[ *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.

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.

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 " :

| | | | 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 ?"