20080430, 18:12  #1 
May 2004
New York City
2^{3}·23^{2} Posts 
Square of Primes
Construct a 5 x 5 square containing distinct primes
such that each row, column and diagonal sums to a distinct prime. 
20080430, 22:35  #2 
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
4553_{10} Posts 

20080430, 22:51  #3 
May 2004
New York City
2^{3}×23^{2} Posts 
A magic square of primes (where every sum is the same) is
solved elsewhere (although it would be a perfectly good puzzle to resolve). Here every sum is a different prime. 
20080513, 18:07  #4 
May 2004
New York City
10210_{8} Posts 
The original problem was perhaps too computationally simple
to be interesting. The following additional condition adds an iota of complexity: The 25 distinct primes in the square should be the first 25 odd primes {3,5,7,...,97,101}. (I have a solution which wasn't hard to find by trial and error, so there must be many solutions; but plain brute force on the 25! such possible squares is obviously too computationally costly.) Last fiddled with by davar55 on 20080513 at 18:08 
20080517, 02:23  #5 
Oct 2007
Manchester, UK
2^{2}·5·67 Posts 
Here's one:
Code:
239
/
/
/
3, 5, 7, 11, 17  43
13, 19, 29, 23, 43  127
31, 67, 61, 47, 71  277
53, 59, 41, 73, 37  263
79, 83, 89, 97, 101  443
     \
     \
     \
179 233 227 251 269 257
Last fiddled with by lavalamp on 20080517 at 02:26 
20080517, 06:12  #6 
Sep 2006
Brussels, Belgium
2^{2}×7×59 Posts 
There is an error in your calculations : the last row total is of by 6. But the right number is prime so the solution stands :)
Jacob 
20080517, 11:59  #7 
Oct 2007
Manchester, UK
10100111100_{2} Posts 
Hm, I think I worked the total out right, but wrote it down wrong.
It wasn't just a fluke, honest! ;) 
20080519, 14:01  #8 
Feb 2007
2^{4}·3^{3} Posts 
Does the sequence
a(n) = number of square matrices containing the first (2n+1)x(2n+1) odd primes, such that row, column and diagonal sums are distinct primes exist on OEIS ? 
20080519, 14:57  #9 
Oct 2007
Manchester, UK
1340_{10} Posts 
It would appear that there are an awful lot of these out there, so perhaps the challange should be to find a square with the lowest standard deviation of column/row/diagonal totals.
I'll start the ball rolling with a slightly modified version of the last square I posted, with an s.d. of 84.51: Code:
239
/
/
/
3, 5, 7, 11, 17  43
13, 19, 29, 23, 43  127
31, 67, 61, 47, 71  277
53, 59, 89, 73, 37  311
79, 83, 41, 97, 101  401
     \
     \
     \
179 233 227 251 269 257
Last fiddled with by lavalamp on 20080519 at 15:06 
20080521, 12:54  #10 
May 2004
New York City
2^{3}×23^{2} Posts 
Here's another solution:
041 005 007 071 003 013 023 029 031 067 059 053 043 047 037 019 011 089 061 017 079 101 083 097 073 Rows: 127,163,239,197,433 Columns: 211,193,251,307,197 Diagonals: 241,167 (Standard deviation: 76.7) An alternative measure is simply minimax: minimize the largest sum. By that measure, lavalamp's solution is a better one. 
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