Reciprocal of zero ?? R. On Wed, 15 Oct 2014, Keith F. Lynch wrote:
Lee Sallows <Lee.Sal@inter.nl.net> wrote:
An unsolved problem that MathFunsters may like to ponder is as follows:
I'm curious how you know whether it's unsolved.
Does there exist an n x n magic square using n^2 distinct numbers that remains magic when every number is replaced by its reciprocal?
Or failing that, how about a semimagic square (magic on rows and columns only)?
I've found the latter, but not yet the former. Here's my semi-magic square:
0 18 1 19 2 20 12 30 13 31 14 32 24 6 25 7 26 8 3 21 4 22 5 23 15 33 16 34 17 35 27 9 28 10 29 11
The numbers are the exponents of cos(pi/18) + i sin(pi/18), the "first" 36th root of 1. All 36 36th roots appear once each. The semi-magic sum is zero. The semi-magic sum of the reciprocals is also zero.
I chose to work with numbers of unit absolute value since all such numbers have their complex conjugate for their reciprocal, hence any (semi-)magic square consisting only of such numbers is guaranteed to remain a (semi-)magic square if all the numbers are replaced by their reciprocals.
I conjectured that for a finite set of unit-absolute-value numbers that includes 1 to sum to 0, it must contain all the Nth roots of 1 for some integer N>1, and analogously if it doesn't include 1. If so, the order of a semi-magic square must be the product of at least two distinct primes. Note that on each row, numbers are next to their negative, i.e. their antipodes on the unit circle, and that on each column, numbers are grouped in threes, forming equilateral triangles on the unit circles.
I did the above entirely by hand, no computers.
To make a true magic square by this method, the order of the magic square would have to be divisible by three distinct primes. So I'd try order 30, using the 900th roots of 1, with groupings by 2 on the rows, by 3 on the columns, and by 5 on the diagonals. Tricky. And completely infeasible by a brute-force search.
But is the conjecture correct? It's not obvious to me, perhaps because I'm not a visual thinker. So I tested it with a semi-brute-force computer search. I found what appear to be exceptions, starting with the 30th roots of 1. Unless there's a glitch in my program, 0 1 7 13 19 20 add up to zero, as do 0 6 7 17 18 24. (Again, these are powers of cos(pi/15) + i sin(pi/15), or points on a regular 30-gon.) Can anyone confirm or refute this exception to my conjecture? If the exception is valid, what is the rule for finding numbers of unit absolute value that add up to zero? Thanks.
If my conjecture is wrong, it may be possible to make a smaller magic square than order 30.
I just did a Google search on "0 1 7 13 19 20," and I got
On the chromatic number of integral circulant graphs - ScienceDirect
Vertices from C={0,1,7,13,19,20} C = { 0 , 1 , 7 , 13 , 19 , 20 } generate a clique in the given integral circulant graph, ...
so maybe it is a thing.
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