Hi all, and thanks for your responses to the magic square problem. First, I knew it was an unsolved problem because I invented it myself but couldn't solve it. But I'm no big shakes as a mathematician so recently I tried it out on my friend Hendrik Lenstra, who also tried it out on some of his students, none of whom came up with a solution. The problem arose as follows. Suppose we take an ordinary n x n magic square and replace the numbers with resistors of same ohmic value. Then connecting the n resistors in any row, column or diagonal in series yields the same total resistance in each case. I asked myself whether a set of resistors could be found that yield a constant total resistance when connected in PARALLEL. That was easy, my example can be seen here: http://www.leesallows.com/index.php?page_menu=Magic . The next step was obvious: can we find a set of resistors that work both in series and in parallel? Keith's solution is cool, my congratulations. Rich's clarifications were helpful, thanks. Richard's query about the reciprocal of 0 surely doesn't apply here? But Keith's solution is hopefully only a good start rather than the best that can be done. It makes me think I should have stipulated a solution in reals. Moreover, I'm disappointed in myself for not coming up with his solution myself. The notion of using roots of unity in a magic square had crossed my mind before, as seen here: http://www.leesallows.com/index.php?page_menu=Magic . And as he says, the reciprocal = complex conjugate is an obvious come-on. Btw, recall that a multimagic square is defined as one in which replacing all numbers with their squares, cubes, etc, preserves magic. So this problem is addressing the case of a negative exponent. A such Keith's solution ought to be recorded on Christian Boyer's most excellent site http://www.multimagie.com/indexengl.htm. So, is there a solution in reals? Lee At 08:20 AM 10/16/2014, you wrote:

Keith is using a little shorthand. The elements are really w^k, with w an Nth root of 1. His upper left corner entry is w^0, i.e. 1. His main trick is that reciprocals automatically work, if the row or column sum is real, since reciprocals of roots of unity are also conjugates.

One possibly useful idea for finding/proving that certain w^k sums are 0 is to note that the sum must be a multiple of the corresponding cyclotomic polynomial, of degree phi(n). The cyclotomic polys often have both positive & negative coefficients, and for N>=105, may have some |coeff| >= 2, which will complicate trying to create poly multiples with only 0,1 coeffs. When N is even, the trick of converting -w^k into w^(k+N/2), which Keith uses in his square, helps a lot.

Rich

----------- Quoting rkg <rkg@cpsc.ucalgary.ca>:

...

Reciprocal of zero ?? R.

On Wed, 15 Oct 2014, Keith F. Lynch wrote:

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Lee Sallows <Lee.Sal@inter.nl.net> wrote:

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An unsolved problem that MathFunsters may like to ponder is as follows:

I'm curious how you know whether it's unsolved.

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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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