# »  Solutions to puzzles in my VDARE.com monthly Diary

## February 2021

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In the Math Corner of my February Diary I offered the following brainteaser …

The Parker Square-Sum Problem.

Write down the numbers from 1 to 15:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.

Now rearrange them into a new order such that any two adjacent numbers add up to a perfect square.

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This was, as I wrote when posing it, one of Matt Parker's puzzles from the Numberphile YouTube vlog. The solution is given there.

Matt shows that this is really a problem in Graph Theory. He adds some interesting comments on extended versions of the problem: to 16, 17, 18, … Rearrangement isn't always possible, he says; for example, none is possible for N = 24. He hints, though, that for N ≥ 25 there is always a solution.

As stated, with the numbers to be rearranged limited from 1 to 15, the problem can be solved intuitively, without appealing to graph theory. Sample, from a reader:

I reasoned that most of the numbers in the list would have to form squares when summed to each of two neighbors. Hence, it would be in my interest to identify a number that could sum to a square with only one other number in the set; that number would become the beginning of the sequence. I guessed that the mid-point of the set, 8, might be such a number, I don't know why; but it seems I was lucky.

Working from 8, 1 was the unique successor. It could be followed by either 3 or 15. I guessed 15 would be a better choice to provide a more uniform distribution of the initial terms, reserving 3 as a replacement if 15 failed.

After a few terms, I noticed a curious pattern: the sums of successive pairs are 9, 16, 25, 16, 9, 16, 25, and so on, cycling through these three numbers. I would like to know why that is.

Matt gives solutions all the way out to N = 23 (illustrated with a graph at 7m57s on the video). The actual rearranged sequence there is: 18, 7, 9, 16, 20, 5, 11, 14, 2, 23, 13, 12, 4, 21, 15, 10, 6, 19, 17, 8, 1, 3, 22.

The corresponding sums of successive pairs there are: 25, 16, 25, 36, 25, 16, 25, 16, 25, 36, 25, 16, 25, 36, 25, 16, 25, 36, 25, 9, 4, 25. That can be grouped as 25, [X], 16, 25, [X], [X], [X], 9, 4, 25, where X represents the subsequence 16, 25, 36, 25. That's as repetitive as I can make it. I don't see any other pattern. Something might show up for N ≥ 25 …