Rational Distance Sets on Conic Sections

Transcription

1 Rational Distance Sets on Conic Sections Megan D Ly Loyola Marymount Shawn E Tsosie UMASS Amherst July 010 Lyda P Urresta Union College Abstract Leonhard Euler noted that there exists an infinite set of rational points on the unit circle such that the pairwise distance of any two is also rational; the same statement is nearly always true for lines and other circles In 004, Garikai Campbell considered the question of a rational distance set consisting of four points on a parabola We introduce new ideas to discuss a rational distance set of four points on a hyperbola We will also discuss the issues with generalizing to a rational distance set of five points on an arbitrary conic section 1 Introduction A rational distance set is a set whose elements are points in R which have pairwise rational distance Finding these sets is a difficult problem, and the search for rational distance sets with rational coordinates is even more difficult The search for rational distance sets has a long history Leonard Euler noted that there exists an infinite set of rational points on the unit circle such that the pairwise distance of any two is also rational In 1945 Stanislaw Ulam posed a question about a rational distance set: is there an everywhere dense rational distance set in the plane []? Paul Erdös also considered the problem and he conjectured that the only irreducible algebraic curves which have an infinite rational distance set are the line and the circle [4] This was later proven to be true by Jozsef Solymosi and Frank de Zeeuw [4] First, we provide the formal definition of a rational distance set: Definition 11 A rational distance set S is a set of elements P i = (x i, y i ) R, 1 i n, such that for all P i, P j S, P i P j is a rational number We are primarily concerned with finding rational distance sets of rational points The distance between two points on a graph is (x i x j ) + (y i y j ) It will prove useful to rewrite this formula as ( ) yi y j x i x j 1 + (1) x i x j Now, consider the following lemma: 1

2 Lemma 1 The rational solutions to the equation α = 1 + β () can be parameterized by α = m + 1 m, β = m 1, for m Q, m 0 m For the proof, refer to Campbell [1, Lemma 1] Main results 1 Line Theorem 1 For any line L: ax + by + c = 0, for a, b, c Q, there exists a dense rational distance set with rational coordinates if and only if there exists m Q such that a + b = m Proof Suppose L has a dense rational distance set with rational coordinates Then we can express the line L as y = ax c Consider two points on L, (x b b i, y i ) and (x j, y j ) Using Lemma 1, we can write the distance between these two points as ( a ) x i x j 1 + (3) b The distance is rational only if 1 + a = n for n Q That is, if a + b = b n b Hence, we may set m = bn Suppose a +b = m for m Q Then line L can be written as y = ax c b The b distance between two rational points on L, (x i, y i ) and (x j, y j ), is x i x j 1 + ( ) a b Note that from Lemma 1 we can write a + b = m as 1 + a b 1 + a = m, which is rational b b Circle = m b Hence, we see To show that there exists a rational distance set of rational points on a circle, we identify the cartesian plane R with the complex plane C Theorem Let r Q with r 0 and let C : z z 0 = r be a circle in the complex plane with z 0 = x+iy, x, y Q Then, there exists a dense rational distance set on C consisting of rational points

3 Sketch of Proof Consider the vertical line L: Re z = 1 r Möbius transformation defined by Let f : C C be the f(z) = (z 0 r)z 1 z Then f maps the line L into the circle C The following figure shows this for a circle with radius r < 1 centered at z 0 A dense rational distance set of rational points on L is given by { S = z = 1 ( ( )) } s i s Q, s 0 r s Notice that Lemma 1 implies that each z S has rational norm z The codomain f(s) = {f(z) z S} will then be a dense set of points on C with rational coordinates Finally notice that f is a translation composed with the map z 1/z Then, the rationality of the distance between the points on the codomain follows from the identity 1 z 1 z w w = z w 3 Rational Distance Set of Three Points on a Parabola Consider the parabola y = ax + bx + c where a, b, c Q We provide a geometric intuition of the problem An important note for this construction is that while the points are at rational distance, the points themselves do not have to be rational We present a method for constructing a rational distance set of three points on a parabola Note that we can find a rational distance set using 3

4 (a) Two concentric circles with rational radii are centered at some point on the parabola (b) Moving the construction along the parabola, we note that the length of segment P P 3 changes Figure 1: While P 1 P and P 1 P 3 are rational, we cannot guarantee that P P 3 is also rational Since the length of P P 3 changes, the segment P P 3 must have a rational length at some point on the parabola this method on any conic section First, we choose some point on the parabola and construct two circles of rational radii around it, as shown in Figure 1 In Figure 1, we see that the line segments P 1 P and P 1 P 3 are rational All we require now is that P P 3 be rational This segment is not necessarily rational We can, however, make it rational If we allow the construction to move along the parabola, we notice that the length of segment P P 3 changes We see from this construction that the length of segment P P 3 is a continuous function on an interval Therefore, we can find some point in the parabola where the segment P P 3 is rational Now we turn to the more difficult problem of finding rational distance sets of rational points on the parabola By equation 1, the distance d ij between two rational points P i and P j on a parabola is By Lemma 1, ax i + ax j + b = m ij 1 m ij the following theorem: d ij = x i x j 1 + (ax i + ax j + b) We let g(m) = 1 a ( m 1 m b), which leads us to Theorem 3 Given a parabola y = ax +bx+c, let S = {P 1, P, P 3 } S is a rational distance set of rational points if and only if 1 i < j 3 we can find a nonzero 4

5 rational value m ij such that x 1 = g(m 1) + g(m 13 ) g(m 3 ) x = g(m 1) g(m 13 ) + g(m 3 ) x 3 = g(m 1) + g(m 13 ) + g(m 3 ) Proof For each 1 i < j 3 where x i + x j = g(m ij ), we have three equations: x 1 + x = g(m 1 ) x 1 + x 3 = g(m 13 ) x + x 3 = g(m 3 ) This is equivalent to the following row reduced augmented matrix: g(m )+g(m 13 ) g(m 3 ) g(m ) g(m 13 )+g(m 3 ) g(m 1)+g(m 13 )+g(m 3 ) which gives us our desired values 4 Rational Distance Set of Four Points on a Parabola We can follow a similar argument to find a rational distance set of four rational points on a parabola Theorem 4 Given a parabola y = ax + bx + c, let S = {P 1, P, P 3, P 4 } S is a rational distance set of rational points if and only if there are rational values m ij, 1 i < j 4 such that x 1 = 1 (g(m 1) + g(m 13 ) g(m 3 )) x = 1 (g(m 1) g(m 13 ) + g(m 3 )) x 3 = 1 ( g(m 1) + g(m 13 ) + g(m 3 )) x 4 = 1 ( g(m 1) g(m 13 ) + g(m 3 ) + g(m 14 )) and g(m 13 ) + g(m 4 ) = g(m 3 ) + g(m 14 ) = g(m 1 ) + g(m 34 ) 5

6 Proof For each 1 i < j 4 where x i + x j = g(m ij ), we have six equations: x 1 + x = g(m 1 ) x 1 + x 3 = g(m 13 ) x 1 + x 4 = g(m 14 ) x + x 3 = g(m 3 ) x + x 4 = g(m 4 ) x 3 + x 4 = g(m 34 ) This is equivalent to the following row reduced augmented matrix: g(m )+g(m 13 ) g(m 3 ) g(m ) g(m 13 )+g(m 3 ) g(m )+g(m 13 )+g(m 3 ) g(m ) g(m 13 )+g(m 3 )+g(m 14 ) g(m 13 ) + g(m 4 ) g(m 3 ) g(m 14 ) g(m 1 ) + g(m 34 ) g(m 3 ) g(m 14 ) which gives us our desired equations We know that any three points lie on a common circle However, in the case of four points, this does not always occur Definition 5 A set of points is concyclic if they lie on a common circle Similarly, a set of points is nonconcyclic if one cannot construct a common circle through them We will now examine the cases of concyclic and nonconcyclic points on a parabola 5 Concyclic Points on a Parabola Proposition 6 Let P i = (x i, ax i + bx i + c) for 1 i 4 be four rational points Then they are concyclic if and only if x 1 + x + x 3 + x 4 = b a Proof Let C α,β,ρ be the circle (x α) + (y β) = ρ, where α, β, ρ R This circle intersects the parabola y = ax + bx + c at the points whose x-coordinates are the roots of (x α) + (ax + bx + c β) ρ = a x 4 + abx 3 + (a(c β) + b + 1)x + b(c β)x + (c β) + α ρ = a (x x 1 )(x x )(x x 3 )(x x 4 ) Since the coefficient of x 3 is ab, then a (x 1 + x + x 3 + x 4 ) = ab Thus, x 1 + x + x 3 + x 4 = b a Conversely, if we have x i, 1 i 4 such that x 1 + x + x 3 + x 4 = b, we can a solve: a (x x 1 )(x x )(x x 3 )(x x 4 ) = 6

7 a x 4 + abx 3 + (a(c β) + b + 1)x + b(c β)x + (c β) + α ρ for some α, β, and ρ Theorem 7 Suppose that P 1, P, P 3, and P 4 are rational and concyclic points on the parabola y = ax + bx + c Then, these points are at rational distance if and only if there are nonzero rational values m 1, m 13, and m 3 such that the equations of Theorem 4 hold If we have this condition, then x 4 = 1 (g(m 1) + g(m 13 ) + g(m 3 ) + 4b a ) Proof By Proposition 6, x 4 = (x 1 + x + x 3 + b ) Therefore, we can directly a solve for x 4 using the values given for the x 1, x, x 3 Inputting our values, we get x 4 = 1(g(m 1) + g(m 13 ) + g(m 3 ) + 4b) a 6 Non-Concyclic Points on a Parabola We now explore the more difficult case of finding a rational distance set of four nonconcyclic rational points on the parabola We therefore need to examine the last condition of Theorem 4: g(m 13 ) + g(m 4 ) = g(m 3 ) + g(m 14 ) = g(m 1 ) + g(m 34 ), in order to find our six m ij values Consider the surface g(m ij ) + g(m kl ) = t where t is an arbitrary rational number We can make the following substitutions: m ij = X + X Y T X, m kl = X + 1 Y T X, t = T b, a to obtain the elliptic curve E : Y = X 3 + (T + )X + X Thus if we find rational points on the elliptic curve we can find rational m ij values that satisfy the last condition 7 Examples In order to illustrate this process, we include the following example: Given the parabola y = x, let T = 13 6 in order to get the following elliptic curve E : Y = X X + X We chose T = 13 because it is a small rational value that 6 gives us an elliptic curve of positive rank We get the rational points Q 1 = (6: 19: 18), Q = (3: 13: 36), Q 3 = (30: 169: 750) on E Now our m ij have the following values: m 1 = 3/10, m 13 = 1/, m 3 = 4/3, m 14 = 4, m 4 = 6, m 34 = 15/ Which yields the following rational distance set: {( , 9449 ) (, , 361 ) ( 17, , 1619 ) ( 757, , )} We illustrate this in the following figure: We now show an example of a rational distance set on the parabola y = x + x If we let T = 13 6 then we get the same elliptic curve as in Example 1: E : Y = 7

8 Figure : Rational distance set of four points on a parabola y = x X X + X Again, we chose T = 13 because it is a small rational value that 6 gives us an elliptic curve of positive rank We get the rational points Q 1 = (6: 19: 18), Q = (3: 13: 36), Q 3 = (30: 169: 750) on E Now our m ij have the following values: m 1 = 3/10, m 13 = 1/, m 3 = 4/3, m 14 = 4, m 4 = 6, m 34 = 15/ This gives us different values for g(m ij ): g(m 1 ) = 11 60, g(m 13) = 11 4, g(m 14) = 41 4, g(m 3 ) = 1 8, g(m 4) = 11 1, g(m 34) = These values give us the following rational distance set: {( , ) (, 107 ) ( , 39911, , 4889 ) ( 109, , 931 )} 6400 We can see this in the following figure: 8 Rational Distance Set of Five Points on a Parabola Next, we provide the necessary conditions for a rational distance set of five rational points on a parabola Theorem 8 Given a parabola y = ax + bx + c, let S = {P 1, P, P 3, P 4, P 5 } S is a rational distance set of rational points if and only if there are rational values m ij, 8

9 Figure 3: Rational distance set of four points on a parabola y = x + x 1 i < j 5 such that x 1 = g(m 1) + g(m 13 ) g(m 3 ) x = g(m 1) g(m 13 ) + g(m 3 ) x 3 = g(m 1) + g(m 13 ) + g(m 3 ) x 4 = g(m 1) g(m 13 ) + g(m 3 ) + g(m 14 ) x 5 = g(m 1) g(m 13 ) + g(m 3 ) + g(m 15 ) as well as, g(m 13 ) + g(m 4 ) = g(m 14 ) + g(m 3 ) g(m 13 ) + g(m 5 ) = g(m 15 ) + g(m 3 ) g(m 1 ) + g(m 34 ) = g(m 14 ) + g(m 3 ) g(m 1 ) + g(m 35 ) = g(m 15 ) + g(m 3 ) g(m 1 ) + g(m 13 ) + g(m 45 ) = g(m 14 ) + g(m 15 ) + g(m 3 ) 9

10 Proof For each 1 i < j 5 where x i + x j = g(m ij ), we have ten equations: x 1 + x = g(m 1 ) x 1 + x 3 = g(m 13 ) x 1 + x 4 = g(m 14 ) x 1 + x 5 = g(m 15 ) x + x 3 = g(m 3 ) x + x 4 = g(m 4 ) x + x 5 = g(m 5 ) x 3 + x 4 = g(m 34 ) x 3 + x 5 = g(m 35 ) x 4 + x 5 = g(m 34 ) This is equivalent to the following row reduced augmented matrix: g(m )+g(m 13 ) g(m 3 ) g(m ) g(m 13 )+g(m 3 ) g(m )+g(m 13 )+g(m 3 ) g(m ) g(m 13 )+g(m 3 )+g(m 14 ) g(m ) g(m 13 )+g(m 3 )+g(m 15 ) g(m 13 ) g(m 14 ) g(m 3 ) + g(m 4 ) g(m 13 ) g(m 15 ) g(m 3 ) + g(m 5 ) g(m 1 ) g(m 14 ) g(m 3 ) + g(m 34 ) g(m 1 ) g(m 15 ) g(m 3 ) + g(m 35 ) g(m 1 ) + g(m 13 ) g(m 14 ) g(m 15 ) g(m 3 ) + g(m 45 ) which gives us our desired equations 9 Rational Distance Sets of Three Points on a Hyperbola At this point, we would like to examine a rational distance set of three rational points on a hyperbola Theorem 9 Let a, b, c, d Q such that a(ad bc) 0 Then there exists a rational distance set of three rational points, S = {P 1, P, P 3 } on the hyperbola axy + bx + cy + d = 0 Proof The proof proceeds in several parts First, we rewrite the distance formula as in Equation 1 So, y i y j x i x j can be parameterized as y i y j x i x j = m ij 1 m ij Then, we solve for y in our hyperbola equation to get the expression: y = bx+d ax+c Substituting y into our modified distance formula yields (ad bc) (ax i + c)(ax j + c) = y i y j x i x j = m ij 1 m ij 10

11 Define D = (ad bc) and let m 1 = m Dn Making the substitutions m = X and D n = Y gives us the elliptic curve E (D) : Y = X 3 D X Note that we can also DX obtain the expression m 1 = m Dn from E (D) with the substitutions X = Dm and Y = D mn Letting Q i = (X i : Y i : 1) be rational points on E (D), we define the following rational points: n 3 = Y 1 DX 1, n 13 = Y DX, n 1 = Y 3 DX 3 in order to allow the following relation to hold: D (ax i + c)(ax j + c) = Dn ij = m ij 1 m ij We can now define an explicit formula for a rational distance set of three rational points on a hyperbola: ( n3 P 1 = c an 1 n 13 a, b a D ) n 1 n 13 a n 3 ( n13 P = c an 1 n 3 a, b a D ) n 1 n 3 a n 13 ( n1 P 3 = c an 13 n 3 a, b a D ) n 13 n 3 a n 1 10 Example In order to illustrate this process, we include the following example: Given the hyperbola xy + x + y + = 0, we get the following elliptic curve E (6) : Y = X 3 36X From this curve, we get the rational points Q 1 = ( 3: 9: 1), Q = (50: 35: 8), Q 3 = ( 58719: 31057: 50653) Now our n ij have the following values: n 3 = 1/, n 13 = 7/60, n 1 = 1551/1709 This yields the following rational distance set: {( , ), We show this in the following figure: ( , ), ( , )} 11 Rational Distance Set of Four Points on a Hyperbola The system of equations we derive to find a rational distance set of four rational points on a hyperbola will be overdetermined, as in the four point parabola case In order to handle the extra constraints, we must consider a new result Before lauching into the proposition, however, we should first state some relevant defitions: 11

12 Figure 4: The hyperbola xy + x + y + = 0 with a rational distance set of rational points Definition 10 Let X, Y, Z be groups and f : X Z and g : Y Z be two homomorphisms Then the fibered of f and g is the subgroup X Y consisting of all pairs (X, Y ), x X, y Y such that f(x) = g(y )[3] Definition 11 Given two curves E : f(x, y) = 0 and D : g(u, v) = 0 a map φ: E D is called a rational map if there exist polynomials r(x, y), s(x, y), and t(x, y) such that u = r(x,y) s(x,y) and v = t(x,y) t(x,y) Moreover, we say that E and D are birationally equivalent if there exist two rational maps φ: E D and ψ : D E such that the following two conditions hold: a ψ φ = id E, that is the composition of ψ and φ is the identity on E b φ ψ = id D, that is the composition of φ and ψ is the identity on D With these definitions in hand, we introduce the following proposition: Proposition 1 Fix a rational number T, and consider the curve { E T = (X i, Y i, X j, Y j ) A 4 Yi = Xi 3 D X i Y j = Xj 3 D and D X ix j X j Y i Y j } = T (4) a The curve E T π 1 ({T }) is a fiber over the product E (D) π E (D) associated to the morphism π : E (D) E (D) P 1, which sends ((X i : Y i : 1), (X j : Y j : 1)) D X ix j Y i Y j b Additionally, E T is birationally equivalent to the curve E T : y 4 = x(x + D )(x + D T 4 )(x D T ) (x + D T ) 1

13 Proof Part (a) is clear from the definition of a fiber product For part (b), we consider a point (x, y) on the curve, and define X i = 1 T y Y i = 1 T 3 (x + D )(x + D T )(x D T ) y x + D X j = D Dy (x D T )(x + D T )(x + D x + D 4 T 4 ) (x D T ) (x + D T ) Y j = D Dy (x D T )(x + D T )(x + D x + D 4 T 4 ) (x D T ) 3 (x + D T ) 3 From the substitutions, we see that this point is on E T Conversely, we consider a point (X i, Y i, X j, Y j ) on E T, and define x = 4D 3 X3 i Xj 3 (X j D) Yi Y j 4 y = 8D 5 X3 i X 3 j (4D 3 X 3 i X 3 j 8D X 3 i X 4 j + 4DX 3 i X 5 j Y i Y 4 j ) Yi Y j 7 Again, from the substitutions we see that this point is on the curve E T We are now ready to examine a method for finding a rational distance set of four rational points on a hyperbola Theorem 13 Let a, b, c, d Q such that a(ad bc) 0 There exists a rational distance set of four rational points on the hyperbola axy + bx + cy + d = 0 if and only if we can find three rational points on the curve E T : y 4 = x(x + D )(x + D T 4 ) (x D T ) (x + D T ) Proof This proof is similar to that of a rational distance set of three rational points on a hyperbla First, we express the distance formula as in Equation 1 So, y i y j x i x j can be parameterized as y i y j x i x j = m ij 1 m ij Then, we solve for y in our hyperbola equation to get y = bx+d Substituting y ax+c into our modified distance formula yields (ad bc) (ax i + c)(ax j + c) = y i y j x i x j = m ij 1 m ij Define D = (ad bc) and let m 1 = m Dn Making the substutions m = X and D n = Y gives us the elliptic curve E (D) : Y = X 3 D X Note that we can also DX obtain the expression m 1 = m Dn from E (D) with the substitutions X = Dm and Y = D mn 13

14 Letting Q i = (X i : Y i : 1) be rational points on E (D), we define the rational values n 13 = Y 1 DX 1, n 4 = Y DX, n 3 = Y 3 DX 3, n 14 = Y 4 DX 4, n 1 = Y 5 DX 5, n 34 = Y 6 DX 6, such that the following relation holds: D (ax i + c)(ax j + c) = Dn ij = m ij 1 m ij Therefore, the four rational points in the rational distance set must be in the form: ( n3 P 1 = c an 1 n 13 a, b a D ) n 1 n 13 a n 3 ( n13 P = c an 1 n 3 a, b a D ) n 1 n 3 a n 13 ( n1 P 3 = c an 13 n 3 a, b a D ) n 13 n 3 a n 1 ( n3 P 4 = c an 4 n 34 a, b a D ) n 4 n 34 a n 3 Additionally, we have the constraint n 13 n 4 = n 3 n 14 = n 1 n 34 In order to satisfy this constraint, we want to find six rational points on E (D) satisfying the relation: Y 1 DX 1 Y DX = Y 3 DX 3 Y 4 DX 4 = Y 5 DX 5 Y 6 DX 6 To solve any pair of such equations, we look for rational points on E T, defined in Equation 4 Recall Propostition 1, which states that E T is birationally equivalent to the curve E T : y 4 = x(x + D )(x + D T 4 )(x D T ) (x + D T ) Thus in order to find a six rational points on E (D), it suffices to find a set of three rational points on the curve E T One way to find rational points on the curve E T is to consider the superelliptic curve E T : w 4 = D z(z + T 4 )(z + z + T 4 )(z + T 4 z + T 4 ) Definition 14 A superelliptic curve is a curve of the form w n = f(z), where f(z) is a polynomial with no repeated roots and the exponent n is relatively prime to the degree of f(z) Let (z p : w p : 1) and (z q : w q : 1) be two rational points on the superelliptic curve Upon defining the rational numbers 14

15 n ij = Y ij = 1 z p T 4 DX ij T w p n kl = Y kl = w p DX kl D z p T 4 n ik = Y ik = 1 z p T 4 DX ik T w p n jl = Y jl = w p DX jl D z p T 4 we have the relation n ij n kl = Y ij DX ij Y kl = DX kl DT = Y ik DX ik Y jl DX jl = n ik n jl Finally, we want to show that finding three rational points on E T corresponds to finding at least three rational points on E T Proof Using the morphism φ : E T E T defined by (z : w : 1) (D (T 4 + z )z : D 3 (T 4 z )w : 4z ) along with the substitutions above, we have that (X ij, Y ij, X kl, Y kl ), where X ij = 1 w T z(z + z + T 4 ) Y ij = D w(z T 4 ) T 3 z(z + z + T 4 ) X kl = w D(z 4 + 4z 3 + 6T 4 z + 4T 4 z + T 8 ) (z T 4 ) Y kl = w w D(z 4 + 4z 3 + 6T 4 z + 4T 4 z + T 8 ) (z T 4 ) 3 is a rational point on E T The rest follows This result asserts that one way to find three rational points on E T is to find three rational points on E T There is a simple way of finding all rational points R = (z : w : 1) on E T A theorem by Gerd Faltings states that if the genus of a curve is greater than one then E T (Q) is a finite set for any fixed rational T where T 1, 0, 1 Our curve has a genus of nine Hence one can use software such as SAGE to find all of them 3 Acknowledgments This work was conducted during the 010 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 H ) We also thank Dr Edray Goins, Dr Luis Lomelí, Dr Duane Cooper, Katie Ansaldi, Ebony Harvey, and the MSRI staff 15

16 References [1] Campbell, Garikai Points on y = x at Rational Distance Mathematics of Computation [] Erdös, Paul Ulam, the Man and the Mathematician Journal of Graph Theory [3] Lang, Serge Algebra Graduate Texts in Mathematics [4] Solymosi, Jozsef and Frank De Zeeuw On a Question of Erdös and Ulam Discrete Computational Geometry 16