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:

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

9. 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

***