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

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.

7	31	23		13	23	29
43	20	3		43	20	3
17	11	37		11	19	31

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.

13	37	41
67	29	5
17	23	53

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 ?

				S+K
P1	P2	P3		S+X1
P4	P5	P6		S+X2
P7	P8	P9		S+X3
S+Y3	S+Y2	S+Y1		S+T

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) :

41	89	83	19	103
113	71	29	17	109
59	53	101	67	47
79	73	31	107	43
37	61	97	127	23

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 ?"