Problems & Puzzles: Puzzles

Puzzle 264.  Antimagic Prime Squares

For sure you know that an antimagic square of order n is an arrangement of nxn distinct numbers such that the 2n+2 sums of  rows, columns and diagonals are consecutive numbers.

If you use exclusively distinct prime numbers inside the square, then the 2n +2 sums are not consecutive numbers but consecutive odd numbers (if n is odd) or consecutive even numbers (if n is even)

During this week I asked - in a personal communication - to J.C. Rosa if he was thinking that an antimagic prime square 3x3 was possible with an additional condition: the nine primes used are consecutive primes.

Well, J. C. Rosa, responded not only 'yes' but sent four examples. Here goes his smaller one:

 220781 73583 73589 73613 220785 73637 73597 73561 220795 73571 73607 73609 220787 220791 220793 220783 220789

J.C. Rosa is not sure if this is the smallest ever possible.

Questions

1. Is this the smallest 3x3 ever possible?

2. Find an example 5x5.

Solution: Faride Firoozbakht wrote (April 2004):

I found a smaller solution as follows:

21569 21613 21589
21601 21587 21577
21599 21563 21611

the eight sums of rows, columns and diagonals are {64763, 64765, 64767, 64769, 64771, 64773, 64775, 64777}. But I don't know what is the smallest 3x3 ever possible,

***

J. C. Rosa wrote (May 2, 2004):

The smallest 4x4 antimagic square , composed only 16 consecutive
primes is :

Step 2:

3        41       43        13
53       7        19        31
17       47       37       11
29      23         5        59

The sums go from 100 up to 118 (step 2

***

J. C. Rosa also found a solution to Q2:

About the question 2 of the Puzzle 264 I have found the
following solution ,a  5x5 antimagic square composed only 25
consecutive primes :
(243)
5     97    73    53     13       (241)
47     7     83    67     41       (245)
89    71    23    11     59       (253)
37    61    29    101    31      (259)
79    19    43    17     103     (261)

(257)(255)(251)(249) (247)    (239)

The sums go from  239 ( prime ! ) up to 261
note: The  sum of the 25 consecutive primes is 1259 ( prime )

***

J. C. Rosa sent the following approach in order to see if a smaller solution than the obtained by Farideh exist, for the case 3x3:

3x3 antimagic square composed by 9 consecutive primes.

Let the following square:

P1        P2      P3    à S+X     P3+P5+P7=S+K

P4        P5      P6     à S+Y

P7        P8     P9     à  S+Z

¨^          ^       ^

S+W      S+V    S+U              P1+P5+P9=S+T

be an antimagic square with {0,2,4,6,8,10,12,14}={X,Y,Z,W,V,U,T,K}, and    X+Y+Z+W+V+U+T+K=56.

Let S=the minimal sum of the eight  possible.

Let T(1),T(2),T(3),…..,T(9) the nine consecutive primes with: T(1)<T(2)<….<T(9).

Let  ST=sum of the nine primes.

We have : ST=3S+(X+Y+Z)=3S+(U+V+W) from where X+Y+Z=W+V+U. Let SP=X+Y+Z=W+V+U, SP is an even number. 2SP=56-(T+K) from where : SP=28-(T+K)/2 . SP being an even number therefore   (T+K)=0 modulo 4.

We have : ST=3S+SP  ( 1) from where S=(ST-SP)/3 ,thus ST=SP modulo 3. Moreover S and ST are odd numbers. We have also :   ST+3P5=4S+T+K+V+Y  (2). The equalities (1) and (2) give 3P5=S+T+K+V+Y-SP

Let  STK=T+K ; SVY=V+Y

1. STK goes from 4 to 24 step 4
2. SP=28-STK/2
3. S=(ST-SP)/3
4. SVY goes from 2 to 26 step 2
5. P5=(S+STK+SVY-SP)/3
6. Control that T(1)<=P5<=T(9) and P5 is prime.
7. Choise T,K,V,Y  so that  T+K=STK and V+Y=SVY
8. Choise P1,P9, P3,P7,P2,P8,P4,P6 so that P1+P5+P9=S+T

and  P3+P5+P7=S+K ; P2+P5+P8=S+V ; P4+P5+P6=S+Y

1. Control that the eight sums are odd consecutive numbers.

***

C. Rivera materialized in a Ubasic code the algorithm proposed above by Rosa and verified exhaustively that:

a) The solution obtained by Farideh is the smallest one.

b) The solution obtained by Rosa is the second smallest possible.

c) Using primes smaller than 10^6 there only 15 distinct solutions (including the already obtained by Farideh & Rosa), if we define a solution as distinct just for the set of the 9 consecutive primes used, disregarding rotations and reflections of the squares.

C.Rivera points out two unexpected characteristics in two of the 15 solutions:

a) The solution #7 shows the following two variations (despite that both have the same set of primes, the same centre and the same set of distinct sum of 8 rows, these variations are not convertible to the other by rotations and/or reflections)

 1237855 1237855 7a 412589 412637 412627 1237853 7b 412591 412651 412609 1237851 412651 412619 412591 1237861 412639 412619 412603 1237861 412609 412603 412639 1237851 412627 412589 412637 1237853 1237849 1237859 1237857 1237847 1237857 1237859 1237849 1237847

b) The solution #13 shows the following two variations (despite that both have the same set of primes, they have distinct centre, distinct set of distinct sums of 8 rows; needles to say that these variations are not convertible to the other by rotations and/or reflections)

 2233175 2233183 13a 744371 744431 744377 2233179 13b 744371 744391 744409 2233171 744391 744389 744397 2233177 744431 744397 744353 2233181 744409 744353 744407 2233169 744377 744389 744407 2233173 2233171 2233173 2233181 2233167 2233179 2233177 2233169 2233175

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