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Fermat’s Last Theorem Samina Saleem Math5400. Introduction The Problem The seventeenth century mathematician Pierre de Fermat created the Last Theorem.

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Presentation on theme: "Fermat’s Last Theorem Samina Saleem Math5400. Introduction The Problem The seventeenth century mathematician Pierre de Fermat created the Last Theorem."— Presentation transcript:

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 3 2 + 4 2 = 5 2 and 5 2 + 12 2 = 13 2. 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 2 and 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| y 4 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 u 3 +6u v 2 =2u(u 2 +3 v 2 )

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

10 Continued….. Now if g.c.d(2u,u 2 +3 v 2 ) =1 then by unique factorization theorem 2u,and u 2 +3 v 2 are cubes. That is 2u= a 3 and u 2 +3 v 2 = b 3 According to key lemma we can write u=c 3 -9cd 2 and v=3 c 2 d-3 d 2 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 a 3 =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= -n 3, c-3d= l 3 c+3d= m 3 Where l, m, n are distinct and pairwise relatively prime. therefore we conclude that n 3 + l 3 + m 3 =0,but |z |> |m | because | z 3 |= |2u(u 2 +3 v 2 )|= | n 3 (c 2 9 d 2 ) (u 2 +3 v 2 )|≥ | n 3 | Because c 2 - 9 d 2 = l 3 m 3 ≠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/teac hers/activities/2414_proof.html http://www.pbs.org/wgbh/nova/teac hers/activities/2414_proof.html  http://www.pbs.org/wgbh/nova/teac hers/ideas/2414_proof.html http://www.pbs.org/wgbh/nova/teac hers/ideas/2414_proof.html


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