Pythagorean vectors and their companions. Lattice Cubes


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1 Lattice Cubes Richard Parris Richard Parris received his mathematics degrees from Tufts University (B.A.) and Princeton University (Ph.D.). For more than three decades, he has lived at Phillips Exeter Academy (NH), where he teaches mathematics, coaches, and supervises a dormitory. For fifteen years, much of his spare time was spent working on the American Mathematics Competitions for the MAA, but now he has his hands full maintaining his publicdomain software, available in seventeen foreign languages. It is a simple matter to construct a lattice square in two dimensions given an arbitrary integer vector [a, b], the perpendicular vector [ b, a] completes the task. The analogous challenge in three dimensions is to construct a lattice cube, a set of three mutually perpendicular integer vectors with a common length. It is natural to wonder whether a given integer vector belongs to such a set. The reader is invited to consider this question for the vector [62, 34, 25], which will be discussed later in the paper. By describing a process for constructing such a cube, this paper will show that the answer is yes for precisely those integer vectors whose lengths are also integers. It turns out that the cube is essentially unique when the given vector is primitive, meaning that its components do not have a common prime divisor. Pythagorean vectors and their companions It does not take long to realize that an arbitrarily selected initial vector is unlikely to belong to a lattice cube. Consider [3, 5, 6]. There is no integer vector [a, b, c] of length 70 perpendicular to [3, 5, 6], so this vector does not even belong to a lattice square. Even if it did, extending such a square to a lattice cube might be impossible, as the perpendicular vectors [4, 5, 7] and [ 8, 5, 1] show. Although these vectors have 90 as their common length, there is no lattice cube that extends this lattice square (see below). Lattice cubes are nevertheless abundant. The examples [2, 3, 6] [3, 6, 2] [6, 2, 3] (1) [1, 4, 8] [8, 4, 1] [4, 7, 4] (2) [2, 5, 14] [10, 10, 5] [11, 10, 2] (3) [1092, 156, 65] [ 119, 408, 1020] [120, 1015, 420] (4) doi: /college.math.j THE MATHEMATICAL ASSOCIATION OF AMERICA
2 illustrate the possibilities. In each example, notice that the common length of the three vectors is actually an integer. This Pythagorean property is not a required feature of lattice squares, but it is for lattice cubes. Pythagorean Lemma. If u, v, and w are three mutually perpendicular integer vectors that share a common length L, then L must be an integer. Proof. It is evident that L 2 is an integer. The cross product u v is an integer vector that is parallel to w, and the length of u v is L 2. The ratio of corresponding components of u v and w is therefore either L or L. It follows that L is rational, which requires that L be an integer. An integer vector whose length is a positive integer will be called Pythagorean. If the nine components of a lattice cube have no common divisor, the cube is called primitive. Example (3) shows that a primitive lattice cube can include imprimitive vectors, and example (4) shows that all the vectors in a primitive lattice cube can be imprimitive. Formulas exist (see [4]) for generating threecomponent Pythagorean vectors in profusion, and the following states that they all belong to lattice cubes. The proof (presented later) is algorithmic, once the prime factors of the length of the vector are known. Theorem 1. Every Pythagorean vector u belongs to a lattice cube. There may be several such cubes. For example, belongs to u = [1020, 420, 65] = 5[204, 84, 13] [1020, 420, 65] [425, 1008, 156] [0, 169, 1092] [1020, 420, 65] [119, 120, 1092] [408, 1015, 156] [1020, 420, 65] [255, 740, 780] [340, 705, 780]. On the other hand, a primitive Pythagorean u belongs to a lattice cube that is essentially unique: Theorem 2. Let u, v, and w be Pythagorean vectors with a common length. If u is primitive, and if v and w are both perpendicular to u, then v and w are either perpendicular or parallel. Perpendicular Pythagorean vectors will be called companions. As the following lemma implies, finding a lattice cube that includes a given u is a problem that is solved by finding a single companion of the same length as u. Divisibility Lemma. If the positive integer e is a divisor of the lengths of the perpendicular Pythagorean vectors u and v, then w = u v is e times a Pythagorean vector. Proof. Let u = [a, b, c] and v = [k, m, n], so that w = [bn cm, ck an, am bk]. Let u = se and v = te, where s and t are positive integers. Because VOL. 42, NO. 2, MARCH 2011 THE COLLEGE MATHEMATICS JOURNAL 119
3 u is perpendicular to v, the length of w is ste 2. Thus s 2 t 2 e 4 = (bn cm) 2 + (ck an) 2 + (am bk) 2 = (bn cm) 2 + c 2 k 2 2ackn + a 2 n 2 + a 2 m 2 2abkm + b 2 k 2 = (bn cm) 2 + (b 2 + c 2 )k 2 + (m 2 + n 2 )a 2 2(ackn + abkm) = (bn cm) 2 + (s 2 e 2 a 2 )k 2 + (t 2 e 2 k 2 )a 2 2(ackn + abkm) = (bn cm) 2 + (s 2 k 2 + t 2 a 2 )e 2 2ak(ak + bm + cn) = (bn cm) 2 + (s 2 k 2 + t 2 a 2 )e 2, from which it is evident that bn cm is divisible by e. Similar remarks apply to the other components of w, and the result follows. In particular, companion vectors of the same length belong to a lattice cube. Familiar examples Before showing that every Pythagorean vector has a companion of the same length, three special cases will be analyzed. The striking aspect of examples (1) and (2) is that companions for u can sometimes be found by simply rearranging the components of u and changing some signs. Let u = [a, b, c] be primitive, and let d be its length. During this discussion, there is no loss of generality in assuming that 0 < a b c < d, for permuting the rows or columns of a 3 3 matrix affects neither perpendicularity nor length, nor does making uniform sign changes to a row or a column. First consider the cyclic permutation [b, c, a] of the components of u. There are seven ways to introduce sign changes into [b, c, a], but only two of them have any chance of producing perpendicularity. Notice first that [ b, c, a] cannot produce an example, because the perpendicularity condition (a + b)c = ab implies that c < a. In a similar fashion, the cases [b, c, a], [b, c, a], and [ b, c, a] are ruled out, and [ b, c, a] is trivially excluded. This leaves two equivalent cases of interest: [b, c, a] and [ b, c, a]. The perpendicularity condition is ba + ca = bc in either case. Solving ba + ca = bc is facilitated by writing it in the factored form a 2 = (b a)(c a). It is then clear that solutions are in onetoone correspondence with divisor pairs for a 2. In other words, if pq = a 2, with p q, then b = a + p and c = a + q determine a Pythagorean quadruple, because d 2 = a 2 + (a + p) 2 + (a + q) 2 = a 2 + a 2 + 2ap + p 2 + a 2 + 2aq + q 2 = a 2 + pq + 2ap + p 2 + pq + 2aq + q 2 = (a + p + q) 2. Notice that u = [a, a + p, a + q] is perpendicular to v = [a + p, (a + q), a]. It is evident that u is primitive if and only if p and q are relatively prime, which requires that each be a perfect square (since their product is a 2 ). Replacing p by m 2 and q by n 2 leads to the vector u = [mn, m(m + n), n(m + n)], whose length is d = m 2 + n 2 + mn = b + c a. 120 THE MATHEMATICAL ASSOCIATION OF AMERICA
4 It is straightforward to verify that the other cyclic permutation [c, a, b] of the components of u produces essentially the same cubes. The lattice cube determined by a primitive u = [a, b, c] that satisfies 0 < a b c < d and bc = ab + ac is therefore completed by the vectors v = [b, c, a] and w = [c, a, b]. The predictable dw = u v is easily verified, using the relationship d = b + c a. Any lattice cube determines a minimal enveloping box whose faces are parallel to the coordinate planes, and for some lattice cubes this box is itself a cube. This property actually characterizes the cyclic examples just discussed. The reader is invited to prove this statement. Next consider transposing the components of u to obtain [a, c, b], [c, b, a], and [b, a, c]. Obtaining a vector perpendicular to u by making sign changes leads to the three conditions a 2 = 2bc for [ a, c, b]; b 2 = 2ac for [c, b, a]; and c 2 = 2ab for [b, a, c]. The first one is excluded because it contradicts a b, but the other two have solutions. First, if b 2 = 2ac, then u = [a, b, c] is perpendicular to v = [c, b, a], and d = a + c. The lattice cube can be completed by w = [b, c a, b] (which is primitive whenever u is). This family of solutions has relatively prime parameters a and c; the defining condition is that 2ac be a perfect square. Example (2) is of this type, as is the familiar [1, 2, 2] [2, 2, 1] [2, 1, 2]. Next, if c 2 = 2ab, then u = [a, b, c] is perpendicular to v = [b, a, c], and d = a + b. The lattice cube can be completed by w = [c, c, b a] (which is primitive whenever u is). This family of solutions has relatively prime parameters a and b; the defining condition is that 2ab be a perfect square. The ubiquitous is of this type, as is [1, 2, 2] [2, 2, 1] [2, 1, 2] [8, 9, 12] [9, 8, 12] [12, 12, 1]. Figure 1 shows example (2), which can be viewed as a rotated image of the congruent Figure 1. VOL. 42, NO. 2, MARCH 2011 THE COLLEGE MATHEMATICS JOURNAL 121
5 cube defined by [0, 0, 9] [0, 9, 0] [9, 0, 0]. The rotation axis lies in a coordinate plane. It is not difficult to prove that this property characterizes the transposed examples above. A third collection of examples has the form u = [0, b, c], with 0 < b < c < d. It is clear that [0, b, c] [0, c, b] [d, 0, 0] is a lattice cube that includes u. The reader is invited to establish its uniqueness. Building a cube As example (3) suggests, a companion vector having the same length as a given u cannot always be found by rearranging its components and changing some signs. It is therefore not yet clear that u has any companions. The remainder of this paper describes a process for calculating a lattice cube for any Pythagorean u. The process is initiated by the Existence Lemma, which provides a formula for a trial cube. This set of vectors solves the latticecube problem for a positive integer multiple of u. Subsequent steps rely on the Shrinking Lemma, which reduces the size of this integer multiple, one prime factor π at a time, by rotating the cube around its uaxis through special angles chosen to make a new set of companion vectors divisible by π. Existence Lemma. For any Pythagorean vector u, there exist Pythagorean vectors v and w, and a positive integer h, such that hu, v, and w form a lattice cube. Proof. Given u = [a, b, c] and d = u, let v = [ac, bc dc, db a 2 b 2 ] (see [3]). It is routine to verify that v is perpendicular to u. The calculation v 2 = a 2 c 2 + b 2 c 2 2bdc 2 + d 2 c 2 + d 2 b 2 2db(a 2 + b 2 ) + (a 2 + b 2 ) 2 = (a 2 + b 2 )(c 2 + a 2 + b 2 ) 2bd(a 2 + b 2 + c 2 ) + d 2 c 2 + d 2 b 2 = (a 2 + b 2 2bd + c 2 + b 2 )d 2 = (d 2 2bd + b 2 )d 2 = (d b) 2 d 2 shows that v = (d b)d is an integer, so h = d b. The Divisibility Lemma reveals that u v is d times a Pythagorean vector w. Because d w = u v = u v = d v, it follows that w = v. The vectors v and w solve the latticecube problem for hu. Although the edge length of this cube is divisible by h, it need not be true that h divides v. If that were true, the desired companion for u would have been found. (For those cases in which d = b + 1, the Existence Lemma formula yields the desired companion directly.) It will now be shown that a trial cube for u in which 1 < h can be used to obtain a smaller trial cube for u. Any companion vector for u must be a rational linear combination of v and w. Moreover, x 2 + y 2 = 1 implies that xv + yw = v = h u, because v and w are 122 THE MATHEMATICAL ASSOCIATION OF AMERICA
6 perpendicular. Thus the desired companion for u is expressible as (x/h)v + (y/h)w, where x and y are rational solutions to x 2 + y 2 = 1. The obvious question is how to find x and y so that (x/h)v and (y/h)w are integer vectors. It will now be shown that there is a canonical solution when h is prime. For example, let u = [62, 34, 25] be the vector that appeared in the introductory paragraph. Its length is d = 75. The Existence Lemma provides a trial cube for u, consisting of 41u, v = [1550, 1025, 2450], and w = [ 769, 2542, 1550]. It is inviting to use an integer solution to λ 2 + τ 2 = 41 2 to define x = λ/41 and y = τ/41, in the hope that z = (x/41)v + (y/41)w will be a companion for u. The length of z is of course 75, but λ and τ must be chosen so that λv + τw is divisible by There are only two possible combinations to try, 40v ± 9w, and it is easy to verify that 40v 9w = [68921, 63878, 84050] = 41 2 [41, 38, 50]. Thus z = [41, 38, 50] is a companion for u. This illustrates the following key result. Prime Lemma. Let πu, v, and w be a trial cube for the Pythagorean vector u, where π is prime. Then there are integers λ and τ satisfying π 2 = λ 2 + τ 2, for which λv + τw = π 2 z, where z is an integer vector. Proof. Let u = [a, b, c] and v = [k, m, n]. It is given that a 2 + b 2 + c 2 = d 2 (5) k 2 + m 2 + n 2 = π 2 d 2 (6) ak + bm + cn = 0 (7) There is no loss of generality in assuming that u is primitive. Notice also that λ = π and τ = 0 solve those examples in which v is already π times the desired companion vector. This includes the case π = 2, because v is even in this case, requiring the sum (6) of three integer squares to be divisible by 4. This can happen only if all three squares are even, and v is therefore divisible by 2. Let π be congruent to 1 modulo 4 (as in the preceding example), and let w = [p, q, r]. The row vectors of the integer matrix πa πb πc M = k m n p q r are mutually perpendicular, and πd is their common length. Because M is an integer multiple of an orthogonal matrix, it follows that the columns of M have the same length πd. In particular, π 2 a 2 + k 2 + p 2 = π 2 d 2. It is well known (see [1, p. 105]) that there are positive integers λ and τ (unique except for order) that satisfy π 2 = λ 2 + τ 2. Hence (λk) 2 (τ p) 2 = λ 2 k 2 (π 2 λ 2 )p 2 = λ 2 (k 2 + p 2 ) π 2 p 2 = λ 2 (π 2 d 2 π 2 a 2 ) π 2 p 2 = π 2 (λ 2 d 2 λ 2 a 2 p 2 ), which shows that π 2 divides (λk + τ p)(λk τ p). If π 2 did not divide one of the factors, then π would divide each factor, as well as their sum and difference, forcing VOL. 42, NO. 2, MARCH 2011 THE COLLEGE MATHEMATICS JOURNAL 123
7 π to divide both k and p. Corresponding remarks can be made for m and q, and for n and r. Because π does not divide v, however, assume (without loss of generality) that k is not divisible by π, and that π 2 divides the factor λk + τ p. It will now be shown that π 2 divides the other two components of λv + τw. Recall the trialcube hypothesis v w = (πd)πu. In terms of components, this means that mr nq = π 2 da np kr = π 2 db kq mp = π 2 dc. In particular, the last equation says that kq and mp are equivalent modulo π 2. The intuitive content of this equation is that m and q are proportional to k and p, hence that λm + τq is equivalent (modulo π 2 ) to a multiple of λk + τ p. In fact, the multiplier mk works, where k is a reciprocal for k, provided by the Euclidean algorithm. In other words, there are integers k and β for which kk = 1 + βπ 2. The reader is now invited to show that λm + τq mk (λk + τ p) is indeed divisible by π 2, by applying kq mp = π 2 dc and kk = 1 + βπ 2. Because π 2 divides λk + τ p, it follows that π 2 also divides λm + τq. In a similar fashion, it follows that π 2 divides λn + τr. Hence λv + τw is divisible by π 2. Now assume that π is congruent to 3 modulo 4. It will be shown that v is divisible by π, by using a wellknown property of such primes: π must divide each of two integers if it divides the sum of their squares. (See [1, p. 105].) Recall the equation π 2 a 2 + k 2 + p 2 = π 2 d 2 for the square of the length of the first column of matrix M above. Combine this equation with a 2 + b 2 + c 2 = d 2, and simplify to obtain k 2 + p 2 = π 2 (b 2 + c 2 ). Thus π divides k 2 + p 2. It follows that π divides k, and (6) then shows that π divides m 2 + n 2. Hence π divides m and n, hence π divides v. Shrinking Lemma. Let hu, v, and w form a trial cube for the Pythagorean vector u, where 1 < h. Then u has a trial cube h 1 u, v 1, w 1, in which h 1 < h. Proof. Let π be a prime divisor of h. Define h 1 = h/π and u = h 1 u. Notice that πu, v, and w form a trial cube for u. Apply the Prime Lemma, to obtain integers λ and τ satisfying π 2 = λ 2 + τ 2, and an integer vector z for which λv + τw = π 2 z. From π 2 z = λv + τw = v λ 2 + τ 2 = hdπ it follows that z = h 1 d. Now apply the Divisibility Lemma to the perpendicular vectors u and z, whose lengths are both divisible by d. It is found that u z = dz, for some Pythagorean vector z. Notice that z = z. Hence v 1 = z and w 1 = z define the promised cube. Proof of Theorem 1. The Existence Lemma establishes a trial cube for any Pythagorean vector u. If the vectors of a trial cube for u are longer than u, then the Shrinking Lemma replaces this cube by a smaller trial cube. Theorem 1 is actually a special case of the HallRyser orthogonal completion theorem. See [2, p. 81], for example. Proof of Theorem 2. Let d = u and θ be the angle between v and w. Assume that v and w are not parallel, so that sin θ = 0. Then v w is a nonzero integer vector 124 THE MATHEMATICAL ASSOCIATION OF AMERICA
8 that is parallel to u. The length L = d 2 sin θ of v w is a rational multiple of d. Thus L is an integer. Because u is primitive, i = d sin θ must be an integer. Notice that cos θ = v w/d 2 is also rational, which implies that j = d cos θ is an integer, and i 2 + j 2 = d 2. This Pythagorean triple might not be primitive, so let i = i 1 g, j = j 1 g, and d = d 1 g, where i 1 and d 1 are relatively prime. It will now be shown that u is divisible by every prime divisor of d 1, which can happen only if d 1 = 1, implying that v and w are perpendicular. Let u = [a, b, c], v = [k, m, n], and w = [p, q, r], so that v w = [mr nq, np kr, kq mp] = i[a, b, c]. A straightforward calculation (as in the proof of the Divisibility Lemma) shows that d 2 i 2 = (mr nq) 2 + (k 2 + p 2 )d 2 2kpd j d 2 i 2 = i 2 a 2 + (k 2 + p 2 )d 2 2kpd j d 2 1 i 2 = i 2 1 a2 + (k 2 + p 2 )d 2 1 2kpd 1 j 1 from which it is now evident that d 1 divides i 2 1 a2. Similar calculations apply to the other components of u. Summary. Given a segment that joins two lattice points in R 3, when is it possible to form a lattice cube that uses this segment as one of its twelve edges? A necessary and sufficient condition is that the length of the segment be an integer. This paper presents an algorithm for finding such a cube when the prime factors of the length are known, and shows that the cube is essentially unique if the given segment does not contain another lattice point. References 1. W. LeVeque, Elementary Theory of Numbers, AddisonWesley, London, M. Newman, Integer Matrices, Academic Press, New York, A. Osborne and H. Liebeck, Orthogonal bases of R 3 with integer coordinates and integer lengths, Amer. Math. Monthly 96 (1989) doi: / R. Spira, The Diophantine equation x 2 + y 2 + z 2 = m 2, Amer. Math. Monthly 69 (1962) doi: / Selfevidence If this sentence is true, then Santa Claus exists. It is easily seen that if the above sentence is true, then Santa Claus exists, and since the sentence asserts just that, it must be true. from Logical Labyrinths by Raymond Smullyan, reviewed on page 159 VOL. 42, NO. 2, MARCH 2011 THE COLLEGE MATHEMATICS JOURNAL 125
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