Arithmetic Progressions Over Quadratic Fields

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1 Arithmetic Progressions Over Quadratic Fields Alexander Diaz, Zachary Flores, Markus Vasquez July 2010 Abstract In 1640 Pierre De Fermat proposed to Bernard Frenicle de Bessy the problem of showing that there is no arithmetic progression of four squares over Q. The solution to this problem was published posthumously in 1780 by Euler. However, this does not hold over Q( D) where D is a squarefree integer. Arithmetic progressions of four squares over Q( D) correspond to rational points on the elliptic curve E : y 2 = x 3 + 5x 2 + 4x and this is related to the quadratic twist E (D) : y 2 = x 3 + 5Dx 2 + 4D 2 x. We use methods of elliptic curves to discuss the existence of arithmetic progressions of squares over Q( D). 1 Introduction Since 1640 s people have been trying to find arithmetic progressions. That is, finding a sequence of numbers a 2, b 2 and c 2 such that b 2 a 2 = c 2 b 2. Fermat sent a letter to Frenicle in 1640 in which he stated that there are no non-constant four squares arithmetic progression over Q, but he did not share the proof. Later in 1780, Euler proved the same statement above, although it was published by his assistants. After that, some other mathematicians, like Weil and Itard, have worked on the same problem. In Section 2 of this paper, we show how to find three squares arithmetic progressions over Q by finding rational points on the circle x 2 + y 2 = 2 and how to find three squares arithmetic progressions with fixed common difference n by finding rational points on the elliptic curve y 2 = x 3 n 2 x. In Section 3, we introduce a method that is crucial in order to prove the main theorems of this paper. In Section 5, we prove that there are no non-constant four squares arithmetic progressions with fixed common difference n over Q. Finally, in Section 6, we present how to find such progressions in quadratic extensions Q( D) by finding rational points on the elliptic curve y 2 = x 3 + 5D + 4D 2. 2 Three Squares Arithmetic Progressions over Q By definition, a three squares arithmetic progression (a 2, b 2, c 2 ) satisfies b 2 a 2 = c 2 b 2. We can rewrite this equation to be 2b 2 = a 2 + c 2. If b = 0 then a = 0 and c = 0, which is the trivial three squares arithmetic progression (0, 0, 0). Assume then 1

2 that b 0 and divide the equation by b 2. We obtain the equivalent formula (a/b) 2 + (c/b) 2 = 2. Then finding a three squares arithmetic progression is equivalent to finding rational points on the circle x 2 +y 2 = 2. This leads us to our first proposition. Proposition 1. A three squares arithmetic progression (a 2, b 2, c 2 ) can be obtained by the following formulas: a 2 = (m 2 2m 1) 2, b 2 = (m 2 +1) 2 and c 2 = ( m 2 2m+1) 2. Moreover, if m is rational, then a 2, b 2 and c 2 are rationals. Proof. In order to find an arithmetic progression, we parametrize the points on the circle x 2 + y 2 = 2. Pick the point P = (1, 1) on the circle and draw a line through P and a point (x 1, y 1 ). The equation of the line is defined as y = mx + b where m is the slope and b the y-intercept. Since the line passes through P, we have b = 1 m. To find the coordinates of the second point (x 1, y 1 ), substitute y = mx + (1 m) into the equation of the circle. This give us the following equation: x 2 + (mx + (1 m)) 2 = 2. Factoring we obtain: ( ) 2m(1 m) x 2 + x + m2 2m 2 = 0. m m Since we know the roots of this polynomial are x = 1 and x = x 1 then the multiplication of the roots gives us the last term of the polynomial, that is: x 1 = m2 2m 2. m Using the equation of the line, we obtain: x 2 = m2 2m + 1. m We have parametrized the points on the circle x 2 + y 2 = 2 and now we can set a/b = x 1 and c/b = x 2, which will give us the desired formulas. It is clear that a 2, b 2 and c 2 are rationals if m is rational. The following proposition will be useful to prove that finding a three squares arithmetic progression with common difference n is equivalent to finding rational points on the elliptic curve y 2 = x 3 n 2 x. Proposition 2. For fixed n 0, finding a three squares arithmetic progression of rational squares with common difference n is equivalent to finding a rational point (m, s) in the equation ns 2 = m 3 m. Proof. To prove the statement we show that there exists a bijection between three squares arithmetic progressions (a 2, b 2, c 2 ) and the rational points on the equation. Pick a rational arithmetic progression (a 2, b 2, c 2 ) with common difference n. Then ((a/b) 2, 1, (c/b) 2 ) is an arithmetic progression with common difference n/b 2. So, n/b 2 = 1 (a/b) 2. Now by Proposition 1, (a/b) 2 = (m 2 2m 2) 2 /(m 2 + 1) 2 2

3 and m = (c b)/(a b) where m is the slope of the line through (1, 1) and (a/b, c/b). Then ( ) n m 2 2 b = 1 2m 2 2 m ( ) 2 n 2 b = (m 3 m). 2 m Let s = (m 2 + 1)/2b then ns 2 = m 3 m. Now, conversely, pick a rational point (m, s) on the equation ns 2 = m 3 m. Then, set a = (m 2 2m 1)/2s, b = (m 2 + 1)/2s and c = ( m 2 2m + 1)/2s. We obtain the arithmetic progression (a 2, b 2, c 2 ) with common difference n = (m 3 m)/s 2. Proposition 3. For fixed n 0, finding a three squares arithmetic progression of rational squares with common difference n is equivalent to finding a rational point (x, y) in the elliptic curve y 2 = x 3 n 2 x. Proof. Proposition 2 show that there is a bijection between an arithmetic progression (a 2, b 2, c 2 ) and the equation ns 2 = m 3 m. Now we show there is a bijection between this last equation and the elliptic curve y 2 = x 3 n 2 x. First, multiply ns 2 = m 3 m by n 3 and we get: (n 2 s) 2 = (nm) 3 n 2 (nm). Make the substitution x = nm, y = n 2 s and we get y 2 = x 3 n 2 x. On the other hand, divide y 2 = x 3 n 2 x by n 3 and we obtain ny 2 n 4 = x3 n 3 n2 x n 3. Make the substitution m = x/n, s = y/n 2 and we get ns 2 = m 3 m. Our next proposition is crucial in order to find infinitely many three squares arithmetic progressions with common difference n. Proposition 4. For fixed n 0, the only rational points with finite order on the elliptic curve y 2 = x 3 n 2 x are (n, 0), ( n, 0) and (0, 0). Proof. Let n = dt 2 where d is the squarefree part of n. Then there is a bijection between the curves y 2 = x 3 n 2 x and y 2 = x 3 d 2 x given by (x, y) (x/t 2, y/t 3 ) that sends the points (n, 0), ( n, 0) and (0, 0) to (d, 0), ( d, 0) and (0, 0). Then we can assume n is a squarefree integer. Pick a rational point P = (x 1, y 1 ) on the curve y 2 = x 3 n 2 x with finite order and assume y 1 0. Then since P has finite order by Nagell-Lutz theorem, the coordinates of P are integers. Let r be the x-coordinate of P P. Then ( ) x 2 r = 1 + n y 1 3

4 Using Nagell-Lutz theorem we know r is an integer. Note that ( ) x 2 r n = 1 2x 1 n n 2 2 2y 1 and ( ) x 2 r + n = 1 + 2x 1 n n 2 2 2y 1 which means we have created a three squares arithmetic progression namely (r n, r, r + n) with common difference n. Then we know an even difference between two squares is divisible by 4 because if you look at (r + n) (r n) = 2n mod 4, the only possible value of 2n is 0 mod 4. Since 2n is divisible by 4, then n is even. Using the same argument we conclude n 0 mod 4 which is a contradiction since n is squarefree. Then the only rational points with finite order are the points with y-coordinate zero, namely (n, 0), ( n, 0) and (0, 0). Now we can state our next proposition: Proposition 5. If there exists a three squares arithmetic progression with common difference n, then there exist infinitely many three squares arithmatic progressions with the same difference n. Proof. Let say we have a three squares arithmetic progression (a 2, b 2, c 2 ), then by Proposition 3 this sequence is map to a rational point on the elliptic curve y 2 = x 3 n 2 x. By Proposition 4, this point has to have infinite order. Then you can add the point infinitely many times to get an infinite amount of points which is equivalent to an infinite amount of three squares arithmetic progression with common difference n. 3 2-Descent Method In order to determine if an elliptic curve has rational points we define a method called 2-Descent that will help us answer that question. The main idea of the 2-Descent method is to calculate the rank of an elliptic curve and therefore determine how many rational points are on an elliptic curve. At the end of the section, we calculate the rank of the elliptic curve y 2 = x 3 + 5x 2 + 4x, which is crucial in Section 5. Before explaining the method, we recall three very important theorems in Elliptic Curve s theory. Theorem 1. (Fundamental Theorem of Finitely Generated Abelian Groups). Let G be a finitely generated abelian group under. Denote the set G tors = {a G [m]a = e for some nonzero integer m} as the torsion subgroup of G. 1. G tors = ZN1 Z N2 Z Ns is a finite subgroup of G. 2. G/G tors = Z r is a free abelian group of rank r. 3. G = G tors Z r. 4

5 Proof. See [1]. Define E(Q) to be the set of rational points on the elliptic curve E and the operation in which you take two points of E(Q) and do the chord-tangent method to get a third point. Theorem 2. (Mordell s Theorem). Let E be an elliptic defined over Q. Then E(Q) is a finitely generated abelian group. In particular, Proof. See [3]. E(Q) = E(Q) tors Z r. Theorem 3. (Mazur s Theorem). Let E be a rational elliptic curve, and 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 i.e., given one of these fifteen finite groups T, there exists a rational elliptic curve E such that E(Q) tors = T. Proof. See [2]. Proposition 6. Consider the elliptic curve E : y 2 = (x e 1 )(x e 2 )(x e 3 ) with e i Q. Then, E(Q) tors = Z2 Z 2N, where N = 1, 2, 3, 4. Proof. It is easy to verify that the rational points (e 1 : 0 : 1), (e 2 : 0 : 1) and (e 3 : 0 : 1) have order two. By Theorem 3, the only possibilities for E(Q) tors with the above description are Z 2 Z 2N, N = 1, 2, 3, 4. Define the m-multiplication map as [m] : E(Q) E(Q) by sending (0 : 1 : 0) if m = 0 P [m]p = P P P if m > 0 [ 1]P [ 1]P [ 1]P if m < 0 Proposition 7. Let E be an elliptic curve in the form y 2 = (x e 1 )(x e 2 )(x e 3 ) with e i Q and 2E(Q) be the image of the [2] map. Then ( ) E(Q) # = 2 r+2. 2E(Q) Proof. Using Theorem 2 and Proposition 6, E(Q) = E(Q) tors Z r = Z2 Z 2N Z r. Then the image under the multiplication by [2] map is: 2E(Q) = Z N (2Z) r E(Q) 2E(Q) = (Z 2 ) r+2. 5

6 Therefore, ( ) E(Q) # = 2 r+2. 2E(Q) Consider E : y 2 = (x e 1 )(x e 2 )(x e 3 ) with e i k (for our purposes we will have e i Z). Define the Tate Paring: ɛ 2 : E(Q) Qx E[2] 2E(Q) (Q x ) 2 (P, T i ) (X e i ) mod (Q x ) 2. Where E[2] = {(e 1 : 0 : 1), (e 2 : 0 : 1), (e 3 : 0 : 1), (0 : 1 : 0)} and T i = (e i : 0 : 1). Proposition 8. The Tate Paring map ɛ 2, defined above, is a perfect pairing i.e., 1. (Non Degeneracy.) If ɛ 2 (P, T ) = 1 for all T E[2] then P 2E(Q). 2. (Bilinearity.) F or all P, Q E(Q) and 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 ). Proof. Pick P = (p 1 : p 2 : p 3 ) E(k), and say that ɛ 2 (P, T ) for all T E[2]. To show P 2E(k) it suffices to exhibit P E(k) such that P = [2]P. If P = O we may choose P = O as well, so assume p 0 0. Upon considering T = (e : 0 : 1), we see that f i = p1 p 0 e i k for i = 1, 2, 3; we choose the signs so that p 1 p 0 = f 1 f 2 f 3. It is easy to check that the desired k-rational point is ( P (e1 e 3 )(e 2 e 3 ) = (f 1 f 3 )(f 2 f 3 ) + e 3 : (e ) 1 e 3 )(e 1 e 2 )(e 2 e 3 ) (f 1 f 3 )(f 1 f 2 )(f 2 f 3 ) : 1. We show ɛ 2 (P Q, T ) = ɛ 2 (P, T ) ɛ 2 (Q, T ). If T = O there is nothing to show since ɛ 2 (P, T ) = ɛ 2 (Q, T ) = ɛ 2 (P Q, T ) = 1, so assume that T = (e : 0 : 1). Choose two points P = (p 1 : p 2 : p 0 ) and Q = (q 1 : q 2 : q 0 ) in E(k). Draw a line through them, say ax 1 + bx 2 + cx 0 = 0 and assume that it intersects E at a third point R = (r 1 : r 2 : r 0 ). The projective curve E is defined by the homogeneous polynomial F (x 1, x 2, x 0 ) = x 2 2x 0 (x 1 e 1 x 0 )(x 1 e 2 x 0 )(x 1 e 3 x 0 ) so the intersection with the line ax 1 + bx 2 + cx 0 = 0 admits the factorization p 0 q 0 r 0 F (x 1, x 2, x 0 ) = (p 1 x 0 p 0 x 1 )(q 1 x 0 q 0 x 1 )(r 1 x 0 r 0 x 1 ). When (x 1 : x 2 : x 0 ) = (be : ae c : b) is the point where the lines ax 0 +bx 1 +cx 0 = 0 and x 1 ex 0 = 0 intersect, we have the equality ( ) ( ) ( ) b 3 p1 q1 r1 e e e = F (be, ae c, b) = (ae + c) 2 b. p 0 q 0 r 0 6

7 This implies the congruence ɛ 2 (p, T ) ɛ 2 (Q, T ) ɛ 2 (R, T ) 1(mod (k x ) 2 ). We conclude that ɛ 2 (P Q, T ) = ɛ 2 (P, T ) ɛ 2 (Q, T ). We show ɛ 2 (P, T 1 T 2 ) = ɛ 2 (P, T 1 ) ɛ 2 (P, T 2 ). If T 1 = T 2 then ɛ 2 (P, T 1 T 2 ) = ɛ 2 (P, O) = 1 ɛ 2 (P, T 1 ) 2 = ɛ 2 (P, T 1 ) ɛ 2 (P, T 2 ). If T 1 T 2, we may assume T 1 = (e 1 : 0 : 1) and T 2 = (e 2 : 0 : 1). (If either T 1 or T 2 is O there is nothing to show). Then T 1 T 2 = (e 3 : 0 : 1). The identity ( ) ( ) ( ) ( ) 2 p1 p1 p1 p2 e 1 e 2 e 3 = p 0 p 0 p 0 implies the congruence ɛ 2 (P, T 1 ) ɛ 2 (P, T 2 ) ɛ 2 (P, T 1 T 2 ) 1(mod (k x ) 2 ). We conclude that ɛ 2 (P, T 1 T 2 ) = ɛ 2 (P, T 1 ) ɛ 2 (P, T 2 ). p 0 Define S i = {p p (e i e i+1 )(e i e i 1 )} and Q(S i, 2) = {d Q /(Q ) 2 d ( 1) m 0 p m 1... p mn n }, where m i = 0 or 1 and p j S i. Consider the map δ E : E(Q) 2E(Q) where ɛ 2 is the Tate Pairing. Qx (Q x ) Qx 2 (Q x ). 2 P (ɛ 2 (P, T 1 ), ɛ 2 (P, T 2 )), Proposition 9. Let δ i be the map defined above. Then δ i is an injective group homomorphism. Furthermore the image of δ i lying in the finitely generated abelian group generated by the set Q(S i, 2). Proof. First we prove δ i is an injective group homomorphism. For P and Q in E(Q)/2E(Q), then δ i (P Q) = (ɛ 2 (P Q, T 1 ), ɛ 2 (P Q, T 2 )) = (ɛ 2 (P, T 1 ) ɛ 2 (Q, T 1 ), ɛ 2 (P, T 2 ) ɛ 2 (Q, T 2 )) = (ɛ 2 (P, T 1 ), ɛ 2 (P, T 2 )) (ɛ 2 (Q, T 1 ), ɛ 2 (Q, T 2 )) = δ i (P ) δ i (Q). Then δ i is a group homomorphism. Say δ i (P ) = (1, 1). Then ɛ 2 (P, T 1 ) = ɛ 2 (P, T 2 ) = 1 = ɛ 2 (P, T 3 ) = ɛ 2 (P, T 1 ) ɛ 2 (P, T 2 ) = 1. Hence ɛ 2 (P, T 1 ) = 1 for all T E[2]. P 2E[k]. This shows δ i is injective. Since ɛ 2 is non-degenerate, we must have 7

8 Let x = a/b and y = c/d with (a, b) = 1 and (c, d) = 1 so that we may rewrite y 2 = (x e 1 )(x e 2 )(x e 3 ) as b 3 c 2 = (a be 1 )(a be 2 )(a be 3 )d 2. (1) Then δ i (P, T i ) p 1... p n for some distinct primes p j thus we have a/b e i = p 1... p n u 2, where u Q. Write u = e/f with (e, f) = 1 and rearranging we have f 2 (a be i ) = p 1... p n be 2. Then if we have p j f 2 we have p j f so write f = p j m so that we have a be i = bp p j... p n e 2 m 2 but p j 1 p j mod (Q ) 2, hence we may assume that none of the p j divde f 2. Thus we have p j a be i so write a be i = kp j a = be i + kp j so then Equation 1 becomes b 3 c 2 = kp j (kp j + b(e i e i 1 ))(kp j + b(e i e i+1 ))d 2 (2) and from p j a be i we cannot have p j b otherwise p j a contradicting (a, b) = 1. Hence p j c 2 and hence p j c, then let c = tp j then let (e i e i 1 ) = z 1 and (e i e i+1 ) = z 2 so we rewrite Equation 2 as Expanding we obtain b 3 c 2 = kp j (kp j + bz 1 )(kp j + bz 2 )d 2 b 3 t 2 p j = kd 2 (kp j + bz 1 )(kp j + bz 2 ) = k 3 d 2 p 2 j + k 2 d 2 p j b(z 1 + z 2 ) + kd 2 b 2 z 1 z 2, which implies p j kd 2 b 2 z 1 z 2 and we know that (p j, b) = (p j, d) = 1 hence we have p j kz 1 z 2. Thus if we have p j k then write k = p r js where p j does not divide s. Hence we have a be i = sp r j bp 1... p n mod (Q ) 2 h 2 sp r j = bp 1... p n g 2 where (h, g) = 1 then we have h 2 s = bp 1... p 1 2r j... p n g 2 s bp 1... p 1 2r j... p n bp 1... p j... p n mod (Q ) 2 then as in the previous argument, we may assume that if sh 2 0 = p 1... p n g 2 0 then (p j, h 0 ) = 1 so that but then p j sh 2 0 but p j does not divide s and h 2 0, contradicting the choice of p j as prime, hence we must have p j dividing z 1 z 2 which completes the proof. Since δ E is a group homomorphism and using Propositions 7 and 9 ( ) 2 r+2 = E(Q) E(Q) 2E(Q) = δ E Q(S, 2) 2 = 2 2 S. 2E(Q) This give us an upper bound on the rank, namely r 2 S 2. Also, the rank is exactly the order of the image of δ E. Proposition 10. Let E(k) : Y 2 = (X e 1 )(X e 2 )(X e 3 ). There exists a bijection between points (d 1, d 2 ) on the image of δ E and rational points (u, v, w) on the curves 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 2 ). 8

9 Proof. Pick a point P = (X : Y : 1) E(k) and consider the image (d 1, d 2 ) = δ E (P ). Then there exist u and v in k such that Looking at the curve E(k) we have Let w = d 1 u 2 = X e 1 d 2 v 2 = X e 2. Y 2 d 1 d 2 u 2 v 2 = X e 3. Y, then X e (X e 1 )(X e 2 ) 3 = d 1 d 2 w 2. Combining the three equations we get d 1 u 2 d 2 v 2 = (e 2 e 1 ) and d 1 u 2 d 1 d 2 w 2 = (e 3 e 1 ). Hence (u, v, w) C d (k). Now, pick (z 1 : z 2 : z 3 : z 0 ) C d (k). We have d 1 z 2 1 d 2 z 2 2 = (e 2 e 1 )z 2 0 d 1 z 2 1 d 1 d 2 z 2 3 = (e 3 e 1 )z 2 0. d 1 z 2 1 = (X e 1 )z 2 0 d 2 z 2 2 = (X e 2 )z 2 0 d 1 d 2 z 2 3 = (X e 3 )z 3 0. precisely when (X : 1) = (d 1 z e 1 z 2 0 : z 2 0). Upon taking the product of the first three equations, we find an involving Y, which we solve to give Finally, since we have (X : Y : 1) = (d 1 z 2 1z 0 + e 1 z 3 0 : d 1 d 2 z 1 z 2 z 3 : z 3 0). X e 1 = d 1 z 2 1 z 2 0 we see that δ E ((X : Y : 1)) = (d 1, d 2 ). and X e 2 = d 2 z 2 2 z Method to discard points on Q(S 1, 2) Q(S 2, 2) Recall that one of our goals is to find bounds on the ranks of the curves E D. We know that δ E is an injective homomorphism from E D (Q) to Q(S 1, 2) Q(S 2, 2). We also know that a point d = (d 1, d 2 ) is in the image of δ E if and 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 2 ). Our approach to finding bounds on the ranks will be to find the d Q(S 1, 2) Q(S 2, 2) that are not in the image of δ E by showing that there are no rational points on C d. Proposition 11. Let E (D) : y 2 = x 3 + 5Dx 2 + 4D 2 x, let m, N Z with m N, and let A (d1,d 2 ) be the set of solutions (a, b, c, d) ( Z mz )4 to the system d 1 a 2 d 2 b 2 Dd 2 (mod m) and d 1 a 2 d 1 d 2 c 2 4Dd 2 (mod m). Let a, b, c, and d be the smallest representatives of a, b, c, and d respectively. If gcd(a, b, c, d, m) 1 for all (a, b, c, d) A (d1,d 2 ), then (d 1, d 2 ) δ E (D 0 )( E(D 0 ) (Q) 2E (D 0 ) (Q) ) for any D 0 such that D 0 D(modN). 9

10 Proof. Let D 0 D(mod N) and let m N. Suppose that for all (a, b, c, d ) A (d1,d 1 ), gcd(a, b, c, d, m) = r 1, where a, b, c, d are the smallest representatives of a, b, c, and d respectively, and that (d 1, d 2 ) δ E (D 0 )( E(D 0 ) (Q) ). Then, by the previous theorem there is a rational point (u, v, w) on the curve C (d1,d 2 ). We write 2E (D 0 ) (Q) u = a, v = b, and w = c with a, b, c, d Z and gcd(a, b, c, d) = 1.Then, d d d d 1a 2 d 2 b 2 Dd 2 (mod m) and d 1 a 2 d 1 d 2 c 2 4Dd 2 (mod m). Now, a a (mod m), b b (mod m), c c (mod m), d d (mod m). Then, r a, r b, r c, and r d, a contradiction. In the present case, we are interested in showing that these curves C d fail to have rational solutions for all D in an equivalence class modulo 24. Thus, we will take N = 24 and m = 3 or m = 8. Example 1. Consider the elliptic curve E (D) : y 2 = x 3 + 5Dx 2 + 4D 2 x.we show that for D 2(mod 3), (2, 1) δ E (D)( E(D) (Q) 2E (D) (Q) ). Proof. Suppose that (2, 1) δ E (D)( E(D) (Q) ). Then, there is some rational point on the 2E (D) (Q) curve given by: 2u 2 v 2 = D and 2u 2 2w 2 = 4D. Write u = a, v = b, and w = c d d d with a, b, c, d Z and gcd(a, b, c, d) = 1. Then, 2a 2 b 2 = Dd 2 and 2a 2 2c 2 = 4Dd 2 Looking at these equations modulo 3, we have 2a 2 + 2b 2 d 2 (mod 3) and 2a 2 + c 2 d 2 (mod 3). Suppose a 2 0(mod 3). Then, b 2 c 2 d 2 0(mod 3). It follows that 3 divides each of a,b,c and d, contradiction our choice of a,b,c, and d. Suppose a 2 1(mod 3). Then, c 2 1(mod 3) and d 2 0(mod 3). But then 2b 2 1(mod 3), which is impossible. This method of showing that the curves C d cannot have a rational point is not the only method of showing that d is not in the image of the connecting homomorphism. We can also use the fact that Im(δ E ) is a group and thus closed under multiplication. We can then argue as follows. Suppose that a, b Q(S 1, 2) Q(S 2, 2) and that b Im(δ E ). If ab Im(δ E ), then it follows that a Im(δ E ). Because of the amount of computation involved with eliminating points this way, we decided that having a computer do most of the work was the prudent thing to do. For the sake of completeness, we provide the computer program in Section 7. 5 Four Squares Arithmetic Progressions over Q The four-tuple (a 2, b 2, c 2, d 2 ) is an arithmetic progression of four squares over a field k if b 2 a 2 = c 2 b 2 = d 2 c 2. In the main theorem of the section we prove that there are no non-constant four squares arithmetic progressions over Q using the fact that there are no rational points with infinite order on the elliptic curve y 2 = x 3 +5x 2 +4x. We use the 2-Descent method (Section 3) to prove the above statement. Proposition 12. There exists a bijection between four squares arithmetic progression and E(Q), the set of rational points on E : y 2 = x 3 + 5x 2 + 4x. 10

11 Proof. Given a four squares arithmetic progression (a : b : c : d) (1 : 1 : 1 : 1), the following point is an element of E(Q): ( ( ) ( ) ) a 3b 3c + d a + b c d 2 : 6 : 1 a + 3b + 3c + d a + 3b + 3c + d and if (a : b : c : d) = (1 : 1 : 1 : 1), then we get the point (0 : 0 : 1). On the other hand, given a point (x : y : z) (0 : 0 : 1) E(Q), then (a : b : c : d) is an arithmetic progression of four squares over Q, where a = 6x + 3x 2 2y + xy, b = 2x x 2 2y xy, c = 2x x 2 + 2y + xy, d = 6x + 3x 2 + 2y xy. The difference n between the terms of the progression is 16xy 4x 3 y. If (x : y : z) = (0 : 0 : 1), then we get the arithmetic progression (1 : 1 : 1 : 1). Proposition 13. If E : y 2 = x 3 + 5x 2 + 4x then E(Q) tors = Z2 Z 4 Proof. With the substitutions x = X 2 X+2 y = 8 Y X = 2 1+x Y = 1 x 6y 3 (X+2) 2 (1 x) 2 we find that E is isogeneous to the quartic curve y 2 = (1 x 2 )(1 k 2 x 2 ) where k = 1/3. We know from [5] that the torsion subgroup of y 2 = (1 x 2 )(1 k 2 x 2 ) is Z 2 Z 4. Proposition 14. Points in E(Q) tors correspond to constant arithmetic progressions (±1 : ±1 : ±1 : ±1 :). Proof. We first calculate the 8 points in E(Q) tors. Using the bijection on Proposition 12 we calculate the corresponding arithmetic progression for each point. P oint P rogression (2 : 6 : 1) (1 : 1 : 1 : 1) (2 : 6 : 1) (1 : 1 : 1 : 1) ( 1 : 0 : 1) ( 1 : 1 : 1 : 1) ( 2 : 2 : 1) ( 1 : 1 : 1 : 1) ( 4 : 0 : 1) (1 : 1 : 1 : 1) ( 2 : 2 : 1) (1 : 1 : 1 : 1) (0 : 1 : 0) ( 1 : 1 : 1 : 1) (0 : 0 : 1) (1 : 1 : 1 : 1). Lemma 1. The rank of E : y 2 = x 3 + 5x 2 + 4x is zero. 11

12 Proof. We can compute the rank via 2-Descent: We have y 2 = x 3 + 5x 2 + 4x = 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 and (e 2 e 1 )(e 2 e 3 ) = 3. Therefore, S 1 = {2}, S 2 = {3} and We consider the curves Q(S 1, 2) = {±1, ±2} Q(S 2, 2) = {±1, ±3}. C d : d 1 z 2 1 d 2 z 2 2 = 1, d 1 z 2 1 d 1 d 2 z 2 3 = 4 and D d1 : W 2 = d 1 + 5Z d 1 Z 4. Since E(Q) tors is generated by ( 4 : 0 : 1) and (2 : 6 : 1) we have δ E (E(Q) tors ) = {( 1, 3), (2, 3), ( 2, 1), (1, 1)}. We see from the equation d 1 z1 2 d 2 z2 2 = 1 that the points (1, 1), (2, 1), (1, 3) and (2, 3) cannot be in the image of δ E since d 1 z1 2 d 2 z2 2 = 1 < 0. Then since δ E is a homomorphism of groups we may also conclude that ( 1, 3), ( 2, 3), ( 1, 2), ( 2, 1), ( 2, 3), (1, 3), (2, 1), and ( 1, 1) are not in the image. Then, we only have four points on the image of δ E, which implies 2 r+2 = 4, hence r = 0. Theorem 4. There are no non-constant arithmetic progressions of four squares over Q. Proof. From the previous Lemma we know that E has rank 0 over Q and hence that E(Q) = Z 2 Z 4 and from Proposition 14 we know that all of the points in E(Q) tors are mapped to constant arithmetic progressions, hence no non-constant arithmetic progression can exist over Q. 6 Four Squares Arithmetic Progressions over Q( D) Unlike Q, it is possible to find arithmetic progressions of four squares over Q( D) where D is a squarefree integer. We show how arithmetic progressions over Q( D) are related to the quadratic twist E (D) : y 2 = x 3 + 5Dx 2 + 4D 2 x of the curve E : y 2 = x 3 + 5x 2 + 4x, give a method for explicitly calculating such progressions and provide examples of such calculations. Proposition 15. Fix a nonzero integer D and 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. Note that the bijection given in Proposition 12 applies to any field k of characteristic 2, 3 hence it applies to the field Q( D). Given a rational point P = (x : y : 1) on E (D) then we have that (x/d : y/ D 3 : 1) is a point on the curve E : y 2 = x 3 + 5x 2 + 4x, hence using the bijection in Proposition 12 we obtain an arithmetic progression of four squares in Q( D). 12

13 For Table XX listed in [4] where there is a? or a yes, we use this method to compute an explicit example of an arithmetic progression of four squares over Q( D). In Table XX, we show the upper bound of the rank for each particular case and we give an example of a curve matching the bound. Table XX summarizes our results on narrowing the possible ranks of the elliptic curves E D. The first number is the upper bound that was found using the computer program. The numbers in parentheses are a D value and the rank of the corresponding E D. These provide a lower bound on the rank. 7 Computer Program 8 Acknowledgments This work was conducted during the 2010 Mathematical Sciences Research Institute Undergraduate Program (MSRI-UP), a program supported by the National Science Foundation (grant No. DMS ) and the National Security Agency (grant No. HB ). We also thank Dr. Cooper, Dr. Goins and Ebony Harvey for all their help during the preparation of this paper. References [1] David S. Dummit and Richard M. Foote Abstract Algebra. Prentice Hall Inc., Englewood Cliffs, NJ, [2] B. Mazur. Modular Curves and the Eisenstein ideal., 2nd Ed., Springer-Verlag, New York, [3] L. J. Mordell. Diophantine equations., Academic Press, London, [4] E. Gonzalez-Jimenez and Jorn Steuding. Arithmetic Progressions of Four Squares Over Quadratic Fields [5] E. Goins lecture notes 13

14 p mod 24 D = p D = 2p D = 3p D = 6p D = p D = 2p D = 3p D = 6p 1 2, (73, 2) 2, (146, 2) 2, (579, 2) 3, (1438, 1) 2, (-97, 2) 2, (-146, 2) 2, (-939, 2) 2, (-582, 2) 5 0, (5, 0) 1, (10, 1) 2, (15, 0) 2, (30, 1) 2, (-5, 1) 1, (-10, 1) 2, (-15, 1) 2, (-30, 0) 7 1, (7, 0) 0, (14, 0) 2, (21, 1) 2, (42, 1) 1, (-7, 0) 2, (-14, 2) 2, (-21, 1) 2, (-42, 0) 11 1, (11, 1) 2, (22, 1) 2, (33, 0) 2, (66,1) 1, (-11,0) 1, (-22, 1) 2, (-33, 1) 3, (-66, 2) 13 1, (13, 1) 0, (26, 0) 2, (39, 2) 2, (78, 1) 2, (-13, 0) 1, (-26, 0) 2, (-39, 1) 2, (-78, 0) 17 1, (17, 1) 2, (34, 1) 2, (51, 0) 3, (102, 1) 2, (-17, 1) 1, (-34, 1) 2, (-51, 0) 2, (-102, 0) 19 1, (19,0) 2, (134, 2) 2, (57, 0) 2, (114, 1) 1, (-19,1) 1, (-38, 0) 2, (-57, 1) 3, (-114, 2) 23 1, (23, 1) 1, (46, 1) 2, (69, 1) 2, (138, 1) 1, (-23, 1) 1, (-46, 1) 2, (-69, 1) 2, (-1002, 2) 14

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