How does this trick work? ?

Question

http://www.youtube.com/watch?v=_jVadg9JTk8&feature=related

Never saw a trick like this one before.

Answer 1

Hum? I learned that when I was in fifth grade!

The solution doesn't even need to be a normal magic square. You can do the same trick with arbitrary arithmetic sequences: for example, instead of 1, you can put down Sqrt(2), and instead of 2, 3, 4, ... you can put down Sqrt(2) + π, Sqrt(2)+2π... etc.

Simple proof of concept: let N be an odd natural number. Consider the integers 0 ... N^2-1, written in base N. We are to assign these numbers to the magic square. Each square then receives a 2 numbers between 0 and N-1, representing the 1's place and the N's place of the number in base N. For example, in the 5x5 case, the square to receive the decimal number 16 will be receiving the two digits 3 and 1, 3 in the 5's place, and 1 in the 1's place.

It is easy to see that a solution of the magic square is an arrangement such that on each row and each column, the number for the N's place and the number fot the 1's place each only appears in a unique square. For example, {12,23,34,40,01} is allowed, since 0,1,2,3,4 each only appears once in the 5's place, and each only appeared once in the 1's place; {24,13,02,31,40} is also allowed, but {01,21,31,11,41} is not allowed, since 1 appears multiple times in the 1's place.

Start with

00 01 02 03 04

10 11 12 13 14

20 21 22 23 24

30 31 32 33 34

40 41 42 43 44

This is obviously bad. But if we shift the 5's place down one row for each column, and the 1's place to the right one column for each row, we will be good:

00 11 22 33 44

41 02 13 24 30

32 43 04 10 21

23 34 40 01 12

14 20 31 42 03

In other words, we send the box at position (x,y) (where 0<=x,y<= N-1)

to

(y-x mod N, x+y mod N)

Simple exercise for the reader: show that this map is bijective only when N is odd!

This way we obtain a semi-magic square. (All the columns and all the rows work out, but not the diagonal.) By shuffling the rows a bit, you see that if we can arrange it such that one of the diagonals have (N-1)/2 in the 1's place, and the other has it in the N's place. Indeed, check that

(x,y) -> ( y- (x-(N-1)/2) mod N, y+ (x-(N-1)/2) mod N )

gives a magic square.

(Hint, one of the diagonals is x+y = N-1, the other is x = y)

The final step is to check that this map can be recovered with the move up 1 and then move right 1 procedure described in the video.

[ Sorry if the exposition is not very clear; it is a bit hard to explain this without being able to draw pictures all the time. ]