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 Square pairs (Posted on 2005-01-09)
Back in An Arrangement of 15 you were asked to place the numbers 1 to 15 in a line so that any two adjacent numbers summed to a square number.

Now, try to arrange the numbers from 1 to 32 in a circle, so any two adjacent numbers again sum a square number.

 See The Solution Submitted by e.g. Rating: 4.3333 (12 votes)

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 Non-Computer-Aided Solution | Comment 5 of 12 |

There are several numbers that must be in a sequence. (4,32,17) (5,31,18) (6,30,19) (7,29,20) (8,28,21) (9,27,22) (10,26,23) (11,25,24) (7,18,31) and (9,16,20). This is because there they can only make two squares.

As you can see, 9 must be next to both 16 and 27, which means we have the following mini-sequence: (22,27,9,16,20).

And 20 must be next to 29 so we extend it to get (22,27,9,16,20,29,7).

7 must be next to 18 so we get (22,27,9,16,20,29,7,18,31,5).

The only combinations for 19 are (6,19) (17,19) and (30,19). But we must have (6,30,19) so 6 and 19 cannot be next to each other. Hence (17,19,30).

Combining (17,19,30) with (4,32,17) and (6,30,19), we have a second mini-sequence (4,32,17,19,30,6)

2 can only be paired up with 7, 14, and 23, but 7 is already between 29 and 18). So we have (14,2,23). Along with (10,26,23), we have (14,2,23,26,10).

Then 13 can only be paired with 3, 12, and 23. Since 23 has already been used, it must be (3,13,12).

8 can only be matched with 1, 17, and 28. 17 has already been used so we have (1,8,28). Along with (8,28,21), we have (1,8,28,21).

15 can only be paired with 1, 10, and 21. But because of the previous mini-sequence, 1 and 21 are too far apart so we have either (1,15,10) or (10,15,21).

We have the following mini-sequences:

(22,27,9,16,20,29,7,18,31,5)
(4,32,17,19,30,6)
(14,2,23,26,10)
(1,8,28,21)
(11,25,24)
(3,13,12)
(1,15,10) or (10,15,21)

Now all 32 numbers are represented above. I can't think of any more eliminations. So I just did a trial and error. 11 can be connected to 5 and 14. By trial and error, I found that 11 connected to 14 does not lead to an answer. If there is an answer 11 must be connected to 5.

After some more trial and error, I get (12,13,3,6,30,19,17,32,4,21,28,8,1,15,10,26,
23,2,14,22,27,9,16,20,29,7,18,31,5,11,25,24,12)

 Posted by np_rt on 2005-01-09 20:16:43

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