It works for any countable (multi-)set S with the following properties: -- Every element of S is positive. -- Every element of S is strictly smaller than the desired total K. -- There exists a sequence of elements in S that converges to zero. As for generalising beyond N x N, we just need an incidence structure of `points' and `lines' (sets of points) with the following property: -- Given any line L and finite set L_1, L_2, L_3, ..., L_N of lines, the difference L \ (L_1 union ... union L_N) is infinite. It is sufficient, for example, for every line to be infinite and every intersection of two lines to be finite. So in particular, we can choose Z^n where we insist that all straight Euclidean lines passing through two (thus infinitely many) points have the same sum; this is much more general than just horizontal, vertical and diagonal lines. (And yes, I agree that the infinite case is so much easier than the finite case. My argument is simply a `just-do-it' approach, similar to those championed by Professor Béla Bollobás and his descendants.) https://gowers.wordpress.com/2008/08/16/just-do-it-proofs/ Sincerely, Adam P. Goucher

Sent: Saturday, December 06, 2014 at 7:38 PM From: "Warren D Smith" <warren.wds@gmail.com> To: math-fun@mailman.xmission.com Subject: [math-fun] infinite magic squares

Adam Goucher's argument should indeed work not only for the sequence 1/N, but indeed any decreasing positive sequence with infinite sum; and also for e.g. constructing infinite magic cubes, 4-dimensional magic hypercubes etc.

What I like about it is, it shows infinite magic is actually way easier to attain than finite magic (for that, you actually have to work).

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun