is too narrow to contain. “

  3. Although x 2 + y 2 = z 2 has an infinite number of solutions ,but x n + y n = z n has no nontrivial whole number solutions for n > 2. For example, we can easily check to see that 32 + 42 = 52 and 52 + 122 = 132. Try as you may you cannot come up with integer values for x, y, and z such that x 3 + y 3 = z 3.

  4. Proofs for special cases • Fermat has given the proof for n=4 • He tried to prove that the area of a right triangle with rational sides is never a perfect square, a condition that is equivalent to the claim that there are no integer solutions to x 4 + y 4 = z 2and hence no solutions to x 4 + y 4 = z 4

  5. Fermat's Proof for n=4 • Fermat studied the following equation x 4+ y 4 = z 2 and showed that this equation has no solution in integers all different from zero.

  6. Continued... Suppose x 4 + y 4 = z 2 has solution. Let (x,y, z) be a positive integers such that x 4 + y 4 = z 2 then g.c.d (x,y)=1 Assume p divides x and y then x=pk and y=pk ' Then p|x and p|y p| x 4 , p| y4 p|( x 4 + y 4 ) it proves p| z 2 Now Let x=px' and y=py ' then x 4 + y 4 = p 4 x '4 + p 4 y '4 = p 4 (x '4 +y '4 )= p 4 z '2 and therefore x '4 + y '4 = z '2

  7. Continued…….. • We assumed that x is the smallest number for which the equation holds. Now found • 0<x'<x and the equation holds for x' which is a contradiction. Hence Equation has no solutions. Fermat further connected the above results by a corollary to prove that x 4+ y 4 = z 4 has no solution in integers, all different from zero

  8. Euler Proof for n=3 Assume that x, y, z are non zero pair wise relatively prime and |z| is the smallest possible. x+ y=2u x - y=2u 2x=2(u+v) X=u+ v 2y=2(u-v) y=u- v (-z 3)= x 3 + y 3 =(u+ v) 3 + (u - v) 3 =2 u3 +6u v2 =2u(u2 +3 v2 )

  9. Continued…… • It implies that (-z 3)= 2u(u2 +3 v2 ) where u2 +3 v2 is odd and z is even. Therefore g.c.d(2u, u2 +3 v2 ) =1 when u is not divisible by 3 =3 when u is divisible by 3

  10. Continued….. Now if g.c.d(2u,u2 +3 v2) =1 then by unique factorization theorem 2u ,and u2 +3 v2 are cubes. That is 2u= a3 and u2 +3 v2 = b3 According to key lemma we can write u=c3 -9cd2 and v=3 c2 d-3 d2 This is where the Euler’s had the gap.

  11. Now if we consider key lemma of this theorem then d is odd and c is even since v is odd then a3 =2u =2c(c-3d)(c+3d). Now because the three factors on the right hand side are pairwise coprime they must all be a three power. 2c= -n3 , c-3d= l3 c+3d= m3 Where l, m, n are distinct and pairwise relatively prime. therefore we conclude that n3 + l3 + m3 =0 ,but |z |> |m | because | z3 |= |2u(u2 +3v2)|= | n3(c2 9 d2) (u2 +3v2)|≥ | n3| Because c2 - 9 d2 = l3 m3 ≠0 and v≠0 since it is odd contradicts the minimality of Z. A contradiction

  12. Infinite descent method • This method is known as method of “infinite descent” created by Fermat in which you assume that you have the smallest solution for a problem and show that there exist still smaller solutions. • In other words there is no smallest solution and if you continue working on it there is no solution at all. Although Fermat proved the theorem for n=4 by using the “infinite descent” method, this method doesn’t work for values n>4.

  13. Sophie Germain • In early 1800’s a French Mathematician Sophie Germain adopted a new strategy to approach FLT According to Sophie Germain if n and 2n+1 are primes then x n + y n = z n implies that one of x, y, and z is divisible by n. Hence Fermat’s last theorem could be broken into two cases. • Case I: none of x y, z is divisible by n • Case II: one and only one of x, y, z is divisible by n. • She has proved the theorem for Sophie Germain primes <100

  14. Further History • In 1825 ,Legendre had followed the Sophie Germain's criterion, and proved the first case of Fermat’s last theorem holds for every prime p<197 He was also able to proof second case. • In 1832 Peter Lejenue-Dirichlet proved the theorem for n=14. • In 1839 Gabriel Lame proved the result for n=7. • Specific cases are proven in 1909, 1915, and 1953. But still no one is able to come up with the complete proof to FLT.

  15. In September 1955, a 28-year-old Japanese mathematician, Taniyama–Shimura claimed that all elliptical curves are modular. • In 1985 Gerhard Frey a German mathematician considered the unthinkable. What would happen if Fermat was wrong and there was a solution to this equation? Frey’s idea became known as the epsilon conjecture.

  16. Frey showed how starting with a fictitious solution to Fermat’s last equation could make an elliptic curve with some very weird properties. That elliptic curve seemed to be not modular . But Taniyama–Shimura said that every elliptic curve is modular .

  17. So it means if there is a solution to this equation it creates such a wired elliptic curve it defies Taniyama–Shimura. In other words if Fermat is false so did the Taniyama–Shimura or if Tanyama Shimura is correct then Fermat’s last theorem is also true. In fact, Frey was the first mathematician who built the bridge between Taniyama–Shimura conjecture and Fermat’s last theorem.

  18. Ken Ribet, a professor at University of California, with the help of Barry Mazur, later proved this result. Now FLT is linked to the Taniyama–Shimura conjecture. • In June 1993, Andrew Wiles has proved FLT. • In September 1993 an error has been found. • On September 19, 1994 Andrew wiles resolved the problem with his former student and Fermat’s last theorem has been proved.

  19. Curriculum link • Principles of Mathematics, Grade 10, Academic (MPM2D), Foundations of Mathematics, Grade 10, Applied (MFM2P) solve problems involving right triangles, using the primary trigonometric ratios and the Pythagorean theorem; • Functions and Relations, Grade 11, University Preparation • Connecting Algebra and Geometry . In the Investigations of Loci and Conics strand, students will extend their inquiry of loci into a study of conics.

  20. Use in Teaching Mathematics • The Story • The video • Activities • http://www.pbs.org/wgbh/nova/teachers/activities/2414_proof.html • http://www.pbs.org/wgbh/nova/teachers/ideas/2414_proof.html

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