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Triangulating a Square (Posted on 2009-08-26) Difficulty: 3 of 5

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Two solutions (or more?) Comment 2 of 2 |
Following Charlieís posting, Iíve been trying to prove that his was the only solution, but Iíve found another in the process, just outside his search area. Iím 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 Pellís 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 ÖÖ.

so m+2 can take the values:       1          7          41         Ö
                        giving m =         -1         5          39         Ö

with corresponding p/2 values:    1          5          29         Ö
                        giving p =          2          10         58         Ö

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 Ö}
p[r] = 6 p[r-1] - p[r-2],  giving p = {2, 10, 58, 338, 1970, 11482, 66922, Ö } (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 havenít yet found it.

All ideas welcomed.
Really good problem.

  Posted by Harry on 2009-08-31 15:11:01
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