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=(STSP)/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+YSP
Let STK=T+K ; SVY=V+Y
 STK goes from 4 to 24 step
4
 SP=28STK/2
 S=(STSP)/3
 SVY goes from 2 to 26 step
2
 P5=(S+STK+SVYSP)/3
 Control that T(1)<=P5<=T(9)
and P5 is prime.
 Choise T,K,V,Y so that
T+K=STK and V+Y=SVY
 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
 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 
***