Sophie Germain: mathématicienne extraordinaire - Scripps College

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The First Attempts 36 with Unique Factorization again in section 4.3.2. □ Returning to our proof we note that x 2 = 4ab(a 2 + b 2 ) is a square. Recall that a and b are of opposite parity. Then ab is even and a 2 + b 2 is odd. By our lemma, since a, b, and a 2 +b 2 are relatively prime and ab(a 2 + b 2 ) is a square, then a is a square, b is a square, and (a 2 + b 2 ) is a square. We write, a = X 2 and b = Y 2 . Then a 2 + b 2 = X 4 + Y 4 is a square, call it Z 2 . We now have Z 2 = X 4 + Y 4 . Since a > b > 0, we know that Z 2 > 0. Beginning this proof, we used the fact that z 4 0 was a square, ignoring its quartic properties. That is, we said, for integers x 0 and y 0 , the integer x 4 0 + y4 0 is a square which gives us new integers X and Y such that X4 + Y 4 is also a square. However, 0 < Z 2 = X 4 + Y 4 = a 2 + b 2 = s < s 2 + t 2 = z0 2, that is, Z 2 < z0 2. So begining with a triple (x 0, y 0 , z0 2) such that x4 0 + y4 0 = (z2 0 )2 , we obtained a triple (X, Y, Z 2 ) such that X 4 + Y 4 = (Z 2 ) 2 and z 2 > Z 2 > 0. So given any integer solution (x 0 , y 0 , z 0 ) with z 0 > 0, we get another integer solution (x 1 , y 1 , z 1 ) such that z 0 > z 1 > 0. By the prinicple of infinite descent, there can be no initial solution. Thus the sum of two fourth powers can never be a square, much less a fourth power. Therefore, Fermat’s Last Theorem for fourth powers (n = 4) is true. □ Now given any equation of the form x 4m + y 4m = z 4m , we may rewrite it as (x m ) 4 + (y m ) 4 = (z m ) 4 .
The First Attempts 37 If we substitute X = x m , Y = y m , and Z = z m , we get X 4 + Y 4 = Z 4 . This gives us the following corollaries: Corollary 1. For any m ∈ N, the equation x 4m + y 4m = z 4m has no non-trivial integer solutions. Moreover, we consider any x m + y m = z m for even m. If m is even then either 4|m or m ≡ 2 (mod 4). If 4|m, we have shown that there are no integer solution to the equation. If 4 ∤ m but 2|m and m = 2d for some odd integer d, we rewrite x 2d + y 2d = z 2d as (x 2 ) d + (y 2 ) d = (z 2 ) d . Corollary 2. If the equation x 2p + y 2p = z 2p has a non-trivial integer solution x 0 , y 0 , and z 0 , then X p + Y p = Z p must also have integer solutions. In particular, the solutions must be X 0 = x 2 0 , Y 0 = y0 2, and Z 0 = z0 2. Thus, we need only consider Fermat’s Last theorem for n = d ≥ 3, an odd integer. If p is a prime and n = pm for some m ≥ 1, then consider any solutions x 0 , y 0 , and z 0 to the equation x n + y n = z n . We can rewrite this equation as (x m ) p + (y m ) p = (z m ) p , with solutions x m 0 , ym 0 , and zm 0 . Thus it suffices to consider non-trivial integer solutions to the equation x p + y p = z p for all primes p > 2.
Page 1 and 2: Sophie Germain: mathématicienne ex
Page 3 and 4: Contents Abstract Acknowledgments i
Page 5 and 6: Chapter 1 Introduction Marie-Sophie
Page 7 and 8: Chapter 2 La Femme 2.1 Sa Biographi
Page 9 and 10: Sa Biographie 5 maths et ils lui in
Page 11 and 12: Sa Biographie 7 Florens Friedrich C
Page 13 and 14: Modestie supposée 9 2.2 Modestie s
Page 15 and 16: Modestie supposée 11 2.2.1 Lalande
Page 17 and 18: Modestie supposée 13 (Bucciarelli
Page 19 and 20: Modestie supposée 15 sa préséanc
Page 21 and 22: Modestie supposée 17 Il est évide
Page 23 and 24: Modestie supposée 19 délai [. . .
Page 25 and 26: Chapter 3 Ses Oeuvres, Ses Travaux
Page 27 and 28: Le philosophe 23 C’est un peu iro
Page 29 and 30: Le philosophe 25 nomènes qu’elle
Page 31 and 32: La scientifique 27 la théorie d’
Page 33 and 34: La scientifique 29 crédibilité, e
Page 35 and 36: Introduction 31 a truly remarkable
Page 37 and 38: The First Attempts 33 4.2 The First
Page 39: The First Attempts 35 4.1.2. Withou
Page 43 and 44: The First Proof 39 the previous 3 c
Page 45 and 46: The First Proof 41 We plug these eq
Page 47 and 48: The First Proof 43 us 5 3 r 4 + 2
Page 49 and 50: The First Proof 45 Returning to the
Page 51 and 52: The First Proof 47 write P 0 = A 16
Page 53 and 54: The First Proof 49 are positive int
Page 55 and 56: Chapter 5 Sophie Germain 5.1 Case 1
Page 57 and 58: Sophie Germain’s Theorems 53 Cube
Page 59 and 60: Sophie Germain’s Theorems 55 Clai
Page 61 and 62: Sophie Germain’s Theorems 57 We h
Page 63 and 64: Sophie Germain’s Theorems 59 fied
Page 65 and 66: Sophie Germain’s Theorems 61 if n
Page 67 and 68: What then 63 Germain prime, discove
Page 69 and 70: What then 65 This means that if P i
Page 71 and 72: Chapter 6 Conclusion 6.1 And Beyond
Page 73 and 74: Conclusion 69 the Frey curve must n
Page 75 and 76: Bibliography [1] Bucciarelli, Louis
Page 77 and 78: Bibliography 73 [10] James, Ioan. R
Page 79: Bibliography 75 1990. v. 41 no. 2,

Sophie Germain: mathématicienne extraordinaire - Scripps College

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