[ May 5, 2002 ]
Antimagic Squares using Primes with sums in Arithmetic Progression
Jean Claude Rosa [email]

Our antimagic square is an arrangement of different prime numbers
in a square matrix such that the row, column and diagonal sums
form a sequence of integers in arithmetic progression.

If the sums need to be consecutive then a central composite
cell is needed. In the following two examples the eight surrounding
cells are prime. The sums go from 60 up to 67.

The next nice antimagic square is composed of nine primes
with its eight sums in arithmetic progression.
The sums go from 87 up to 101 in steps of +2.

Let us search for full prime solutions with larger n*n matrices
Remember that the sums may be in any arithmetic progression.

Who can find the first antimagic square with sums
being eight consecutive primes
or eight consecutive palindromic primes ?

[ June 21, 2004 ]
The answer to the above question is : nobody ! Proof (by JCR) of
" A 3*3 antimagic square composed of primes such that the eight sums
form prime numbers in Arithmetic Progression can not exist " :

The numbers K, T, X1, X2, X3, Y1, Y2, Y3
are all distinct and belong to the set :
{0, PAS, 2xPAS, 3xPAS, 4xPAS, 5xPAS, 6xPAS, 7xPAS}
( PAS is the reason of the arithmetic progression, it is an even number
since the numbers S+X1, S+X2, ..., are all prime numbers ) We have :
P1+P2+P3+P4+P5+P6+P7+P8+P9 = 3*S+X1+X2+X3 = 3*S+Y1+Y2+Y3
Hence X1+X2+X3 = Y1+Y2+Y3
Already we knew : X1+X2+X3+Y1+Y2+Y3+K+T = 28*PAS
If we assign SP with the value of the sum of X1+X2+X3
we have : SP+SP+K+T = 28*PAS
From which follows : SP = 14*PAS-(K+T)/2 .
On the other hand we have :
P1+P5+P9 = S+T
P2+P5+P8 = S+Y2
P3+P5+P7 = S+K
P4+P5+P6 = S+X2
After a few calculations we get : P5 = (S+T+K+X2+Y2-SP)/3 .
If S is a prime number then S@3 = 1 or S@3 = 2 .
If the eight sums are prime numbers (palindromic or not)
in arithmetic progression then we have necessarily PAS@3 = 0
and consequently T, K, X2, Y2 and SP are multiples of 3.
Hence the expression : S+T+K+X2+Y2-SP = 1 or ... = 2 modulo 3
and so is never divisible by 3
and the number P5 is never an integer.
Thus a 3*3 antimagic square with 8 sums in
prime arithmetic progression does not exist.

For a 4*4 antimagic square composed with prime numbers it is
clear that the sums cannot be prime since they are all 'even'.
Hence, we are left with examining the matrices of 5*5, 7*7, etc.

Sorry for these negative results... but to compensate things
here is a 5*5 antimagic square composed of 25 consecutive primes
(going from 17 all the way up to 127 with the
twelve sums going from 325 up to 347)
containing a genuine classic magic square in the upperleft part.
(with magic sum equal to 213).
(I sent another solution to Carlos Rivera, see PP&P Puzzle 263) :

Remark : I haven't found yet a solution with a magic square
positioned at the middle of a 5*5 antimagic square.

Epilogue from Jean Claude
"Je pense que les carrés antimagiques peuvent être aussi
intéressants que les carrés magiques. Qu'en penses-tu ?"