# 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 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
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  
20        7
7   (350,227)   496797  1112300  1218203  
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       18
(1147,136)  1297113   311984   334105  
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

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.

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

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
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 :
(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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