Do there exist rectangular solids with integer edge lengths such that all three geodesic distances between opposite vertices are also integers? Herman Baer noted the three solutions (108,357,368), (564,748,1425), and (348,975,2380), and asked for more information about such solutions. This is an interesting problem - one that you'd think would have been considered before - but I haven't been able to find any reference to this particular set of Diophantine equations. Dave Rusin responded to Baer with some additional solutions, such as (243984,675500,689613). I was curious to know how many primitive solutions exist between the three small examples and this larger one. To review, the problem is to find a rectangular box with integer dimensions AxBxC such that all the (locally) geodesic paths on the surface of the box from one corner to the diagonally opposite corner, as shown below, have integer lengths. (We exclude the infinitely many geodesic paths that enter any face more than once.) ___________end / /\ / / \ / / \ B/ / / / / / / / / / / / /__________/ / \ \ / A\ \ / \__________\/ start C Clearly any geodesic path on the surface must be a straight line on any given wall, so if we "unfold" the box in various ways and lay all the sides out flat, the geodesic paths are just the straight lines connecting opposite corners, as illustrated below (where "s" indicates the starting corner and e1,e2,e3 are the three basic positions of the ending corner for the three different ways of unfolding the box. | A | _ __________e1 | | A | | _ |__________|_____e2 | | | | | | | | | B | | | | | | | | | | | | _s|__________|_____|____________ | | | A | | | _ |__________|__________________| e3 | C | B | From this we see that the three locally geodesic path lengths x,y,z are related to the edge lengths A,B,C according to the equations x^2 = (A+B)^2 + C^2 y^2 = (A+C)^2 + B^2 z^2 = (B+C)^2 + A^2 So, the task is to find integers A,B,C,x,y,z that satisfy all three of these equations simultaneously. If we re-write these equations in the form 2ab + D2 = x^2 2ac + D2 = y^2 2bc + D2 = z^2 where D2 is the squared length of the main diagonal of the box (which we don't require to be a square integer), this Diophantine problem bears some resemblence to a class of problems considered by Diophantus himself, namely, to find three numbers such that their products in pairs increased by a constant are each squares. In other words, Diophantus considered the set of simultaneous equations (in rational numbers) ab + k = x^2 ac + k = y^2 bc + k = z^2 The case k=1 in particular has been studied extensively, and extended to sets of four, and even five (rational) numbers whose products in pairs are 1 less than squares. This was studied by Euler and many others. However, the Diophantine box problem is a bit trickier, because it asks us to find three integers such that TWICE each pair- wise product, increased by the sum of of the squares, is a square. I'm not sure if the classical methods used to solve Diophantus' original system of equations can be adapted to treat this problem, but we can say a few things about the integer S = A+B+C that makes searching for solutions a bit easier. Obviously each of the conditions corresponds to a Pythagorean triple which can be parameterized in the usual way, i.e., (ku^2 - kv^2)^2 + (2kuv)^2 = (ku^2 + kv^2)^2 where u,v are coprime (one odd, one even). Now, S = A+B+C is the sum of the two terms on the left, so essentially the conditions of the problem require that S for a solution be expressible in the form (u^2 - v^2) + (2uv) in (at least) three distinct ways. Equivalently, S must have at least three distinct representations of the form (u+v)^2 - 2v^2. Thus, we are dealing with the number field Q[sqrt(2)], which possesses unique factorization. By Fermat's "Christmas theorem" we know that S must be a product of at least two primes of the form X^2 - 2Y^2, and probably three. Also we can see that a primitive solution must not be divisible by any primes NOT of this form. At first it may seem surprising that a value of S having only 2 factors could yield a solution, because S must be expressible in THREE distinct ways as a sum of two legs of a Pythagorean triple, viz, in the quadratic form X^2 - 2Y^2. Thus, we might expect that S must have at least three prime divisors congruent to +1 or -1 (mod 8). However, there exist solutions such as S = (41)(17921) in which S does have just two prime factors. This is quite rare - the only other such solution less than 10 million is S = (89)(88001). The reason such solutions are possible is that not all of the Pythagorean triples are necessarily primitive, and, furthermore, the common divisor "k" for any non-primitive triples must be a divisor of S=pq. As a result, we can get two of our representations of S from the usual composition of the quadratic forms of the two prime factors p,q of S, and then we get the 3rd representation as p times the minimal representation of q. In any case, the acceptable S values must be composed entirely of at least two (and usually three) primes, not necessarily distinct, modulo which 2 is a quadratic residue, i.e., primes congruent to +-1 modulo 8. The first few primes of this type are 7 17 23 31 41 47 71 73 79 89 97 103... etc. Indeed, the three smallest solutions have the S values 833 = (7)(7)(17) 2737 = (7)(17)(23) 3703 = (7)(23)(23) In general, to find values of S that give primitive solutions we can examine each integer and immediately exclude any that are not composed entirely of 2 or more primes congruent to +-1 (mod 8). Now, if two or all three of the Pythagorean triples are primitive, we can infer the solution simply by finding two minimal representations of S in the form 2y^2 - x^2 where 0 < x < y, which can clearly be found by a 1-dimensional search up to sqrt(S). Once we have found these minimal representations, we can take them, two at a time, to see if they give a valid solution. For example, the minimal representations of 833 in the form 2y^2 - x^2 are (x,y) = (7,21), (15,23), and (25,27), and these are converted to (u,v) parameterizations of the respective Pythagorean triples by u = y v = y - x Taking the 2nd and 3rd minimal forms gives the triples u v u^2 - v^2 2uv --- --- --------- ------- 23 8 465 368 27 2 725 108 Recalling that our three triples are of the form (A+B,C), (A+C,B), and (B+C,A), it's clear that if this is a solution then we must have A=108, B=368, and C = 465-108 = 725-368 = 357, and we can easily verify that this is, in fact, a solution. On the other hand, if our solution involves two (or more) non-primitive Pythagorean triples, the above method won't work, because the primitive triples are based on the minimal forms of the entire S, whereas the non-primitive triples are based on the minimal forms of proper FACTORS of S. So, to find such solutions we need to proceed as before, but now we need to include, in addition to the minimal representations of S, the representations of the form n(2y^2 - x^2) where n is any proper divisor of S. For any divisor n we find the minimal form(s) of S/n, convert them to (u,v) paramaters as above, and then the resulting Pythagorean legs are n(u^2 - v^2) and 2nuv. We can then check these purported triples in pairs as above to determine if they give three suitable triples. A table showing all 79 primitive solutions with S less than 50 million is presented in Attachment: Table of Geodesic Boxes Obviously if (A,B,C) is a solution then so is (kA,kB,kC) for any integer k, so we've considered only the "primitive" solutions, i.e., those for which gcd(A,B,C) = 1. Now, each primitive solution leads to the three Pythagorean triples (A+B)^2 + C^2 = x^2 (A+C)^2 + B^2 = y^2 (B+C)^2 + A^2 = z^2 where x,y,z are the three geodesic path lengths. Notice in the attached table that almost all the solutions less than 3 million have two primitive Pythagorean triples. For example, the smallest solution is S A B C (prime factors of S) A+B C x k1 u1 v1 A+C B y k2 u2 v2 B+C A z k3 u3 v3 = 833 108 357 368 (7)(7)(17) 465 368 593 1 23 8 476 357 595 119 2 1 725 108 733 1 27 2 and we note that two of the three Pythagorean triples are primitive, but one is just the simple 3^2 + 4^2 = 5^2 triple multiplied by 119. (Oddly enough, the larger solution found by Rusin by means of elliptic curves also involved a multiple of the 3^2 + 4^2 = 5^2 triple.) Most of the solutions are of this type, i.e., two primitive triples and one non- primitive triple, but there are also solutions in which only one of the Pythagorean triples is primitive, such as 1228687 112812 431955 683920 (79)(103)(151) 544767 683920 874367 103 83 40 796732 431955 906293 1 818 487 1115875 112812 1121563 79 119 6 Furthermore, there are even primitive geodesic boxes for which NONE of the Pythagorean triples is primitive, including 2627639 337824 516260 1773555 (7)(17)(71)(311) 854084 1773555 1968491 7 517 118 2111379 516260 2173579 311 83 10 2289815 337824 2314601 17 368 27 In addition, the table also shows that we have some "ultra-primitive" geodesic boxes, i.e., primitive geodesic boxes such that all three Pythagorean triples are also primitive. The smallest example of this is 2107609 312040 587340 1208229 (7)(17)(89)(199) 899380 1208229 1506221 1 1165 386 1520269 587340 1629781 1 1255 234 1795569 312040 1822481 1 1345 116 which has the dimensions A=312040, B=587340, C=1208229. The geodesic path lengths for this box are given by three primitive Pythagorean triples as x = (1069)(1409) y = (197)(8273) z = (1822481) and the sum of the edge lengths is S = A+B+C = 2107609 = (7)(17)(89)(199) There are a total of 8 ultra-primitive geodesic boxes with S less than 50 million, as shown in the table. Having found three-dimensional boxes of size AxBxC such that the edge lengths and all three geodesic path lengths connecting opposite vertices are integers, it's interesting to consider the analagous question for 4-dimensional boxes. First, note that for a 2D box of size AxB there is essentially just one geodesic distance S along the surface connecting opposite corners, given by S^2 = (A+B)^2 In other words, the distance is simply A+B. For a 3D box of size AxBxC there are three local geodesic distances, given by S1^2 = (A+B)^2 + C^2 S2^2 = (A+C)^2 + B^2 S3^2 = (B+C)^2 + A^2 The usual way of seeing this is to simply unfold the box and draw the straight lines connecting opposite corners, be we could also deduce this more formally by noting that we're required to go from the point [0,0,0] to the point [A,B,C] in the minimum linear segments, each of which is on (at least) one surface of the box. In other words, every segment must have at least one of the components at its min or max limit (0 or A,B,C respectively) for the entire segment. So the shortest path must be of the form [0,0,0] -> [A,0,t] -> [A,B,C] Notice that for the first segment the 2nd coordinate is always 0, whereas for the second segment the 1st coordinate is constantly A, so both segments are on the surface. The total length of this path S = sqrt(A^2 + t^2) + sqrt(B^2 + (C-t)^2) Differentiating with respect to t and setting the result to zero, we find that t = AC/(A+B), which can be substituted back into the above equation to give the expected result S^2 = (A+B)^2 + C^2 Now, for a 4-dimensional box we need to go from [0,0,0,0] to [A,B,C,D] via the most economical path on the surface. In this case the "surface" consists of 8 solid regions in which three of the coordinates are anywhere between their min and max while the fourth coordinate is at its min or max. Again we can visualize this by "unwrapping" the box, placing two of the faces adjacent to each other. For example, if we unfold a face with dimensions AxCxD into the same 3D space as a face of dimensions BxCxD we have a solid with overall dimensions (A+B)xCxD, and the shortest path connecting the two opposite corners has length S given by S^2 = (A+B)^2 + C^2 + D^2 Likewise for the other face combinations. Alternatively, we can proceed more formally by considering the path [0,0,0,0] -> [A,0,u,v] -> [A,B,C,D] whose total length is S = sqrt[A^2 + u^2 + v^2] + sqrt[B^2 + (C-u)^2 + (D-v)^2] Here we have two unknown paramaters, so we differentiate with respect to each of them separately to give the two conditions dS/du = 0 dS/dv = 0 Solving these simultaneous equations for u and v gives u = AC/(A+B) v = AD/(A+B) Substituting these back into the preceeding expression gives the expected result S^2 = (A+B)^2 + C^2 + D^2. In general, if the edge lengths of an n-dimensional rectangular box are E1, E2,..,En then the lengths of the geodesics connecting opposite vertices on the surface are of the form S^2 = (E1+E2)^2 + E3^2 + E4^2 + ... + En^2 for all n(n-1)/2 distinct permutations of the edges. Thus, for the case of a 4-dimensional box we have six distinct geodesic lengths S1^2 = (A+B)^2 + C^2 + D^2 S2^2 = (A+C)^2 + B^2 + D^2 S3^2 = (A+D)^2 + B^2 + C^2 S4^2 = (B+C)^2 + A^2 + D^2 S5^2 = (B+D)^2 + A^2 + C^2 S6^2 = (C+D)^2 + A^2 + B^2 So, the question is whether there exist 4-tuples of integers A,B,C,D such that all six of these lengths are integers. It's interesting that this can be expressed in terms very similar to the ancient Diophantine problem of finding four numbers whose six pairwise products increased by a constant n are all squares. However, in this case we need to consider TWICE the pairwise products, because we have S1^2 = 2AB + K S2^2 = 2AC + K S3^2 = 2AD + K S4^2 = 2BC + K S5^2 = 2BD + K S6^2 = 2CD + K where K = A^2 + B^2 + C^2 + D^2 is the squared main diagonal. The problem involving just the pairwise products (without the factor of 2) was treated by Euler, but unfortunately the method he used is not applicable when the factor of 2 is included (because it relies on being able to set 4 - y^2 to zero with a rational number y, but this term becomes 8 - y^2 when the factor of 2 is included). I haven't found any non-trivial integer solutions to this set of equations. Does anyone know of any solutions? Or a proof that such solutions are impossible? By the way, just for fun, we might consider the general ways in which a sum of squares can be defined on a 4-part partition of a given number S = A+B+C+D. We have the following possibilities Number of distinct values Form under permutations of A,B,C,D ------------------------------- ----------------------------- (A)^2 + (B)^2 + (C)^2 + (D)^2 1 (A+B)^2 + (C)^2 + (D)^2 6 (A+B+C)^2 + (D)^2 4 (A+B)^2 + (C+D)^2 3 (A+B+C+D)^2 1 Since every number is expressible as a sum of four squares, it's clear that the first form, which is invariant under permutations, can be set equal to a square. Also, the last form is invariant and is automatically a square. In addition, it's not hard to find sets of integers [A,B,C,D] such that all 4 of the permutations of the 3rd form are squares, as shown by [189,1032,564,952], which gives (189+1032+564)^2 + (952)^2 = (2023)^2 (189+1032+952)^2 + (564)^2 = (2245)^2 (189+564+952)^2 + (1032)^2 = (1993)^2 (1032+564+952)^2 + (189)^2 = (2555)^2 (Notice that this solution has S = 189+1032+564+952 = 2737, which is the 2nd smallest S value for solutions of the 3D box problem.) In addition, we can find integers A,B,C,D such that all three of the permutations of the 4th form are squares, as shown by [23,47,55,113], which gives (23+47)^2 + (55+113)^2 = (182)^2 (23+55)^2 + (47+113)^2 = (178)^2 (23+113)^2 + (47+55)^2 = (170)^2 So it's easy to find solutions for all of the forms EXCEPT the one that gives the geodesic lengths of a 4D box, which is clearly the most stringent because it imposes SIX Diophantine conditions on four numbers. Regarding the 3D case, all the solutions for S=A+B+C less than 50 million were given above, including eight "ultra-primitive" solutions. Only one solution for each S is listed, and I think in most cases it is the ONLY solution. The only exception to this for S less than 1 million is S=82943, which actually has two solutions 82943 = 6647 + 33228 + 43068 = 13940 + 31155 + 37848 Multiple representations of other forms seem to be more common. For example, the number 1666 can be partitioned into four parts A,B,C,D in 15 distinct ways such that (A+B)^2 + (C+D)^2 is a square for all the permutations of those four parts (although only one of them, 137+353+577+599, involves just prime numbers).

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