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| Triangulating a Square (Posted on 2009-08-26) |
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Submitted by brianjn
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Two solutions (or more?)
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 | Comment 2 of 3 | 
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Following Charlie’s posting, I’ve been trying to prove that his was the only solution, but I’ve found another in the process, just outside his search area. I’m still not sure whether the number of solutions is finite.
In square 1, let the sequence of corner numbers start with the m th triangular number. Each of these is counted twice in the summation, so denoting the central number by 2p2 gives:
2p2 = m(m + 1) + (m + 1)(m + 2) + (m + 2)(m + 3) + (m + 3)(m + 4)
p2 = 2m2 + 8m + 10 (1)
In square 2, let the sequence of corner numbers be the nth, (n+d)th, (n+2d)th and (n+3d)th triangular numbers. The central number is now p2, giving:
p2 = n(n + 1) + (n + d)(n + d + 1) + (n + 2d)(n + 2d + 1) + (n + 3d)(n + 3d + 1)
p2 = 4n2 + 4n(3d + 1) + 14d2 + 6d (2)
A computer search for integer solutions that satisfy both (1) and (2) finds:
m=5, n=1, d=2, p=10, which give the squares shown below, found by Charlie, but is very time consuming for larger values of m, n and d.
15 36 21 1 7 6
51 200 49 29 100 21
36 64 28 28 43 15
From (1) it is clear that p is even, so p/2 is an integer, and (1) can be written as
2(p/2)2 = m2 + 4m + 5
2(p/2)2 = (m + 2)2 + 1
This is Pell’s equation, with m+2 and p/2 therefore being given by alternate numerators and denominators of the convergents of sqrt(2).
Convergents of sqrt(2) are: 1/1, 3/2, 7/5, 17/12, 41/29, 239/169 …….
so m+2 can take the values: 1 7 41 …
giving m = -1 5 39 …
with corresponding p/2 values: 1 5 29 …
giving p = 2 10 58 …
These can also be found from recurrence relations (subscripts in square brackets):
m[r] = 6 m[r-1] - m[r-2] + 8, giving m = {-1, 5, 39, 237, 1391, 8117, 47319 …}
p[r] = 6 p[r-1] - p[r-2], giving p = {2, 10, 58, 338, 1970, 11482, 66922, … } (3)
Successive values of p, from (3) can then be used with equation (2) so that a more focused computer search can be made for values of n and d.
(2) can first be rearranged to: p2 = (2n + 3d + 1)2 + 5d2 - 1
which then gives: n = (1/2)(sqrt(p2 - 5d2 + 1) - 3d - 1)
p=10 gives the solution shown above and found by Charlie.
p=1970 gives n=634, d=214, m=1391 shown below
968136 1937664 969528 201295 561271 359976
1940451 7761800 1940449 1016021 3880900 924429
972315 1943236 970921 814726 1379179 564453
Having tried all the values of p as far p=13250218, only p=10 and p=1970 have produced solutions, and to go further is very time consuming (computer-wise).
I had hoped to find an analytical method that could help to usefully combine (1) and (2) and to rule out further solutions, but have so far failed. There may be a way of filtering the p values before searching, but I haven’t yet found it.
All ideas welcomed.
Really good problem.
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Posted by Harry
on 2009-08-31 15:11:01 |
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