On a Problem of Steinhaus

Transcription

1 MM Research Preprints, MMRC, AMSS, Academia, Sinica, Beijing No. 22, December 2003 On a Problem of Steinhaus DeLi Lei and Hong Du Key Lab of Mathematics Mechanization Institute of Systems Science, AMSS, Academia Sinica Beijing , China Abstract. In this paper, we consider a problem of Steinhaus. We connect this problem with the study of a class of elliptic curves and give results in some cases. 1. Introduction. Steinhaus posed the following problem: whether exist a square of integer edges and a point on the plane such that four distances between the point and the four vertexes are integers. He remarked he did not know the answer to this very difficult problem. (see [4]). In this paper, we consider the following problem which is equivalent to the original problem of Steinhaus: PROBLEM: Given a square in planar Descarts coordinates system whose four vertexes are (0, 0), (0, 1), (1, 0), (1, 1). Whether exists a point on the plane such that the distances between it and the four vertexes are rationals? For a point has the such properties, we call it a Steinhaus point. Suppose (x, y) is a Steinhaus point, we have the following equations: x 2 + y 2 = a 2, (1 x) 2 + y 2 = b 2, x 2 + (1 y) 2 = c 2, (1 x) 2 + (1 y) 2 = d 2, (1.1) (1.2) (1.3) (1.4) where a, b, c, d are nonnegative rationals. To minus the second and the third equations from the first one respectively, we obtain: 2x 1 = a 2 b 2, 2y 1 = a 2 c 2. x, y must be rationals. To seek Steinhaus points or to prove the nonexistence of such points, we can confine to those points of rational coordinates. In the next section, we give some facts related to elliptic curves. Then we connect the current problem to the study of elliptic curves. 2. Some facts of elliptic curves.

2 On a Problem of Steinhaus 187 An elliptic curve over Q is defined by an equation y 2 = f(x), where f(x) is a polynomial of degree three with rational coefficients and with distinct complex roots. The points on such a curve, together with a point at infinity, form an abelian group under a geometric definition of addition. The identity is the point at infinity. Let E(Q) denote the set of rational points on the elliptic curve E. In 1922, L.J.Mordell proved the following remarkable theorem, conjectured by H.Poincare. (see [1],[2],[3]). Mordell s theorem. If E is an elliptic curve over Q, then the abelian group E(Q) is finitely generated. Thus E(Q) = Z r F, where F is a finite abelian group. F is defined as the torsion subgroup. The integer r is the rank of E(Q). For a class of elliptic curves, we can compute the rank of E(Q) by the method described in [3]. Let Γ denotes E(Q). For an elliptic curve of integer coefficient in the following form we define E to be the elliptic curve: E : y 2 = x 3 + ax 2 + bx, y 2 = x 3 + ax 2 + bx, where a = 2a, b = a 2 4b. Let Q be the multiplicative group of non-zero rational numbers, and let Q 2 denote the subgroup of squares of elements of Q. Let α be the map from Γ to Q defined by modq 2 ), α(0, 0) = b(modq 2 ), α((0, 1, 0)) = 1( α(x, y) = x(modq 2 ), x 0. Having made such preparation, we introduce the formula to compute the rank of E(Q): 2 r = α(γ) α(γ). 4 To compute the order of α(γ), we take the integer b and factor it as a product b = b 1 b 2 in all possible ways. For each way of factoring, we write down the equation where N, M, e are the variables satisfying N 2 = b 1 M 4 + am 2 e 2 + b 2 e 4, gcd(m, N) = gcd(m, e) = gcd(n, e) = gcd(b 1, e) = gcd(b 2, M) = 1. Then α(γ) consists of 1, b(modq 2 ), together with b 1 (modq 2 ) such that the equation has a solution with M 0. For details, see [3]. In general, we can determine the torsion subgroup of a particular elliptic curve with the aid of the following theorem.

3 188 D.L.Lei and H.Du Nagell-Lutz Theorem. Let y 2 = f(x) = x 3 + ax 2 + bx + c be an elliptic curve with integer coefficients a, b, c; and let D be discriminant of the cubic polynomial f(x), D = 4a 3 c + a 2 b abc 4b 3 27c 2. Let P = (x, y) be a rational point of finite order. Then x and y are integers; and either y = 0, in which case P has order two, or else y divides D. (see [3].) 3. Two special cases. Because Steinhaus point is of rational coordinates, our scheme is to consider the problem: whether exists Steinhaus points on the line y = λ? λ 0 is a rational number. As before, we have x 2 + λ 2 = a 2, (3.1) (1 x) 2 + λ 2 = b 2, (3.2) x 2 + (1 λ) 2 = c 2, (3.3) (1 x) 2 + (1 λ) 2 = d 2. (3.4) To minus the second equation from the first one, we have Substitute into the first equation, 2x 1 = a 2 b 2, x = a2 b Let U = a + b, V = a b, then: ( a2 b ) 2 + λ 2 = a 2. 2 (UV + 1) 2 + 4λ 2 = (U + V ) 2, (U 2 1)(V 2 1) + 4λ 2 = 0. Let X = U 2 1, Y = XUV, then V 2 = 1 4λ2 X, Y 2 = (XUV ) 2 = X U 2 XV 2 = X(X + 1)(X 4λ 2 ). So, if (x, λ) is a Steinhaus point, we can construct a corresponding rational point on the elliptic curve E λ : Y 2 = X(X + 1)(X 4λ 2 ). Similarly, if 1 λ 0, from the last two equations (3.3) and (3.4), we can construct a corresponding rational point on the elliptic curve E 1 λ : Y 2 = X(X + 1)(X 4(1 λ) 2 ) by the same procedure. Our problem becomes the problem: wether the polynomial system: (U 2 1)(V 2 1) + 4λ 2 = 0, (W 2 1)(Z 2 1) + 4(1 λ) 2 = 0, UV = W Z, (3.1 ) (3.2 ) (3.3 )

4 On a Problem of Steinhaus 189 has a rational solution (U, V, W, Z)? The study of corresponding elliptic curves can give some useful information for this problem. Having made such preparation, we can prove: Theorem 3.1. There is no Steinhaus points on the lines x = 1 2, y = 1 2. To begin with, we study the equation:(u 2 1)(V 2 1) + 1 = 0. This is the case for λ = 1 2. Using fermat s method of descent, we can prove the following theorem which implies theorm 3.1. Theorem 3.2. (X 2 1)(Y 2 1) + 1 = 0 has no rational solutions. Proof. Assume (X 2 1)(Y 2 1) + 1 = 0 has a rational solution (X, Y ). Since 2 is not a rational number, it is clear that XY 0. Set where p, q, s, t are nonzero integers such X = q p, Y = t s, gcd(p, q) = gcd(s, t) = 1. From (( q p )2 1)(( t s )2 1) + 1 = 0, we have Hence we have following possibilities: Case i). In this case we have (q 2 p 2 )(t 2 s 2 ) + p 2 s 2 = 0. i) q 2 p 2 = s 2, t 2 s 2 = p 2, ii) q 2 p 2 = s 2, t 2 s 2 = p 2. q 2 = p 2 + s 2, s 2 = p 2 + t 2. These equations are primitive Pythagorean triples, and we have That is Where p is even. Now we will show that s 4 p 4 = (p 2 + s 2 )(s 2 p 2 ) = (qt) 2. s 4 = p 4 + (qt) 2 X 4 = Y 4 + Z 2 has no integral solution (X, Y, Z) with XY Z 0 and Y to be even. Assume that (X, Y, Z) is a positive integral solution of X 4 = Y 4 + Z 2

5 190 D.L.Lei and H.Du with X 2 to be minimal and Y to be even. It is clear that X 2, Y 2, Z are primitive Pythagorean triple. Hence there exist integers A, B with gcd(a, B) = 1 and AB to be even such that X 2 = A 2 + B 2, Y 2 = 2AB, Z = A 2 B 2. From Y 2 = 2AB,we get A = a 2, B = 2b 2 (or A = 2a 2, B = b 2,then use same argument). Consequently which leads X 2 = A 2 + B 2 = a 4 + (2b 2 ) 2 2b 2 = 2CD, a 2 = C 2 D 2 where gcd(c, D) = 1.Hence, there exist S, T such that C = S 2, D = T 2 (clearly,t must be even), so we have a 2 = S 4 T 4, i.e. S 4 = T 4 + a 2. Note that, X 2 > (2b 2 ) 4 = (2CD) 2 = (2S 2 T 2 ) > S 2 which contradict to the assumption that X 2 is minimal. Case ii). In this case, we have Hence p 2 = s 2 + q 2, t 2 = p 2 + s 2. p 4 = s 4 + (qt) 2 and s is even. So we can treat this case in the same way as case i). Consequently we have proved theorem 3.2. Having made connection with elliptic curves, we can give another proof to theorem 3.1. Lemma 3.1. For elliptic curve E : Y 2 = X 3 X, E(Q) = Z 2 Z 2. The only rational points on the elliptic curve are (0, 0), ( 1, 0), (1, 0), (0, 1, 0). For the proof of this lemma, see [1],[2],[3]. Another Proof of Theorem 3.1. Suppose (x, 1 2 ) is a Steinhaus point, we have the corresponding elliptic curve Y 2 = X 3 X. According to Lemma 3.1, the rational points on this elliptic are (0, 0), (1, 0), ( 1, 0), (0, 1, 0). By the construction we have made, (U 2 1, (U 2 1)UV ) is a rational point on this elliptic curve. U 2 1 must be one of 0, 1, 1. U 2 1 can not equal 0, because (U 2 1)(V 2 1) + 1 = 0. If U 2 1 = 1, then U 2 = 2.

6 On a Problem of Steinhaus 191 U is rational. This is impossible. If U 2 1 = 1, then V 2 1 = 1. V is rational. This is impossible. So, there is no Steinhaus points on the line y = 1 2. Because of symmetry, there is no Steinhaus points on the line x = 1 2. Furthermore, we can prove: Theorem 3.3. There is no Steinhaus points on the edges of the square. At first, we give a lemma needed in the proof of Theorem 2. Lemma 3.2. For elliptic curve E : Y 2 = X 3 3X 2 4X, E(Q) = Z 2 Z 2. The only rational points on the elliptic curve are (0, 0), ( 1, 0), (4, 0), (0, 1, 0). Proof. To begin with, we compute the rank of E(Q). To compute the order of α(γ), we need only consider the following two equations: N 2 = 2M 4 3M 2 e 2 2e 4, N 2 = 2M 4 3M 2 e 2 + 2e 4. For the first equation, N 2 = (2M 2 +e 2 )(M 2 2e 2 ). Suppose there is a prime p gcd(2m 2 + e 2, M 2 2e 2 ), then p 5M 2, p 5e 2. Because gcd(m, e) = 1, p = 5, gcd(2m 2 +e 2, M 2 2e 2 ) = 5. We have 2M 2 + e 2 = 5C 2, M 2 2e 2 = 5D 2, where C, D satisfying gcd(c, D) = 1, N = CD. For any integer a, a 5 2 0, 1, 4(mod5). The 2 above two equations can not hold simultaneously under the assumption gcd(m, e) = 1. For this case, The first equation has no admissible solutions. Suppose gcd(2m 2 + e 2, M 2 2e 2 ) = 1, we have 2M 2 + e 2 = C 2, M 2 2e 2 = D 2, where C, D satisfying gcd(c, D) = 1, N = CD. Then, 2C 2 + D 2 = 5M 2, C 2 2D 2 = 5e 2. The above two equations can not hold simultaneously under the assumption gcd(c, D) = 1. For this case, The first equation has no admissible solutions. We can prove the second equation has no admissible solutions in a similar way. So, α(γ) = 2. To compute the order of α(γ), we have to consider E : y 2 = x 3 + 6x x. we need only consider the following two equations N 2 = 5M 4 + 6M 2 e 2 5e 4, N 2 = 5M 4 + 6M 2 e 2 + 5e 4.

7 192 D.L.Lei and H.Du For the first equation, we have N 2 = 5(M 2 + e 2 ) M 2 e 2. If 16 N 2, then M, e must be odd. But 5(M 2 +e 2 ) 2 20(mod16),so 16 Γ 2D N 2. Because a 2 0, 1, 4, 9 (mod16), so N 2 1, 4, 9 (mod16). In these cases, the above equation does not hold. So, α(γ) 3, 2 r 2 3 4, r = 0. Next, we prove there is no rational points of finite order other than (0, 0), ( 1, 0), (4, 0), (0, 1, 0). Consider the corresponding elliptic curve reduced to F 3 (the field of three elements),y 2 = X 3 X. The number of F 3 points is 4. So, E(Q) = 4, E(Q) = Z 2 Z 2.(see [1],[2],[3].) Proof of Theorem 3.3. Without losing generality, we can assume that (x, 1) is a Steinhaus point. The corresponding elliptic curve is Y 2 = X 3 3X 2 4X. By the construction we have made, (U 2 1, (U 2 1)UV ) is a rational point on this elliptic curve. According to Lemma 3.2, U 2 1 must be one of 0, 1, 4. U 2 1 can not equal 0, because (U 2 1)(V 2 1) + 4 = 0. If U 2 1 = 4, then U 2 = 5. This is impossible, because U is rational. If U 2 1 = 1, then V 2 1 = 4. V is rational. So, there is no Steinhaus points on the line y = 1. Because of symmetry, there is no Steinhaus points on the three lines y = 0, x = 0, x = 1. This completes the proof. 4. Some general results. Theorem 4.1 Let λ = n m, n 0. If 3 Γ 2D n, m, then E λ(q) tors = Z2 Z 2. ( E λ (Q) tors denotes the torsion subgroup of E λ (Q).) Proof. n 2 m 2 1(mod3), if 3 Γ 2D n, m. For E λ : Y 2 = X(X + 1)(X 4λ 2 ), we can make an admissible change of variables so that all the coefficients are in Z. In other words, E λ has the form: Y 2 = X(X + m 2 )(X 4n 2 ). Reducing to F 3, the field of three elements, the corresponding elliptic curve is Y 2 = X(X + 1)(X 1) = X 3 X. The number of F 3 point is 4. E λ (Q) tors = 4. E λ (Q) tors = Z2 Z 2. (see [1],[2],[3].) As a useful criterion, The following theorems may be helpful to treat many cases. Theorem 4.2. Let λ be a rational number, λ 0, 1. If λ satisfies one of the following conditions then there is no Steinhaus points on the lines x = λ, y = λ.(1) E λ (Q) has rank zero and has only four rational points, 1 + 4λ 2 / Q 2. (2) E 1 λ (Q) has rank zero and has only four rational points, 1 + 4(1 λ) 2 / Q 2. Proof. Suppose E λ (Q) has rank zero and has only four torsion rational points. If there is a Steinhaus point on the line y = λ, then U 2 1 must be one of 0, 1, 4λ 2. U 2 1 can not equal 0, because (U 2 1)(V 2 1)+4λ 2 = 0 and λ 0. If U 2 1 = 1, then V 2 1 = 4λ 2. So, either U 2 = 1 + 4λ 2 or V 2 = 1 + 4λ 2. Because U, V are rational numbers and 1 + 4λ 2 / Q 2, U 2 1 1, U λ 2. This completes the proof. Corollary. Let λ be a rational number, λ 0, 1, λ = n 2, n an integer. If E λ(q) or E 1 λ (Q) has rank zero and has only four rational points, then there is no Steinhaus points on the lines x = λ, y = λ. In this case, 1 + 4λ 2 = n / Q 2. Theorem 4.3. Let λ be a rational number, λ 0, 1, λ = n m. Suppose E λ(q) and E 1 λ (Q) have rank zero and 3 Γ 2D n, m, m n, then there is no Steinhaus points on the lines x = λ, y = λ.

8 On a Problem of Steinhaus 193 Proof. By theorem 4.1, E λ (Q) = E 1 λ (Q) = Z 2 Z 2. If there is a Steinhaus point on the line y = λ, then U 2 1 must be one of 0, 1, 4λ 2. U 2 1 can not equal 0, because (U 2 1)(V 2 1) + 4λ 2 = 0 and λ 0. If U 2 1 = 1, then V 2 1 = 4λ 2. So, either U 2 = 1 + 4λ 2 or V 2 = 1 + 4λ 2. Similarly, either W 2 or Z 2 = 1 + 4(1 λ) 2. Namely, 1 + ( 2n m )2 = A 2, 1 + (2 2n m )2 = B 2, where A, B are rational numbers. According to theorem 3.1, the two equations can not hold simultaneously. This completes the proof. Theorem 4.4. Let λ be a rational number, λ 0, 1, λ = n m. Suppose E λ(q) and E 1 λ (Q) have rank zero and they have no points of order three, then there is no Steinhaus points on the lines x = λ, y = λ. Proof. According to a theorem of Mazur (see [1],pp15), the torsion subgroup of E λ (Q) must be Z 2n Z 2, n = 1, 2, 3,or 4. Suppose the ranks of E λ (Q) and E 1 λ (Q) are zero, E λ (Q) and E 1 λ (Q) have no points of order three. If there is a Steinhaus point on the line y = λ, we have a corresponding rational point P of order 2 i on E λ. Let 2 i 1 P = (X 0, Y 0 ). X 0 must be one of 0, 1, 4λ 2. If i > 1, X 0, X 0 + 1, X 0 4λ 2 are squares of rational numbers (see [1], pp85, theorem 4.2). We have X 0 = 4λ λ 2 is a square of rational number. If i = 1, the proof of theorem 4.3 shows the same fact. Similarly, for E 1 λ, 1 + 4(1 λ) 2 is a square of rational number. Now, the proof is the same as that of theorem 4.3. Remark. The problem of Steinhaus seems to be very difficult. If the ranks of E λ (Q) and E 1 λ (Q) are greater than zero, the method of this paper does not take effect. In fact, the assumptions of theorems in this section exclude situations in which E λ (Q) or E 1 λ (Q) have points of order three. It is not clear whether E λ (Q) has point of order three for some λ. We believe there is no Steinhaus points on the plane. References [1] A.W. Knapp, Elliptic curves, (1992) Priceton university Press. [2] N. Koblitz, Introduction to elliptic curves and modular forms, GTM 97, (1984) Springer-verlag. [3] J. H. Silverman and J. Tate, Rational points on elliptic curves, UTM, (1992) Springer-verlag. [4] G. Steibhaus, Another hundred problems in Mathematics (in Chinese), Shanghai educational press (1980). (translated by Zhuang Yadong from Russian)