[math-fun] First known bimagic square of primes
One century ago, Henry E. Dudeney studied the first magic squares of primes even if, at that time, "1" was considered as a prime number, and was used in his squares. Look for example at http://mathworld.wolfram.com/PrimeMagicSquare.html Hot news: this week, the first BIMAGIC square of primes has been constructed (and not using "1"!). And its order is also prime: 11. Reminder. A bimagic square of order n is a nxn magic square remaining magic after each of its n² numbers have been squared. It is the first solved problem on the 10 open problems published in my Math Intelligencer article, Spring 2005. But it means also that 9 problems are still unsolved. A lot of work remains to be done, including "small" unsolved problems: - 3x3 magic square of squares (only semi-magic are known) - 4x4 magic square of cubes (only semi-magic are known) - 5x5 bimagic square (only semi-bimagic are known) Christian.
Hot news: this week, the first BIMAGIC square of primes has been constructed (and not using "1"!). And its order is also prime: 11.
would you share with us the actual square and maybe by which method (algorithm) it was obtained? Christoph
Ooops, sorry, I forgot the "cut and paste". Here is the square that you can check: 137 131 317 47 5 457 541 359 467 353 683 401 277 239 647 23 421 229 181 7 419 653 463 269 701 59 157 257 563 557 179 191 101 593 311 379 503 197 83 53 521 149 619 89 307 617 397 241 571 661 109 107 79 127 281 373 443 29 587 383 61 19 409 631 389 173 73 11 607 433 613 577 263 97 227 313 283 43 599 151 199 509 487 223 163 293 691 139 673 37 113 271 193 31 601 431 331 337 479 67 233 103 439 499 251 547 659 491 41 167 367 569 461 71 347 211 349 13 643 17 449 Characteristics: -121 distinct prime integers, and more precisely 121 consecutive prime integers <= 701, only excluding 2, 3, 523, 641, 677 -bimagic square (11 rows, 11 columns, 2 diagonals) with magic sums S1=3497, S2=1578251 You asked my method. Six main steps: 1) Choose the order. My idea -for the beauty- was to have a prime order. 2 and 3 are impossible. 5, nobody knows a bimagic square using any set of numbers. 7, probably yet to small to have a sufficient number of bimagic series. 11, hmmm. 2) Select carefully a good set of 121 prime numbers, using some modulo reasoning. 3) With this set, compute the S1 and S2 sums. 4) Compute the list of bimagic series of 11 primes, from the selected set, having the good S1 and S2 sums. 5) Use these series, use combinatorics algorithms, and find 22 series that you can cross and correctly organize, producing a semi-bimagic square. 6) Close to be finished? No! As usual for magic squares, one of the most difficult step, sometimes impossible step: find TWO diagonals in rearranging the cells of the semi-magic square. Most of them are unable to create a bimagic square: I needed several hundreds of semi-bimagic squares before to succeed. Christian.
Nice work, Christian ! just one (too?) obvious comment to
121 consecutive prime integers <= 701, only excluding 2, 3, 523, 641, 677
2 has to be excluded from any square (with size > 1) of primes since the row and column with 2 would have a wrong sum modulo 2. and one obvious question: are you sure that no bimagic square of primes can exist with 121 consecutive primes starting from 3? ;-) Christoph
answering my own question
and one obvious question: are you sure that no bimagic square of primes can exist with 121 consecutive primes starting from 3? ;-)
Ok, it canNOT exist: Of course two necessary conditions to hold are order divides Sum[Prime[n],{n,2,order^2+1}] order divides Sum[Prime[n]^2,{n,2,order^2+1}] (here the sum is over the first order^2 odd primes). this does not hold for 11, but even more: besides order = 1 and 2 it does NOT hold for any order <= 5000, interesting. The first condition is however satisfied for order=1,2,12,35,215,225,398,2097,... (tested up to 5000). BTW: this sequence is not in the OEIS... The 2nd is true for order=1,2,4,8,14,16,32,44,172,173,344,430,712,944,2744,... (tested up to 8144), strange 44, isn't it: 44+300,900,2700, but not 8144! again the sequence is not in the OEIS... Christoph
You can't have a bimagic square of primes including 3; every other prime squared is = 1 (mod 3), so the row and column containing 3 will be different modulo 3. Franklin T. Adams-Watters ________________________________________________________________________ Check Out the new free AIM(R) Mail -- 2 GB of storage and industry-leading spam and email virus protection.
In a bimagic square of primes of order 11, if we want to use 121 consecutive primes, then the smallest "good" set starts from 823: 823, 827, 829, 839,... 1669 For the modulo reasons on sums well seen below by Christoph. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Pacher Christoph Envoyé : vendredi 17 novembre 2006 22:04 À : math-fun Objet : RE: [math-fun] First known bimagic square of primes answering my own question
and one obvious question: are you sure that no bimagic square of primes can exist with 121 consecutive primes starting from 3? ;-)
Ok, it canNOT exist: Of course two necessary conditions to hold are order divides Sum[Prime[n],{n,2,order^2+1}] order divides Sum[Prime[n]^2,{n,2,order^2+1}] (here the sum is over the first order^2 odd primes). this does not hold for 11, but even more: besides order = 1 and 2 it does NOT hold for any order <= 5000, interesting. The first condition is however satisfied for order=1,2,12,35,215,225,398,2097,... (tested up to 5000). BTW: this sequence is not in the OEIS... The 2nd is true for order=1,2,4,8,14,16,32,44,172,173,344,430,712,944,2744,... (tested up to 8144), strange 44, isn't it: 44+300,900,2700, but not 8144! again the sequence is not in the OEIS... Christoph
Hi Christian, That's a nice new result you discovered. My congratulations. Lee At 06:23 PM 11/17/2006 +0100, you wrote:
Hot news: this week, the first BIMAGIC square of primes has been constructed (and not using "1"!). And its order is also prime: 11.
would you share with us the actual square and maybe by which method (algorithm) it was obtained? Christoph
Ooops, sorry, I forgot the "cut and paste". Here is the square that you can check:
137 131 317 47 5 457 541 359 467 353 683 401 277 239 647 23 421 229 181 7 419 653 463 269 701 59 157 257 563 557 179 191 101 593 311 379 503 197 83 53 521 149 619 89 307 617 397 241 571 661 109 107 79 127 281 373 443 29 587 383 61 19 409 631 389 173 73 11 607 433 613 577 263 97 227 313 283 43 599 151 199 509 487 223 163 293 691 139 673 37 113 271 193 31 601 431 331 337 479 67 233 103 439 499 251 547 659 491 41 167 367 569 461 71 347 211 349 13 643 17 449
Characteristics: -121 distinct prime integers, and more precisely 121 consecutive prime integers <= 701, only excluding 2, 3, 523, 641, 677 -bimagic square (11 rows, 11 columns, 2 diagonals) with magic sums S1=3497, S2=1578251
You asked my method. Six main steps:
1) Choose the order. My idea -for the beauty- was to have a prime order. 2 and 3 are impossible. 5, nobody knows a bimagic square using any set of numbers. 7, probably yet to small to have a sufficient number of bimagic series. 11, hmmm.
2) Select carefully a good set of 121 prime numbers, using some modulo reasoning.
3) With this set, compute the S1 and S2 sums.
4) Compute the list of bimagic series of 11 primes, from the selected set, having the good S1 and S2 sums.
5) Use these series, use combinatorics algorithms, and find 22 series that you can cross and correctly organize, producing a semi-bimagic square.
6) Close to be finished? No! As usual for magic squares, one of the most difficult step, sometimes impossible step: find TWO diagonals in rearranging the cells of the semi-magic square. Most of them are unable to create a bimagic square: I needed several hundreds of semi-bimagic squares before to succeed.
Christian.
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participants (4)
-
Christian Boyer -
franktaw＠netscape.net -
Lee Sallows -
Pacher Christoph