Problems & Puzzles: Puzzles

Puzzle 252.  Kurchan squares

Let's remember first what a magic square is.

"a magic square is a square array of integer numbers such that the sum of the numbers of each row, column and (main) diagonal is a constant"

For example, for a square 3x3 filled with the first 9 natural numbers (1 to 9), there is only one magic square

8    1    6
3    5    7
4    9    2
 constant = 15

The key word for a magic square is the sum operation. But what are the multiplication values of the elements for the same magic square, for each row, column and (main) diagonal, equal or unequal? The answer is that in general the multiplication values  are not equal to a constant.

               120
8    1    6  48
 3    5    7  105
4    9    2  72
96  45  84 80

Now let's calculate the difference between the maximal and the minimal of these eight products and we will get 75 = 120 -45.

Let's define K(n) for a square array nxn as the difference of the maximal and the minimal products for each row, column and (main) diagonal.

Is there a square array 3x3 such that K(3) is itself a minimal quantity, let's say K°(3), when filled with the first n2 natural numbers?

This exactly the question that was posed by Rodolfo Kurchan (1989) whose answer, given by himself, is K°(3)=72, and the corresponding 3x3 square is this one:

              126
8    1    7  56
  4    6    5  120
3    9    2  54
96  54  70 96

K°(3)=126-54=72

I will call this kind of squares - filled with the first n2 X-type of numbers and having a minimal K°(n) value - Kurchan multiplicative squares or shortly a Kurchan squares (*)

He solved also the same question filling the square with the first n2 prime numbers

            
19     2    13
 5    11    17
 7    23      3

K°(3) = 518

While Kurchan says in his email (10/1/04) that this last answer may be improved, I verified exhaustively his two answers and I can assure that he has gotten the minimal K°(3) solutions for both ways of filling the 3x3 square (natural numbers and prime numbers).

Here is the Kurchan question:

Q1. Find K°(n) for n=4-10  for both ways of filling the squares (the first n2 natural numbers and the first n2 prime numbers)

Now I want to add four (4) questions.

More interested in the method than in the results, and -of course- avoiding the exhaustive approaches...

Q2. ...do you devise a smart approach in order to get the K°(n) values and the corresponding squares?

I have obtained specific squares - filled with the first n2 natural numbers - such that K(4)=188  and K(5)=3680. I'm almost sure that my K(4) is K°(4) and then it can not be improved, but perhaps my K(5) is not yet the proper K°(5), so probably it can be improved.

Q3.  Can you improve my K°(n) values for n=4 and 5, and/or get the specific squares associated?

Two more and last issues related to the Kurchan squares are the following ones:

Q4. Is there a Kurchan nxn square such that it is at the same time a magical nxn square, if the square is filled with the first consecutive n2 a) integers, b) primes, c) X-type numbers?

Q5. Is there a n value such that K°(n)=0, if the square is filled with the first consecutive n2 a) integers, b) primes, c) X-type numbers?

 

Solution:

Contributions came from Luke Pebody, J. K. Andersen and Carlos Rivera.

Luke Pebody confirmed that the K°(4) obtained by C. Rivera is correct (the best possible).

Carlos Rivera improved his own solution for K(5) from 3680 to 2610 (can you find what is this improved arrangement?)

J. K. Andersen wrote, for the question 5:

No for a) and b). Yes for c) with certain X-types, e.g. powers of an arbitrary number.

By definition, K0(n)=0 means there is a square where all rows, columns and diagonals have the same product. Two products are the same if and only if they have the same prime factorization, so every prime must appear the same number of times in the factorization of each row, column and diagonal. This means the multiplicity (exponent) of each prime must form a magic (sum) square. It does not have to be the same magic square for different primes.

Example: Show that K0(3)=0 for X-type = powers of 2.

Start with an arbitrary magic square with numbers from 0 to 3^2-1, e.g.:

7 0 5
2 4 6
3 8 1

Raise an arbitrary integer to these powers, e.g. 2:

2^7 2^0 2^5
2^2 2^4 2^6
2^3 2^8 2^1

And get a square with constant product 2^12=4096:

128 1 32
4 16 64
8 256 2

In general: X-type has K0(n)=0 if and only if the first n^2 X-type numbers have prime factorizations where the multiplicity for each prime factor can form a magic n*n square.

***

Carlos Rivera improved (2/2/04) his own solution for K(5) from 2610 to 2052 (can you find what is this improved arrangement?)

***

Anurag Sahay wrote (May, 2005):

For Q3, I found a solution better than your third best
: k(5) = 3474

10 20 22 25 1
5 4 12 24 19
8 14 23 2 21
15 11 6 7 16
18 9 3 13 17

***

Anurag Sahay wrote (Set. 05):

Q3 of puzzle 252: I improved the value of k(5) to 3168.

10

23

24

1

20

 2

19

17

21

8

18

12

 4

25

5

14

3

11

16

15

22

7

6

13

 9

***

On Set 26, 05, Luke Pebody reported:

k(5)=1744, This is the best solution... I have searched all ranges [m, m+1, ..., n] where m,n are products of five numbers in the range [1-25], n-m<1744 and m^5<25!, n^5>25! for possible squares, and there is no such range.

The square will be published later on Anurag's request.

***
 

 

 

 
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