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# Euler's work on Fermat's Last Theorem

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### Euler's work on Fermat's Last Theorem

1. 1. n=2 n=4 n=3 E-272 Euler’s Proof of Fermat’s Last Theorem (for n = 3) Lee Stemkoski Adelphi Univeristy December 5, 2012 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 1 / 33
2. 2. n=2 n=4 n=3 E-272Outline 1 n=2 2 n=4 3 n=3 4 E-272 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 2 / 33
3. 3. n=2 n=4 n=3 E-272Fun Algebra to Check Thoroughly If p is prime and p | ab, then p | a or p | b. (Euclid’s Lemma) If gcd(r, s) = 1 and r · s = tn , then r = un and s = v n . The set S = {x2 + ny 2 | x, y ∈ Z} is closed under multiplication. (a +nb2 )(c2 +nd2 ) = (ac±nbd)2 +n(ad bc)2 2 (Later in this talk, we will consider n = 3.) Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 3 / 33
4. 4. n=2 n=4 n=3 E-272A Babylonian Tablet Figure: Plimpton 322 Also see: Euclid, Book X, Lemma 1 - Proposition XXIX Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 4 / 33
5. 5. n=2 n=4 n=3 E-272Proving a Pythagorean Parameterization Assume a solution exists: x2 + y 2 = z 2 , with x, y, z ∈ Z+ , relatively prime. Some cases to consider: odd + odd = even ////// +////// = even even/// even////////// even + odd = odd ///// +////// =///// odd/// even//// odd Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 5 / 33
6. 6. n=2 n=4 n=3 E-272(odd)2 + (odd)2 = (even)2 This equation is impossible! (2m + 1)2 + (2n + 1)2 = (2p)2 4(m2 + m + n2 + n) + 2 = 4p2 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 6 / 33
7. 7. n=2 n=4 n=3 E-272(even)2 + (odd)2 = (odd)2 x2 + y 2 = z 2 x2 = z 2 − y 2 x2 = (z + y)(z − y) z + y = even = 2p z − y = even = 2q x = 2r, y = p − q, z = p + q p and q: relatively prime, opposite parity Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 7 / 33
8. 8. n=2 n=4 n=3 E-272Finishing it up... x2 = (z + y)(z − y) r2 = p · q gcd(p, q) = 1 implies p = a2 and q = b2 Putting it all together: √ x = 2r = 2 pq = 2ab y = p − q = a2 − b 2 z = p + q = a2 + b 2 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 8 / 33
9. 9. n=2 n=4 n=3 E-272The Pythagorean Parameterization Theorem If x2 + y 2 = z 2 , with x, y, z relatively prime, then there exist integers a and b, relatively prime and with opposite parity, such that x = 2ab y = a2 − b2 z = a2 + b2 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 9 / 33
10. 10. n=2 n=4 n=3 E-272Fermat Figure: Pierre de Fermat, 1601-1665 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 10 / 33
11. 11. n=2 n=4 n=3 E-272Method of Inﬁnite Descent Prove: if P (x) is true, then there exists y < x with P (y) true, where x, y ∈ Z+ . Obtain an inﬁnite sequence of strictly decreasing positive integers. Contradicts the Well Ordering Principle: S ⊆ Z+ has a smallest element. Conclude: initial assumption is false. Useful for showing solutions do not exist. Fermat’s account of this method: “Relation des nouvelles d´couvertes en la science des nombres” e letter to Pierre de Carcavi, 1659. Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 11 / 33
12. 12. n=2 n=4 n=3 E-272Wanted: Larger Margins Fermat’s annotation of Bachet’s translation of Diophantus’ Arithmetica Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 12 / 33
13. 13. n=2 n=4 n=3 E-272A proof Assume a solution exists: x4 + y 4 = z 4 , with x, y, z ∈ Z+ , relatively prime. Let X = x2 , Y = y 2 , Z = z 2 , then: X2 + Y 2 = Z2 X = 2ab, Y = a2 − b2 , Z = a2 + b2 a and b: relatively prime, opposite parity Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 13 / 33
14. 14. n=2 n=4 n=3 E-272The descent Y = a2 − b2 implies b2 + y 2 = a2 b = 2cd, y = c2 − d2 , a = c2 + d2 c and d: relatively prime, opposite parity X = 2ab implies x2 = 4cd(c2 + d2 ) cd and (c2 + d2 ): relatively prime cd = e2 and c2 + d2 = f 2 and c = g 2 , d = h2 ... let the descent begin... Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 14 / 33
15. 15. n=2 n=4 n=3 E-272Recap Assume there are positive integers such that X 2 + Y 2 = Z 2 , where X = x2 and Y = y 2 Obtain positive integers such that c2 + d2 = f 2 , where c = g 2 and d = h2 f is strictly smaller than Z We obtain an inﬁnite sequence of strictly decreasing positive integers, which is impossible; the original assumption was false. Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 15 / 33
16. 16. n=2 n=4 n=3 E-272Euler Figure: Leonhard Euler, 1707-1783 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 16 / 33
17. 17. n=2 n=4 n=3 E-272Correspondence with Goldbach 196 letters from 1729 to 1764 Goldbach motivates Euler to examine Fermat’s work 1748 - Euler ﬁrst mentions Fermat’s Last Theorem 1753 - Euler announces proof for n = 3 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 17 / 33
18. 18. n=2 n=4 n=3 E-272Euler to Goldbach, 13 February 1748 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 18 / 33
19. 19. n=2 n=4 n=3 E-272Euler to Goldbach, 04 August 1753 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 19 / 33
20. 20. n=2 n=4 n=3 E-272Euler’s Algebra (1770) Assume a solution exists: x3 + y 3 = z 3 , with x, y, z ∈ Z+ , relatively prime. Exactly one of these three numbers are even. Case 1: x, y are odd and z is even. Case 2: y, z are odd and x is even. Proof of Case 1. Assume that x > y. x + y = even = 2p and x − y = even = 2q x = p + q and y = p − q p and q: positive, relatively prime, opposite parity Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 20 / 33
21. 21. n=2 n=4 n=3 E-272The proof continues z 3 = x3 + y 3 = (p + q)3 + (p − q)3 = 2p3 + 6pq = 2p(p2 + 3q 2 ) Note: p2 + 3q 2 is odd. If g = gcd(2p, p2 + 3q 2 ) > 1 then g is odd; g | p, so g | 3q 2 ; since g q, we have g = 3. Therefore, gcd(2p, p2 + 3q 2 ) = 1 or 3. (Two subcases to consider.) Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 21 / 33
22. 22. n=2 n=4 n=3 E-272gcd(2p, p2 + 3q 2) = 1 2p(p2 + 3q 2 ) = z 3 By earlier fact: 2p = u3 and (p2 + 3q 2 ) = v 3 . Know: v ∈ S → v 3 ∈ S. Also: v 3 ∈ S → v ∈ S p2 + 3q 2 = v 3 = (a2 + 3b2 )3 Since gcd(p, q) = 1, we have: p = a3 − 9ab2 , q = 3a2 b − 3b3 , gcd(a, b) = 1. 2p = 2a(a + 3b)(a − 3b) = u3 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 22 / 33
23. 23. n=2 n=4 n=3 E-272A descent appears! 2p = 2a(a + 3b)(a − 3b) = u3 2a, (a + 3b), (a − 3b) are relatively prime, so: 2a = α3 , (a + 3b) = β 3 , (a − 3b) = γ 3 α3 = β 3 + γ 3 Move terms if necessary so all terms are positive. α, β, γ < z, since α3 β 3 γ 3 = 2p < 2p(p2 + 3q 2 ) = z 3 ...a smaller positive solution to FLT, n = 3. Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 23 / 33
24. 24. n=2 n=4 n=3 E-272Excerpt #1, Euler’s Algebra Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 24 / 33
25. 25. n=2 n=4 n=3 E-272Excerpt #2, Euler’s Algebra Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 25 / 33
26. 26. n=2 n=4 n=3 E-272Commentarii... Euler proved FLT(3) in 1770. Euler proved FLT(3) in 1770, but key steps were unjustiﬁed. Euler proved FLT(3) in 1770, but key steps were justiﬁed in 1759/1763 (E-272). The results of E-272 are insuﬃcient to prove FLT(3). Euler had a proof of FLT(3) by 1753, but waited to publish a more polished version. Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 26 / 33
27. 27. n=2 n=4 n=3 E-272Red Flags Revisited Need to fully justify: 1 If p2 + 3q 2 = v 3 (also gcd(p, q) = 1, and v 3 is odd) then v = a2 + 3b2 . 2 In this situation, p2 + 3q 2 = (a2 + 3b2 )3 ⇒ p = a3 − 9ab2 ⇒ q = 3a2 b − 3b2 Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 27 / 33
28. 28. n=2 n=4 n=3 E-272The missing link? E272: Supplementum quorundam theorematum arithmeticorum... (1759/1763) Proves properties of numbers of the form x2 + 3y 2 . Quick tour: 1 If gcd(a, b) = m, then m2 |(a2 + 3b2 ) 2 2 2 If 3|(a2 + 3b2 ), then a +3b = n2 + 3m2 . 3 2 2 3 If 4|(a2 + 3b2 ), then a +3b = n2 + 3m2 . 4 4 If P = p + 3q is prime and P |(a2 + 3b2 ), 2 2 2 2 then a +3b = n2 + 3m2 . P Corollary: a = 3mq ± np and b = mp ± nq Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 28 / 33
29. 29. n=2 n=4 n=3 E-272 5 If Pi = (pi )2 + 3(qi )2 is prime and 2 2 Pi |(a2 + 3b2 ), then a 1+3bk = n2 + 3m2 . P ...P 6 If A = p2 + 3q 2 is prime and A|(a2 + 3b2 ), then there exists a similar B < A. 7 All odd prime factors of a2 + 3b2 , when gcd(a, b) = 1, have the form p2 + 3q 2 . 8 Primes of the form p2 + 3q 2 (except 3) have the form 6n + 1. 9 Primes of the form 6n + 1 have the form p2 + 3q 2 . Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 29 / 33
30. 30. n=2 n=4 n=3 E-272Red Flag # 1 “If p2 + 3q 2 = v 3 (also gcd(p, q) = 1, and v 3 is odd) then v = a2 + 3b2 .” Fully justiﬁed by Euler’s Proposition 7. Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 30 / 33
31. 31. n=2 n=4 n=3 E-272Red Flag # 2 “In this situation, p2 + 3q 2 = (v)3 = (a2 + 3b2 )3 ⇒ p = a3 − 9ab2 and q = 3a2 b − 3b2 ” Can be addressed by Cor. to Euler’s Prop. 4: Applied to a prime a2 + 3b2 , yields uniqueness of representation. (p2 + 3q 2 )(12 + 3 · 02 ) = (a2 + 3b2 ) a = 3 · 0 · q ± 1 · p and b = 0 · p ± 1 · q Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 31 / 33
32. 32. n=2 n=4 n=3 E-272 “In this situation, p2 + 3q 2 = (v)3 = (a2 + 3b2 )3 ⇒ p = a3 − 9ab2 and q = 3a2 b − 3b2 ” Repeat for each prime power factor of v; only one of the two methods of composition will preserve gcd(a, b) = 1. Combine prime power factors; many representations of v. Must use the same representation to calculating v 3 to preserve gcd(p, q) = 1. Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 32 / 33
33. 33. n=2 n=4 n=3 E-272Did Euler consider this? Similar work on sums of two squares Unusual for Euler to not publish refutation of FLT for n = 3? E-255, x3 + y 3 = z 3 + v 3 (pres. 1754) (one year after letter to Goldbach) E-256, x2 + cy 2 conjectures (pres. 1753/4) 1755 letter to Goldbach: convinced Fermat was correct, searching for FLT n = 5 proof E-272, x2 + 3y 2 (pres. 1759) Lee Stemkoski (Adelphi) Euler on Fermat’s Last Theorem December 5, 2012 33 / 33