Diophantic Spider and Fly

   Given a rectangular room with three integer sides A<=B<=C, A+B+C=S, 
  is it possible that all three geodesic distances between extreme corners
  are integers as well ?

   The answer is positive, and the three minimal solutions are for :

    A. S =  833 = 7 *  7 * 17 = 108 + 357 +  368
    B. S = 2737 = 7 * 17 * 23 = 564 + 748 + 1425
    C. S = 3703 = 7 * 23 * 23 = 348 + 975 + 2380

   As is shown by :

    A. 108^2 + 725^2 = 733^2
       357^2 + 476^2 = 595^2
       368^2 + 465^2 = 593^2 (min)

    B.  564^2 + 2173^2 = 2245^2
        748^2 + 1989^2 = 2125^2
       1425^2 + 1312^2 = 1937^2 (min)

    C.  348^2 + 3355^2 = 3373^2
        748^2 + 2728^2 = 2896^2
       2380^2 + 1323^2 = 2723^2 (min)

    According to the schematic unfolding as demonstrated
    for the case S = 7 = 1 + 2 + 4  [shown below]

    The above solutions were arrived at by computational methods,
    starting from an "open" list of primitive pythagorian triples
    ordered by rising values of both their generating parameters.
    Only if S is a composite number there exists several
    pythagorian equations and by combining, for given S, any two
    of them, a whole set of "near-solutions" is received, that
    is : two of the three critical distances  are integers,
    while the third in general is not. But a sufficient
    condition can easily be given under which the third distance
    will be integer as well. This condition is fulfilled sometimes
    for bigger values of S as well, but so far all these solutions
    proved to be only whole multiplies of the first three, and so
    really do not add anything new.

    On the other hand, purely statistical considerations show that
    using a more composite number for S considerably enlarges the
    set of equations, however still more S itself, so that the
    chances the sufficient condition to be fulfilled are decreasing.

    So, maybe, no other positive solutions exist. But contrary to
    this fact, reality may be otherwise. And besides, there well
    may exist solution which dont fulfill the sufficient, but not
    necessary, condition. All this was done with the aid of an H.P.
    scientific pocket-calculator, and may be put on a broader base
    by a program on a more adapted computer. Or maybe the problem
    can be solved, or was even treated before, by classical methods ?


 Scanned shcematic, drawn by Herman Baer

Illustration showing a fly's walk on the box for S=7=1+2+4

Prof. Rusin's reply to Herman Baer. Click Here.
Herman Baer's reply to Prof. Rusin Moledet, 29.VI.98 Dear Prof. Rusin, Through my grandson, Uri Raz, I got your answer to my question concerning the "Diophantic Spider" and I was very please, you were able to decide by "neoclassical" (?) methods, where I had to leave off. In principle I could follow your procedure though about some of the last details (e.g. point of infinite order) I am not well informed. In particular, I could make use of your solution / [a,b,c] = [243984,689613,67500] < S = a+b+c = 1609097 = 7*457*503 \ [q,r,s] = [1149355,1152347,1386745] and it may interest you, how this fits into the method, I developed : For the given S=7*457*503 there exist just 13 (= n) pythagorean triples 2 _2 2 a + a = d v v v which may be formed in the following order : _ generating a a d v factors parameters v v v _ ------- ---------- ------- ------- ------- -- b q 457*503 (2,1) 689613 919484 1149355 <11> <-- 11 16 7*503 (17,6) 890813 718284 1144325 <13> 14 13 7*457 (16,13) 278313 1330784 1359575 (5) 5 22 503 (40,39) 39737 1569360 1569863 (2) 2 25 503 (52,5) 1347537 261560 1372687 (4) 23 4 457 (45,22) 704237 904860 1146613 <12> 12 15 457 (51,10) 1142957 466140 1234357 [7] 20 7 7 (350,227) 496797 1112300 1218203 [8] 8 19 7 (386,125) 933597 675500 1152347 <10> <-- 17 c 10 r (897,886) 19613 1589484 1589605 (1) 1 26 (927,596) 504113 1104984 1214545 [9] 9 18 (1147,136) 1297113 311984 334105 [6] 21 6 (1173,104) 1365113 243984 1386745 (3) <-- 24 a 3 s afterwards we arrange them according to / d >= d >= ... >= d / 1 2 n < \ a <= a <= ... <= a <= a <= ... a \ 1 2 n n+1 2n _ <= _ 2 _2 2 a = a a + a = S a + a = d v 2n+1-v v v v v v and we can identify / a + a + a = S = a + b + c / 3 10 11 < _ \ a + a = S - a = a = a \ 3 10 11 11 16 ~~~~~~~ (anything in pairs) / 2 _2 2 2 / a + a = d = S / 3 3 3 / 2 _2 2 2 < a + a = d = r \ 10 10 10 \ 2 _2 2 2 \ a + a = d = q \ 11 11 11 In this case a,b,c appear as some pythagorean components fulfilling a+b+c=S, which represents exactly the sufficient condition, I found before; together with your exposition, I think, it follows that this condition is necessary as well, as the expressions you start from, must appear within the full set of pythagorean triples I begin with. (?) So both methods, ma way, seem to supplement one another. Thanking you once again for your kind and immediate reply I remain Yours Herman Bar
Prof. Rusin's second reply to Herman Baer. Click here.
Herman Baer's second reply to Prof. Rusin Moledet, 20/Jul/98 Dear Prof. Rusin, My letter to you of end of June was delayed and reached you only nearly a week ago. I hope it was understable even with two or three "misprints" by my grandson, who besides did a very good piece of work. In the meantime I did a bit of search for "a missing link" between the three minimal solutions and your "specimen" as one of the presumably infinite number of existing ones. According to your analysis I wasnt quite sure to find one, but here it is : a=a b=a c=a 1 10 15 S = 11431 = 624 + 4551 + 6256 = 7*23*71 2 2 2 2 2 624 + 10807 = 10825 = d = S a a 10 1 2 2 2 2 2 4551 + 6880 = 8249 = d = r a a 10 10 10 2 2 2 2 2 6256 + 5175 = 8119 = d = q a a 15 15 15 I found it by the same method of a full set of pythagorean triples, which I applied to your solution, using the same designations corespondingly and examining all the pairs a a (u < v) The "conditin" was fulfilled with a + a = a or u v a + a + a = S 1 10 12 1 10 15 Technically the search was made easier by using thhe complex numbers program with _ _____ 2 2 2 _2 _____ a +ia = (alpha + i*alpha ), d = a + a , alpha > alpha , v v v v v v v v v _____ alpha , alpha being the generating parameters. v v To get the full set of pyth. tyiples I often used the algebraic numbers _____ _____ ___ (alpha + alpha ) +/- alpha * / 2 | v v v V I had to try several values for S in vain, but then there were accumulations of cases, in which the condition a + a = a u v w was nearly fulfilled with relatively slight deviations. So I am expecting to hear from you, concerning your fundemental soluion from one side and my practical treatment from the other side, which with some luck, arrived at solutions in smaller numbers, and remain thankfully yours Herman Baer.
Kevin Brown's Geodesic Diophantine Boxes page. Click here.
24.VIII.98 On 5th of August I sent to my grandson Uri for the Internet my letter, begun on July 29th as well as a list of results from from probing Rusin's first polynomial solution, the transmission of which took some more time. There often I met the following question : already from the list two solutions 23x|S= 833=7* 7*17 \ we can build 7x|S=2737=7*17*23 / S=19159=7*7*17*23 in two different ways, and accordingly receive two different solutions for the same S, though neither of them primitive. So it was natural to ask, if there exist also two primitive solutions for the same S. As a first candidate I chose Kevin Brown's list S=82943=7*17*17*41 and really found by the method used before, from the complete set of pythagorean triples for S, a second solution k (u ,v ) v v u a 6647 74296 79585 289(12,11) b 33228 49715 59797 1(234,71) c 43068 39875 58693 1(222,97) ----- ----- S=82943 165886=2S in addition to Brown's a 13940 69003 70397 697(10,1) b 31155 51788 60437 1(214,121) c 37848 45095 58873 1(228,83) ----- ----- S 82943 165886=2S both using two different primitive triples of the four existing ones; the none-primitive third triples both belong to 17 of this kind, completing the whole set. My expectations for other such cases were disappointed, when I tried S=7*23*31*44 and S=7*17*17*409 Only on 21st of Aug I learned from Kevin Brown's letter, that I had no chance, as he established the above to be the only case up to a million. The list extended up to S=50 million is impressive, and it is astonishing, that inspite of the rather low density (~1.6 in a million) the solutions, according to Dave Rusin, manage to be spread (in some way) on an infinite number of different curves on a surface. The generalization of the problem to four dimensions - or even more - is convincing and interesting as well as the general splitting in squares of S=a+b+c+d One week earlier I received Prof. Rusin's letter of August 6th with a row of explanations on the position of the spider-problem among similar investigations; informations I was pleased to receive. On the other hand it was unexpected to hear that you did not understand the method I used to get the first solutions. In my first note to the Internet I really only sketched the way, as I was not sure if there wereanything new or even of interest. But later on in the course of our exchange I described the procedure rather in detail, and had the impression that Kevin Brown, at least, got the gist of it and transfered it to the frame of his computer with much success. At any rate I feel obliged to expose once more my method in detail, in order to be understood. To begin with again I started from an "open" list of primitive pythagorean triples, ordered as follows : _ u>v a a d S (1) (2,1) (3 , 4, 5) 7 (3,2) (5 , 12, 13) 17 (4,1) (15, 8, 17) 23 (4,3) (7 , 24, 25) 31 (5,2) (21, 20, 29) 41 (5,4) (9 , 40, 41) 49 = 7*7 (6,1) (35, 12, 37) 47 (6,5) (11, 60, 61) 71 (7,2) (45, 28, 53) 73 (7,4) (33, 56, 65) 89 (7,6) (13, 84, 85) 97 (8,1) (63, 16, 65) 79 (8,3) (55, 48, 73) 103 (8,5) (39, 80, 89) 119 = 7*17 (8,7) (15,112,113) 127 and so on 2 2 _ 2 2 a = u - v , a = 2uv, d = u + v (u,v) = 1 and of different parity _ 2 2 a + a = S = (u - v ) +2uv = (u + v)(u + v) - 2*v*v (Pell) = (u+v)+/-/*v*root(2) Observing that for S prime there exists only one pythagorian triple I continued with composite S only which provides two or more. In general we obtain for any such S a whole set of pythagorean triples, the minor part of them ( 2^(n-1) ) primitive, the major part suitable multiples : 2 2 2 _2 2 (2) a + (S-a ) = a + a = d v v v v v Let them be ordered according to : (2') d >= d >= d >= ... >= d , and changing where 1 2 3 n need should be, _ _ between a and a so that a < a v _ v v v and putting a = a , this means v 2n+1-v (2'') a <= a <= ... <= a <= a <= ... <= a 1 2 n n+1 2n if we now choose, for instance, a and a and S - (a + a ) 1 2 1 2 as the sides [a,b,c], so we have (3) S = a + a +[S - (a + a )] and 1 2 1 2 (3') / 2 2 2 _2 2 | a + (s - a ) = a + a = d | 1 1 1 1 1 | | 2 2 2 _2 2 < a + (S - a ) = a + a = d | 2 2 2 2 2 | | 2 2 | (a + a ) + [S - (a + a )] = D = d*d +/-delta \ 1 2 1 2 that is : two of the critical distances are integers, while the third in general is not. Instead of a1, a2 we may take any pair of components al, am (l<=m). If by chance al+am equals any av, we are lucky and have found a solution. This was the case for the three first solutions. The condition (4) a + a = a or l m v _ (4') a + a + a = S = a + a + a l m v l m 2n+1-v is sufficient; if the set of pythagorean equations for S (2) is complete, it will be necessary as well, as every pythagorean equations for S is contained in it. To attain this complete set I rely, in the first place, on the list (1), beginning with the representation of p1, p2, p3, ..., pn if S=p1*p2....*pn multiplying by the suitable factors to get S. Afterwards, outside the list, I make use of multiplication of the algebraic numbers (u+b)+/-v*root(2) to get the more complex representations, until I arrive at the primitive ones. For convenience we may prepare a secondary "open" list of all S, which are a product of two prime factors _ u, v ( u+v )= v*root(2)*a*a*d S ( 5, 4) ( 9-4*root(2)) ( 9, 40, 41) 49 = 7*7 ( 8, 5) (13-5*root(2)) ( 39, 80, 89) 119 = 7*17 (10, 1) (11- root(2)) ( 99, 20,101) 119 = 7*17 ( 9, 8) (17-8*root(2)) ( 17,144,145) 161 = 7*23 (11, 2) (13-2*root(2)) (117, 44,125) 161 = 7*23 (11, 6) (17-8*root(2)) ( 85,132,157) 217 = 7*31 (13, 2) (15-2*root(2)) (165, 52,173) 217 = 7*31 If S is a product of n different primes, then there are 2^(n-1) pythagorian representations (from Pell) and 3^n-1 representations in all, multiples included. This number is considerably reduced, when not all the primes are different. Occasionally a further reduction by one (or so) may occur, apparenly caused by special arithmetical relations between prime factors. If N is the total number of represantations, the number of sums of pairs in to be checked is about N*N, which with my calculator means a rather cumbersome task. But for a computer program, I suppose, it is a built-in routine. And also maybe Kevin Brown has more proficient means to his disposal to get solutions. As I remarked earlier, I also used the complex relation _ 2 2 2 2 a+ia = (u+iv), d = u + v By the way i=root(-1) and root(2) are algebraically cancelled by epsilon = (1+i)/root(2), defining the regular octagon, as 2 _______ epsilon = i and epsilon + epsilon = root(2) but it seems, there is no practical use of it for the present problem. To be quite clear let me give a detailed example and apply the procedure to the simplest case S = 833 = 7 * 7 * 17 : completely factored Rv 7 -> ( 2,1) ( 3- root(2)) ( 3, 4, 5) * 7*17 = 119 17 -> ( 3,2) ( 5-2*root(2)) ( 5,12, 13) * 7*7 = 49 7*7 -> ( 5,4) ( 9-4*root(2)) ( 9,40, 41) * 17 7*17 = 119 -> ( 8,5) (13-5*root(2)) (39,80, 89) * 7 7*17 = 119 -> (10,1) (11- root(1)) (99,20,101) * 7 (23,8) (465,368,593) (27,2) (725,108,733) The last two from : as above already ----\ (11-root(2)) (3+/-root(2)) +/- /33\/(35-14*root(2)) (21,14) = 7*7 (3,2)<-/ \ 2/\(31+ 8*root(2)) (23, 8) (465,368,593) (13-5*root(2)) (3+/-root(2)) +/- /39\/(49-28*root(2)) (21,<28) out of bounds \10/\(29- 2*root(2)) (27,2) (725,108,733) So we can put down the list of all possible pythagorean representations for S = 833 : _ _ a a d k u,v a a d v v v 119 ( 2,1) (357,476,595) 6 | 1)108 725 (14 733 49 ( 3,2) (245,588,637) 4 | 2) 140 693 (13 707 17 ( 5,4) (153,680,697) 3 | 3) 153 680 (12 697 7 ( 8,5) (273,560,623) 5 | 4) 245 588 (11 637 7 (10,1) (693,140,707) 2 | 5) 273 560 (10 623 (23,8) (465,368,593) 7 | 6)357 476 (9 595 (27,2) (725,108,733) 1 | 7)368 465 (8 593 \ ^ \---------/ and if we check the 49 pairs al+am, l>m, successively, we find, that only with a + a = a = S - a or 1 6 8 7 a + a + a = S (and as consequence a + a = a , a + a = a ) 1 6 7 1 7 9 6 7 14 the condition (4) is fulfilled, producing the only solution in this case, with 833 = 108 + 357 + 368 / 2 2 2 / 108 + 725 = 733 / / 2 2 2 \ 357 + 476 = 595 \ \ 2 2 2 \ 368 + 465 = 593 (min) Herman Baer.

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