Arithmetic Progressions over Quadratic Fields

Transcription

1 uadratic Fields ( D) Alexer Díaz University of Puerto Rico, Mayaguez Zachary Flores Michigan State University Markus Oklahoma State University Mathematical Sciences Research Institute Undergraduate Program 23 July 2010

2 Outline 1 ( D)

3 Outline 1 2 ( D)

4 Outline ( D)

5 Outline ( D) ( D)

6 ( D) Let k be a field. We say that the n-tuple (a 2 1, a2 2,, a2 n) is an arithmetic progression of n squares over k if a 2 2 a2 1 = a2 3 a2 2 = = a2 n a 2 n 1. Example: Consider the progression (1, 25, 49). In 1640, Pierre de Fermat wrote a letter to French number theorist Bernard Frénicle de Bessy in which he claimed that there are no nonconstant arithmetic progressions of four squares. Leonhard Euler proved this in 1780, though there is some controversy as to whether the proof was by Euler or his students.

7 ( D) We say that a 2, b 2 c 2 form an arithmetic progression of three squares if b 2 a 2 = c 2 b 2. Rewrite this as 2b 2 = a 2 + c 2. Finding (a 2, b 2, c 2 ) is equivalent to finding rational points on the circle x 2 + y 2 = 2 where x = a/b y = c/b. Proposition Let m Q. The following formulas give an arithmetic progression of three squares (a 2, b 2, c 2 ) : (a : b : c) = (m 2 2m 1 : m : m 2 2m + 1).

8 ( D) Proof. Parametrize all the rational points (x, y) on the circle x 2 + y 2 = 2 obtain: x = m2 2m 2 m 2 + 1, y = m2 2m + 1 m Set x = a/b y = c/b, which give the desired formula.

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10 ( D) Consider the curve E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6. Using a substitution we can convert the curve E into the form Y 2 = X 3 + AX + B. This is a nonsingular curve if only if 4A B 2 0. A nonsingular curve in the form defined above is called an elliptic curve.

11 Examples of Example Elliptic curve non-elliptic curve: Figure: Y 2 = X 3 X, D = 4 Y 2 = X 3 + X 2, D = ( D)

12 Theorem (Mordell s Theorem). Let E be an elliptic curve defined. Then E(Q) is a finitely generated abelian group under. In particular, E(Q) = E(Q) tors Z r. Define r to be the rank of the elliptic curve. ( D)

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14 The ( D) Theorem (Mazur s Theorem). Let E be a rational elliptic curve, let E(Q) tors denote its torsion subgroup. This finite group can only be one of the fifteen types: { E(Q) tors Z N for N=1,2,3,4,5,6,7,8,9,10,12; = Z 2 Z 2N for N=1,2,3,4. Moreover, each of these possibilities does occur. Example Consider an elliptic curve E : y 2 = (x e 1 )(x e 2 )(x e 3 ) where e i Q. The three points T i = (e i : 0 : 1) have order two. By Mazur s Theorem, we must have that E(Q) tors = Z2 Z 2N.

15 The Multiplication by 2 Homomorphism ( D) Define the homomorphism [2] : E(Q) E(Q) by P P P. The kernel of the homomorphism, E[2], is the group of points of order two together with the identity. When E : y 2 = (x e 1 )(x e 2 )(x e 3 ) with e i Q, 2E(Q) = Z N (2Z) r. Then, ( ) E(Q) 2E(Q) = Z2 r+2 E(Q) r = log 2 2E(Q) 2.

16 The Tate Pairing ( D) Theorem Consider E : y 2 = (x e 1 )(x e 2 )(x e 3 ) with e i Q. Define the map: ɛ 2 : E(Q)/2E(Q) E[2] Q x /(Q x ) 2 by { (X e i ) if T = T i = (e i : 0 : 1), (P, T ) 1 if T = (0 : 1 : 0). This map is a perfect pairing, by which we mean: (Non-Degeneracy) If ɛ 2 (P, T ) = 1 for all T E[2], then P 2E(Q). (Bilinearity) For all P, Q E(Q) T E[2] we have ɛ 2 (P Q, T ) = ɛ 2 (P, T ) ɛ 2 (Q, T ) ɛ 2 (P, T 1 T 2 ) = ɛ 2 (P, T 1 ) ɛ 2 (P, T 2 ).

17 The Homomorphism δ E ( D) Theorem Consider E : y 2 = (x e 1 )(x e 2 )(x e 3 ) with e i Q. Define the map δ E : E(Q) 2E(Q) Qx (Q x ) 2 Qx (Q x ) 2, P (ɛ 2 (P, T 1 ), ɛ 2 (P, T 2 )) where ɛ 2 is the Tate Pairing. The map δ E is an injective group homomorphism.

18 The Homomorphism δ E (continued) ( D) Proposition Let S i be the set of primes that divide (e i e j )(e i e k ) define { } Q(S i, 2) = d Q (Q ) 2 d ( 1)m 0 p m pn mn mod (Q ) 2, where m j {0, 1} p j S i. The image of δ E lies in the finite abelian group Q(S 1, 2) Q(S 2, 2) = Z 2+#S 1+#S 2 2. Moreover, a point d = (d 1, d 2 ) Im(δ E ) if only if there is a rational point on the curve C d : d 1 u 2 d 2 v 2 = (e 2 e 1 ), d 1 u 2 d 1 d 2 w 2 = (e 3 e 1 ).

19 ( D) Computing Im(δ E ) exactly can be rather difficult, so we must sometimes content ourselves with finding an upper bound on the rank. To find an upper bound, we try to find the points d Q(S 1, 2) Q(S 2, 2) \ Im(δ E ). We have three methods at our disposal to show that a point d Im(δ E ). 1 The curve C d cannot have a rational point because the square of a real number is nonnegative. 2 We use the closure property that Im(δ E ) has as a group. 3 We homogenize the curve C d show that the new curve has no solutions over Z/mZ for some m Z.

20 Examples of Discarding Points from Q(S 1, 2) Q(S 2, 2) ( D) Example In the following examples, consider the elliptic curve E (D) : y 2 = x 3 + 5Dx 2 + 4D 2 x. Here, e 1 = 0, e 2 = D, e 3 = 4D. Let D > 0. Then the point (1,-1) Im(δ E (D)). The curve C d in this case is given by u 2 + v 2 = D, u 2 + w 2 = 4D. Note that u 2 + v 2 0, but D < 0. Example Let D > 0. Then the point ( D, 3D) Im(δ E (D)). We know that the point ( D, 3D) Im(δ E (D)) because it is the image of the point ( 4D : 0 : 1) E (D) (Q) tors. Then, ( D, 3D) ( D, 3D) = (1, 1) Im(δ E (D)). Since Im(δ E (D)) is closed under multiplication, the result follows.

21 Examples of Discarding Points from Q(S 1, 2) Q(S 2, 2) (continued) ( D) Example Let D 2 (mod 3). Then the point (2, 1) Im(δ E (D)). Suppose (2, 1) Im(δ E (E)). Then there is a rational point on the curve C d : 2u 2 v 2 = D, 2u 2 2w 2 = 4D. We write u = a/d, v = b/d w = c/d with gcd(a, b, c, d)=1. We then have 2a 2 b 2 = Dd 2 2a 2 2c 2 = 4Dd 2. Looking at these equations modulo 3, we get 2a 2 + 2b 2 d 2 (mod 3) 2a 2 + c 2 d 2 (mod 3). The only solution to these congruences has a, b, c, d 0 (mod 3), contradicting our assumption that gcd(a, b, c, d) = 1.

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23 A Bijection with E : y 2 = x 3 + 5x 2 + 4x ( D) Proposition Let k be a field with characteristic neither 2 nor 3. There exists a bijection between four squares in arithmetic progression E(k), the set of k-rational points on E : y 2 = x 3 + 5x 2 + 4x. Proof. Given four squares in arithmetic progression (a 2, b 2, c 2, d 2 ) with (a : b : c : d) (1 : 1 : 1 : 1), the following point is an element of E(k): ( ( ) a 3b 3c + d 2 a + 3b + 3c + d ( ) ) a + b c d : 6 : 1 a + 3b + 3c + d if (a : b : c : d) = (1 : 1 : 1 : 1), then we get the point (0 : 0 : 1).

24 A Bijection with E (continued) ( D) Proof. Given a point (x : y : z) (0 : 0 : 1) E(k), then (a : b : c : d) is an arithmetic progression of four squares over k, where Notice that a = 6x + 3x 2 2y + xy, b = 2x x 2 2y xy, c = 2x x 2 + 2y + xy, d = 6x + 3x 2 + 2y xy. b 2 a 2 = c 2 b 2 = d 2 c 2 = 16xy 4x 3 y. If (x : y : z) = (0 : 0 : 1), then we get the arithmetic progression (1 : 1 : 1 : 1).

25 Constant ( D) Proposition The arithmetic progressions of four squares that are mapped to points in E(Q) tors satisfy (a : b : c : d) = (±1 : ±1 : ±1 : ±1). Proof. Nagell-Lutz with some computation gives us the following points in E(Q) tors (with corresponding arithmetic progressions): Point Progression (2 : 6 : 1) (1 : 1 : 1 : 1) (2 : 6 : 1) (1 : 1 : 1 : 1) ( 2 : 2 : 1) ( 1 : 1 : 1 : 1) ( 2 : 2 : 1) (1 : 1 : 1 : 1) ( 1 : 0 : 1) ( 1 : 1 : 1 : 1) ( 4 : 0 : 1) (1 : 1 : 1 : 1) (0 : 1 : 0) ( 1 : 1 : 1 : 1) (0 : 0 : 1) (1 : 1 : 1 : 1).

26 The : y 2 = x 3 + 5x 2 + 4x Lemma The rank of E : y 2 = x 3 + 5x 2 + 4x is zero. ( D) Proof. Compute the rank via. Write E as y 2 = x(x + 4)(x + 1). Let e 1 = 0, e 2 = 1, e 3 = 4 so that (e 1 e 3 )(e 1 e 2 ) = 4 (e 2 e 1 )(e 2 e 3 ) = 3. Then S 1 = {2} S 2 = {3} Q(S 1, 2) = {±1, ±2}, Q(S 2, 2) = {±1, ±3}. We also have δ E (E(Q) tors ) = {( 1, 3), (2, 3), ( 2, 1), (1, 1)}.

27 The (continued) ( D) Proof. Consider the curves C d : d 1 u 2 d 2 v 2 = 1, d 1 u 2 d 1 d 2 w 2 = 4. We immediately see from the equation d 1 u 2 d 2 v 2 = 1 that the points (1, 1), (2, 1), (1, 3), (2, 3) cannot be in the image of δ E. Then since δ E is a homomorphism we may also conclude that ( 1, 3), ( 2, 3), ( 1, 1), ( 2, 1), ( 2, 3), (1, 3), (2, 1), ( 1, 1) are not in the image.

28 The Torsion Subgroup of E : y 2 = x 3 + 5x 2 + 4x ( D) Proposition Let E : y 2 = x 3 + 5x 2 + 4x. Then E(Q) = Z 2 Z 4. Proof. From previous proposition we know that the rank of E is zero. Therefore, E(Q) = E(Q) tors. Several computations will confirm that each of the 8 points in the previous table has finite order. The only torsion subgroup of order eight admitted by Mazur s Theorem that has three elements of order two is Z 2 Z 4.

29 ( D) Theorem Let n 4 be an integer, there are no non-constant arithmetic progressions of n squares. Proof. Pick the first four squares. From the previous Lemma we know that E has rank 0 hence that E(Q) = Z 2 Z 4 from previous propositions we know that all of the points in E(Q) tors are mapped to constant arithmetic progressions. Therefore, the difference between each consecutive terms is 0. Hence, there are no non-constant n squares arithmetic progressions.

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31 of over Q( D) ( D) Unlike Q, it is sometimes possible to find arithmetic progressions of four squares ( D) where D is a squarefree integer. We show that rational points on the curve E (D) give us arithmetic progressions of four squares ( D). This allows us to use the techniques for finding ranks of elliptic curves to discuss the existence of arithmetic progressions.

32 The Elliptic Curve E (D) ( D) Proposition Fix a nonzero squarefree integer D let E (D) : y 2 = x 3 + 5Dx 2 + 4D 2 x. If P = (x : y : 1) is a rational point on E (D) then there is a corresponding arithmetic progression in Q( D). Proof. Given a rational point (x : y : 1) E (D) (Q), then a 2, b 2, c 2 d 2 are four squares in arithmetic progression where a = 6x D + 3x 2 D 2 c = 2x D x 2 D 2 + 2y D 3 + 2y D 3 + xy, b = 2x D 5 D x 2 D 2 xy, d = 6x D 5 D + 3x 2 D 2 + 2y D 3 2y D 3 xy D 5, xy D 5.

33 The Elliptic Curve E (D) (continued) ( D) Proposition Let D = tp, where t {±1, ±2, ±3, ±6} p is a prime congruent to one of 1, 5, 7, 11, 13, 17, 19, 23 modulo 24. There is a non-constant arithmetic progression of four squares ( D) if only if the rank of E (D) is nonzero. If the rank is nonzero, then there exist infinitely many progressions. Proof. See Corollary 2 of of uadratic Fields by Enrique Gonzalez-Jimenez Jorn Steuding.

34 Upper Bounds on the (D) ( D) p 5 prime Upper Bounds on Ranks with Highest Ranks Found p mod 24 D = p D = 2p D = 3p D = 6p 1 2, (73, 2) 2, (146, 2) 2, (579, 2) 3, (22038, 3) 5 0, (5, 0) 1, (10, 1) 2, (15, 0) 2, (30, 1) 7 1, (7, 0) 0, (14, 0) 2, (21, 1) 2, (42, 1) 11 1, (11, 1) 2, (22, 1) 2, (33, 0) 2, (66,1) 13 1, (13, 1) 0, (26, 0) 2, (39, 2) 2, (78, 1) 17 1, (17, 1) 2, (34, 1) 2, (51, 0) 3, (102, 1) 19 1, (19,0) 2, (134, 2) 2, (57, 0) 2, (114, 1) 23 1, (23, 1) 1, (46, 1) 2, (69, 1) 2, (138, 1)

35 Upper Bounds on the (D) (continued) ( D) p 5 prime Upper Bounds on Ranks with Highest Ranks Found p mod 24 D = p D = 2p D = 3p D = 6p 1 2, (-97, 2) 2, (-146, 2) 2, (-939, 2) 2, (-582, 2) 5 2, (-5, 1) 1, (-10, 1) 2, (-15, 1) 2, (-30, 0) 7 1, (-7, 0) 2, (-14, 2) 2, (-21, 1) 2, (-42, 0) 11 1, (-11,0) 1, (-22, 1) 2, (-33, 1) 3, (-66, 2) 13 2, (-13, 0) 1, (-26, 0) 2, (-39, 1) 2, (-78, 0) 17 2, (-17, 1) 1, (-34, 1) 2, (-51, 0) 2, (-102, 0) 19 1, (-19,1) 1, (-38, 0) 2, (-57, 1) 3, (-114, 2) 23 1, (-23, 1) 1, (-46, 1) 2, (-69, 1) 2, (-1002, 2)

36 Some Explicit Four Square ( D) ( D) D P Progression -5 ( ) 2 ( ) (4 : 8 : 1) 2 ( ) ( 48 : 216 : 1) (2 : 36 : 1) ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2 ( ) 2

37 Conclusion Further Research ( D) There are infinitely many three squares in arithmetic progression. There are no non-constant n squares in arithmetic progression, when n 4. There are some squarefree integers D such that there exist non-constant four squares in arithmetic progression over Q( D). If we can find properties of primes modulo 24, we may be able to prove that there exist four squares in arithmetic progression ( D). Let p 13 mod 24, we can find rationals u w that satisfy C d, v = xy = ( 2bm + m 2 a a + b)( bm 2 2ma + b + m 2 a) 1 m + m 2.

38 ( D) Acknowledgements We would like to thank our mentor Ebony Harvey, academic advisor Dr. Edray Goins, program director Dr. Duane Cooper the MSRI Staff. This work was conducted during the 2010 Mathematical Sciences Research Institute Undergraduate Program (MSRI-UP) which is supported by the National Science Foundation (grant No. DMS ) the National Security Agency (grant No. H ).