Genealogy of Pythagorean triangles

Transcription

1 Chapter 0 Genealogy of Pythagorean triangles 0. Two ternary trees of rational numbers Consider the rational numbers in the open interval (0, ). Each of these is uniquely in the form q, for relatively prime positive integers p > q. We call p + q the height p of the rational numbers. The rational numbers in (0, ) with odd heights can be arranged in a ternary tree with root, as follows. For a rational number t of odd heights, the numbers, t, and t are also in (0, ) and have odd heights. We call these the descendants +t +t of t and label them the left (L), middle (M), and right (R) respectively. If we write t = q p, then these three descendants are, p and q, and have greater p p q p+q p+q heights. Thus, the rational number has left descendant, middle descendant, and right descendant. s s = t s = +t s = t +t 0 On the other hand, each rational number s (0, ) \ {, } with odd height is the descendant of a unique rational number t, which we call its parent. In fact, s = n is m t

2 0 Genealogy of Pythagorean triangles (i) the left descendant of = n m if < s <, s n (ii) the middle descendant of = m n if < s <, and n s (iii) the right descendant of = n if 0 < s <. s m n Thus, every rational number in (0, ) of odd height is in the ternary tree with root : The same applies to rational numbers with even heights. They constitute a ternary tree with root : Therefore, each rational parameter s (0, ) \ {, } has a unique genealogy sequence tracing back to the root. For example, 6 L 0 M 0 R L L. Consider one of these ternary trees. If we flatten the entire tree by listing the vertices in order, beginning with the root, going down through each level from left to right, what is the position of a vertex with a known genealogy sequence?

3 0. Genealogy of Pythagorean triangles Suppose this genealogy sequence has k terms, i.e., the vertex is k levels below the root. Convert it into an integer N in base expansion by L 0, M, R respectively. Then the position of the vertex in the list is (k +)+N. For example, the rational number is in position 6 ( + ) = + = 6, with a genealogy sequence 6 L 0 M 0 R L L. Exercise () What is the 000-th vertex in this list from the ternary tree of rational numbers of odd heights, and what is its genealogy sequence? 0 69 R 0 89 M 9 0 R 9 M 9 M R. () Show that the rational numbers t and t belong to different ternary trees. +t How are their genealogy sequences related? 0. Genealogy of Pythagorean triangles The ternary trees in the preceding sections can be translated into a genealogy of Pythagorean triangles. A Pythagorean triangle (or its similarity class) is generated by a positive rational number t = q p of odd height. The tree with root translates into (,, ) (,, ) (, 0, 9) (, 8, ) (,, ) (, 8, ) (9, 80, 89) (, 6, 8) (, 6, 6) (,, (, 8, ) (9, 0, 69) (6,, 9)

4 Genealogy of Pythagorean triangles We find the descendants of a Pythagorean triangle (a, b, c) in terms of the sides The left descendant is generated by a = p q, b = pq, c = p + q. p p q and has sides a l = (p q) p = p pq + q = a b + c, b l = (p q)p = p pq = a b + c, c l = (p q) + p = p pq + q = a b + c. The middle descendant is generated by p p+q and has sides a m = (p + q) p = p + pq + q = a + b + c, b m = (p + q)p = p + pq = a + b + c, c m = (p + q) + p = p + pq + q = a + b + c. The right descendant is generated by q and has sides p+q a r = (p + q) q = p + pq + q = a + b + c, b r = (p + q)q = pq + q = a + b + c, c r = (p + q) + q = p + pq + q = a + b + c. Depending on the value of q, the parent of (a, b, c) is generated by one the p fractions q p p q q,, and. Since these fractions have the same numerator q q p q and denominators, up to permutation and change of signs, they all generate the Pythagorean triangle a = q (q p) = p + pq q = a + b c, b = q(q p) = pq + q = a + b c, c = q + (q p) = p pq + q = a b + c. Consider a right triangle ABC with vertices A = (0, b), B = (a, 0), and C = (0, 0), with semiperimeter s = (a + b + c). The incenter and the excenters are the points I = (s c, s c), I a = (s b, (s b)), I b = ( (s a), s a), I c = (s, s). The circles with these centers and respective radii r = s c, r a = s b, r b = s a, and r c = s are tangents to the sidelines of the triangle. According to the famous Feuerbach theorem, each of these circles is tangent to the nine-point circle, which is ( the circle passing the midpoints of the three sides. This circle has center N = a, ) b and radius c. The following theorem gives a nice geometric interpretation of the genealogy of Pythagorean triangles.

5 0. Genealogy of Pythagorean triangles Theorem 0.. The right triangles with hypotenuses NI a, NI b, NIc and sides parallel to BC and AC are similar to the descendants of ABC. The one with hypotenuse NI (and sides parallel to BC and AC) is similar to the parent of ABC. Proof. The following table shows the sidelengths of the right triangles involved each magnified by a factor : horizontal vertical hypotenuse NI a + b c a + b c a b + c parent NI a a b + c a b + c a b + c left NI b a + b + c a + b + c a + b + c right NI c a + b + c a + b + c a + b + c middle I c A I b N I C B I a

6 Chapter Polygonal numbers. The polygonal numbers P k,n The n-th triangular number is T n = n = n(n + ). The first few of these are,, 6, 0,,, 8, 6,,,.... The pentagonal numbers are the sums of the arithmetic progression (n ) + The n-th pentagonal number is P n = n(n ). Here are the beginning ones:,,,,,, 0, 9,,,... More generally, for a fixed k, the k-gonal numbers are the sums of the arithmetic progression + (k ) + (k ) +. The nth k-gonal number is P k,n = n((k )n (k )).

7 6 Polygonal numbers. The equation P k,a + P k,b = P k,c By a k-gonal triple, we mean a triple of positive integers (a, b, c) satisfying P k,a + P k,b = P k,c. (.) A -gonal triple is simply a Pythagorean triple satisfying a + b = c. We shall assume in the present chapter that k. By completing squares, we rewrite (.) as [(k )a (k )] + [(k )b (k )] = [(k )c (k )] + (k ), (.) and note, by dividing throughout by (k ), that this determines a rational point on the surface S: x + y = z +, (.) namely, P(k; a, b, c) := (ga, gb, gc ), (.) where g = (k ). This is always an integer point for k =,, 6, 8, with corresponding g =, 6,,. For k = (triangular numbers), we shall change signs, k and consider instead the point P (; a, b, c) := (a +, b +, c + ). (.) The coordinates of P (; a, b, c) are all odd integers exceeding.. Double ruling of S The surface S, being the surface of revolution of a rectangular hyperbola about its conjugate axis, is a rectangular hyperboloid of one sheet. It has a double ruling, i.e., through each point on the surface, there are two straight lines lying entirely on the surface. Let P(x 0, y 0, z 0 ) be a point on the surface S. A line l through P with direction numbers p : q : r has parametrization l : x = x 0 + pt, y = y 0 + qt, z = z 0 + rt.

8 . Primitive Pythagorean triple associated with a k-gonal triple Substitution of these expressions into (.) shows that the line l is entirely contained in the surface S if and only if It follows that This means px 0 + qy 0 = rz 0, (.6) p + q = r. (.) r = r (x 0 + y 0 z 0 ) = r (x 0 + y 0) (px 0 + qy 0 ) = (p + q )(x 0 + y 0 ) (px 0 + qy 0 ) = (qx 0 py 0 ). qx 0 py 0 = ǫr, ǫ = ±. (.8) Solving equations (.6) and (.8), we determine the direction numbers of the line. We summarize this in the following proposition. Proposition.. The two lines lying entirely on the hyperboloid S : x + y = z + and passing through P(x 0, y 0, z 0 ) have direction numbers for ǫ = ±. x 0 z 0 ǫy 0 : y 0 z 0 + ǫx 0 : x 0 + y 0 In particular, if P is a rational point, these direction numbers are rational.. Primitive Pythagorean triple associated with a k-gonal triple Let P be the rational point determined by a k-gonal triple (a, b, c), as given by (.), for k and (.) for k = (triangular numbers). We first note that the

9 8 Polygonal numbers coordinates of P all exceed. This is clear for k =, and for k, it follows from the fact that g = (k ) >. The direction numbers of the ruling lines on S k through the point P, as given in Proposition, are all positive. In view of (.), we may therefore choose a primitive Pythagorean triple (p, q, r) for these direction numbers. As is well known, every such triple is given by p = m n, q = mn, r = m + n (.9) for relatively prime integers m > n of different parity. We study the converse question of determining k-gonal triples from (primitive) Pythagorean triples.. Triples of triangular numbers Given a primitive Pythagorean triple (p, q, r) as in (.9), we want to determine a triangular triple (a, b, c) corresponding to it. Given an odd integer z 0 >, we obtain, from (.6) and (.8), x 0 = pz 0 + ǫq r, y 0 = qz 0 ǫp. (.0) r We claim that it is possible to choose z 0 > so that x 0 and y 0 are also odd integers >. By the euclidean algorithm, there are odd integers u and v such that qu + rv =. (Note that v must be odd, since q is even. If u is even, we replace (u, v) by (u r, v + q), in which both entries are odd). Clearly, the integer z 0 = ǫpu is such that qz 0 ǫp = ǫp(qu ) is divisible by r. This makes y 0 an integer. The corresponding x 0 is also an integer. Replacing z 0 by z 0 + rt for a positive integer t if necessary, the integers z 0, x 0, and y 0 can be chosen greater than. From (.0), the integers x 0 and y 0 are both odd, since p and q are of different parity and z 0 is odd. We summarize this in the following theorem. Theorem.. Let (p, q, r) be a primitive Pythagorean triple. There are two infinite families of triangular triples (a ǫ (t), b ǫ (t), c ǫ (t)), ǫ = ±, such that one of the lines l ǫ (P), P = P (; a ǫ (t), b ǫ (t), c ǫ (t)), has direction numbers p : q : r.

10 .6 k-gonal triples determined by a Pythagorean triple 9 Triangular triples from primitive Pythagorean triples (m, n) (p, q, r) (a + (0), b + (0), c + (0)) (a (0), b (0), c (0)) (, ) (,, ) (,, ) (,, 6) (, ) (, 8, ) (9,, 0) (,, 6) (, ) (,, ) (, 9, 0) (,, ) (6, ) (,, ) (0, 6, ) (,, ) (, ) (, 0, 9) (6,, 8) (,, 0) (, ) (,, ) (6, 0, ) (,, 8) (8, ) (6, 6, 6) (, 8, 6) (,, 8) (, ) (, 8, ) (,, ) (9, 6, ) (, ) (9, 0, ) (8,, 6) (9,, ).6 k-gonal triples determined by a Pythagorean triple Now, we consider k. We shall adopt the notation for an integer h. { h if h is odd, h := h if h is even, Theorem.. Let k and g = (k ). The primitive Pythagorean triple (p, q, r) k defined in (.9) by relatively prime integers m > n with different parity corresponds to a k-gonal triple if and only if one of n (m n) and is an integer. g g Proof. As in (.0) above, the rational points through which the surface S contains a line of direction numbers p : q : r are of the form ( pz + ǫq, r qz ǫp, z). (.) r Suppose this corresponds to a k-gonal triple (a, b, c), so that z = rc. From (.), we obtain, for ǫ =, a = b = m + n (k ) (m + n ) [(k ) (m n)c + (k ) n], (.) n (k ) (m + n ) [(k ) mc (k ) (m n)]. (.) Note that (k ) and (k ) are always relatively prime, since gcd(k, k ) = or according as k is odd or even. From these expressions,

11 0 Polygonal numbers a + b c = (k ) n (k ) (m + n ) [(k ) (m n)c + (k ) n]. We claim that n must be divisible by (k ) for a, b, c to be integers. Let d := gcd(n, (k ) ), so that n = d n, for relatively prime integers n and (k ). a + b c = (k ) = d (k ) (k ) n (k ) (m + n ) [(k ) (m n)c + (k ) n ]. Since (k ) is prime to each of (k ) and n, the only possible prime divisor of (k ) is. This means that (k ) is a power of, (possibly ). If (k ) is even, then after cancelling a common divisor, the numerator of a +b c is odd, and the denominator is even. This cannot be an integer. It follows that (k ) =, justifying the claim that n must be divisible by (k ). Since g = (k ), the condition that n be divisible by (k ) is equivalent (k ) to n being an integer. Under this condition, there is a unique positive integer g c 0 < m +n for which a 0 defined by (.) is an integer. Note that a 0 +b 0 c 0 is also an integer. Since b 0 is rational, it too must be an integer. Every k-gonal triple associated with the primitive Pythagorean triple (p, q, r) is of the form a t = a 0 + pt, b t = b 0 + qt, c t = c 0 + rt for a positive integer t. For ǫ =, the treatment is exactly the same, with n replaced by m n. Indeed, we have m n a = (k ) (m + n ) [(k ) (m + n)c (k ) n], m b = (k ) (m + n ) [(k ) nc + (k ) (m n)]. Since m and n are relatively prime, the integer (k ) > cannot divide both n and m n. This means that a primitive Pythagorean triple (p, q, r) corresponds to at most one line on S associated with k-gonal triples (for k ). Indeed, if k = h +, (k ) is the even number h, and cannot divide the odd integer m n. It follows that only those pairs (m, n), with n a multiple of h give (h+)-gonal pairs. For example, by choosing m = h+, n = h, we have p = h +, q = 8h + h, r = 8h + h +, a 0 = h +, b 0 = 8h + h +, c 0 = 8h + h +.

12 .6 k-gonal triples determined by a Pythagorean triple These give an infinite family of (h + )-gonal triples: a t = (h + )(t + ), b t = 8h + h + + (8h + h)t, = 8h + h + + (8h + h + )t. c t (h + ) gonal triples (h,k,g) (m,n) (p,q,r) (a,b,c) (,6,) (,) (,,) (,,) (,) (,0,9) (,,9) (,) (9,0,) (9,8,9) (,) (,8,) (8,,) (,) (,6,6) (,8,) (,6) (,8,8) (,8,8) (9,) (,6,8) (,,) (9,) (6,,9) (,,9) (9, 8) (,, ) (, 0, ) (, ) (,, ) (0, 9, ) (, ) (0, 88, ) (60, 0, 8) (, 6) (8,, ) (68, 0, ) (, 8) (, 6, 8) (8, 6, ) (, 0) (, 0, ) (,, 6) (,0, 8 ) (,) (9,0,) (9,,8) (,) (,6,6) (,,6) (9,) (6,,9) (,,) (9, 8) (,, ) (, 8, 9) (, ) (0, 88, ) (90,, ) (, 8) (, 6, 8) (,, 8) (,, ) (,6) (,8,8) (,9,80) (, 6) (8,, ) (8,, 6)