alt.math.recreational,sci.math

Hello, people. This night I had a terrible headache and while I was trying to get
asleep, the following problem popped in my mind:

So, we have matrix 2n*2n in Z/2Z (or simply saying each cell of the matrix is 0
or 1).More precisely, we have all of them - that's 2^(4*n^2) matrices. We want to
build a "meta-matrix" which contains all of them. So if size of meta-matrix is
m*m, then there are (m-2n+1)^2 different submatrices 2n*2n. From that we conclude
that minimal value of m is 2^(2*n^2)+2*n-1. The question is whether it is
possible to build a meta-matrix for that m. The second question is whether it is
possible to build a torus m*m where m=2^(2*n^2) which also contains all 2n*2n
matrices. ( note that if we can build a torus then corresponding meta-matrix is
also easily constructed).

Well, it's probably hard to understand without an example.
Let n=1. So we have all 2*2 binary matrices:
OO OO XO OX OO XO OX XX
OO OX OO OO XO OX XO XX

XX XO OO OX XX XX OX XO
OO XO XX OX XO OX XX XX

(0 and 1 are O and X so that people without fixed-width font would see it too)

Meta-matrix should be 5*5 and torus 4*4. You could try build them yourself.
Answer after

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That's torus

XXXO
XXOX
OXOO
XOOO

And that's matrix

XXXOX
XXOXX
OXOOO
XOOOX
XXXOX

So, let n be 2. then we have m=259. That's really big matrix. I have wonderful
proof that such matrix exists for every n, but this post is already too long.

Theorem (Shatrov's Last Theorem): for every n both torus and meta-matrix exist. 

Well, in fact I don't have the proof so you could try to find it. I just hope
elliptical curves have nothing to do with that problem...

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