# Prime sums in a 4 x 4 board

Place the integers 1 to 16 in the sixteen cells of a 4 x 4 board so that the sum of any four numbers in a row or column is a different prime.

• Bernardo may be if you add a condition where diagonals must also add to Prime numbers it could be a unique solution?? Not sure it exists
– DrD
Sep 4 '17 at 22:58

Here's one solution I found by hand:

  1  2  3  5  = 11
4  6  8 13  = 31
7  9 10 15  = 41
11 12 16 14  = 53
=  =  =  =
23 29 37 47 

This is how I found it:

I first worked out which primes to use. I needed to find two disjoint sets of four primes that had an average of 34, the 4x4 magic number. I made a list as follows:

  p    p-34
11  -23
13  -21
17  -17
19  -15
23  -11
29  -5
31  -3
--34
37  +3
41  +7
43  +9
47  +13
53  +19
59  +25
61  +27
Then I made pairs of primes, one pair both below 34 or and one pair above 34, with matching surplus/deficits:
 16) 29+23                 + 41+43 / 37+47
20) 29+19 / 31+17         + 41+47
22) 29+17                 + 37+53 / 43+47
26) 31+11 / 29+13 / 19+23 + 41+53
From these I chose two disjoint sets of 4: 29+23 + 37+47, and 31+11 + 41+53. One set are the row sums, the other the column sums. The order does not matter as you can reorder the rows. Then I made a little excel sheet to keep track of the sums as I placed the 16 numbers.

For completeness, I wrote a computer program and below are all 28 solutions it found. There are three sets of row/column sum primes that can occur.

  1  2  3  5 = 11    1  2  3  5 = 11    1  2  3  5 = 11    1  2  3  5 = 11
4  6  7 14 = 31    4  7  9 11 = 31    4  6  9 12 = 31    7  4  8 12 = 31
8  9 11 13 = 41    8  6 12 15 = 41    7  8 10 16 = 41    6 10 11 14 = 41
10 12 16 15 = 53   10 14 13 16 = 53   11 13 15 14 = 53    9 13 15 16 = 53
=  =  =  =         =  =  =  =         =  =  =  =         =  =  =  =
23 29 37 47        23 29 37 47        23 29 37 47        23 29 37 47

1  2  3  5 = 11    1  2  3  5 = 11    1  2  3  5 = 11    3  1  2  5 = 11
4  6  8 13 = 31    4  9  7 11 = 31    4  6 13  8 = 31    4  7  9 11 = 31
7  9 10 15 = 41    8 12  6 15 = 41    7  9 15 10 = 41    6  8 12 15 = 41
11 12 16 14 = 53   10 14 13 16 = 53   11 12 16 14 = 53   10 13 14 16 = 53
=  =  =  =         =  =  =  =         =  =  =  =         =  =  =  =
23 29 37 47        23 37 29 47        23 29 47 37        23 29 37 47

1  3  6  7 = 17    1  2  6  8 = 17
2  4  8  9 = 23    3  4  7  9 = 23
5 10 13 15 = 43    5 11 13 14 = 43
11 12 14 16 = 53   10 12 15 16 = 53
=  =  =  =         =  =  =  =
19 29 41 47        19 29 41 47

1  3  9 10 = 23    1  3  9 10 = 23    1  3  9 10 = 23    1  2  9 11 = 23
2  4 11 12 = 29    2  4 11 12 = 29    2  4 11 12 = 29    3  4 10 12 = 29
6  5 14 16 = 41    8  5 13 15 = 41    6  7 13 15 = 41    5  7 13 16 = 41
8  7 13 15 = 43    6  7 14 16 = 43    8  5 14 16 = 43    8  6 15 14 = 43
=  =  =  =         =  =  =  =         =  =  =  =         =  =  =  =
17 19 47 53        17 19 47 53        17 19 47 53        17 19 47 53

1  2  9 11 = 23    1  2  9 11 = 23    1  2  9 11 = 23    1  2  9 11 = 23
3  4 10 12 = 29    3  4 10 12 = 29    3  4 10 12 = 29    3  4 10 12 = 29
5  7 15 14 = 41    6  8 13 14 = 41    8  6 13 14 = 41    7  5 15 14 = 41
8  6 13 16 = 43    7  5 15 16 = 43    5  7 15 16 = 43    6  8 13 16 = 43
=  =  =  =         =  =  =  =         =  =  =  =         =  =  =  =
17 19 47 53        17 19 47 53        17 19 47 53        17 19 47 53

1  2  9 11 = 23    2  1  9 11 = 23    2  1  9 11 = 23    2  1  9 11 = 23
3  4 10 12 = 29    4  3 10 12 = 29    4  3 10 12 = 29    4  3 10 12 = 29
7  5 13 16 = 41    5  7 15 14 = 41    5  7 13 16 = 41    6  8 13 14 = 41
6  8 15 14 = 43    6  8 13 16 = 43    6  8 15 14 = 43    5  7 15 16 = 43
=  =  =  =         =  =  =  =         =  =  =  =         =  =  =  =
17 19 47 53        17 19 47 53        17 19 47 53        17 19 47 53

1  3  9 10 = 23    1  3  9 10 = 23    1  3  9 10 = 23    3  1  9 10 = 23
4  2 11 12 = 29    4  2 11 12 = 29    4  2 11 12 = 29    2  4 11 12 = 29
5  6 14 16 = 41    7  6 13 15 = 41    5  8 13 15 = 41    5  6 14 16 = 41
7  8 13 15 = 43    5  8 14 16 = 43    7  6 14 16 = 43    7  8 13 15 = 43
=  =  =  =         =  =  =  =         =  =  =  =         =  =  =  =
17 19 47 53        17 19 47 53        17 19 47 53        17 19 47 53

3  1  9 10 = 23    3  1  9 10 = 23
2  4 11 12 = 29    2  4 11 12 = 29
7  6 13 15 = 41    5  8 13 15 = 41
5  8 14 16 = 43    7  6 14 16 = 43
=  =  =  =         =  =  =  =
17 19 47 53        17 19 47 53

By reordering rows or columns it is even possible to let the two diagonals be two other primes, different to the row/column sums. My program found 44 such solutions. Here is just one:

             /31
3  7  1  6 = 17
10 15  5 13 = 43
12 16 11 14 = 53
4  9  2  8 = 23
=  =  =  = \
29 47 19 41  37

And finally, here are solutions for 5x5, 6x6, 7x7 squares:

   1  2  3  4  7  = 17
5  6  8  9 13  = 41
10 11 12 14 20  = 67
15 24 17 25 16  = 97
22 18 19 21 23  = 103
=  =  =  =  =
53 61 59 73 79

1   2   3   4   5   8  = 23
6   7   9  10  11  16  = 59
12  13  14  15  17  18  = 89
19  20  21  22  23  26  = 131
24  27  29  28  32  33  = 173
35  34  31  30  25  36  = 191
=   =   =   =   =   =
97 103 107 109 113 137

1   2   3   4   5   6   8  = 29
7   9  10  11  12  13  17  = 79
14  15  16  18  19  20  25  = 127
21  22  23  24  26  27  30  = 173
28  29  31  32  33  34  36  = 223
35  37  38  42  39  49  41  = 281
45  43  46  48  47  44  40  = 313
=   =   =   =   =   =   =
151 157 167 179 181 193 197

• Any other solutions? Sep 4 '17 at 13:27
• What about the 5 x 5 case with integers 1 to 25? Sep 4 '17 at 13:28
• I don't know if there are other sets of primes, Once I found the ones I used I didn't look further. There are other solutions to the square, e.g. swap 3 and 5 as well as 14 and 16, but probably not very many. Sep 4 '17 at 14:12
• Interestingly, this also has a prime sum on the diagonal. Maybe requiring that makes the puzzle unique? Sep 4 '17 at 19:15
• @Typhon: According to my computer program there are 44 solutions where the rows, columns, and main diagonals add up to distinct prime numbers. I have edited my answer to show one of them. Sep 5 '17 at 12:43

There are too many solutions, so I have showns a few of them then stopped:

1:

$\begin{bmatrix} 1& 2 &7 &3 \\ 6 & 11 &9 &5 \\ 8 &12 &16 &13 \\ 10 & 4 & 15 &14 \end{bmatrix}$

2:

$\begin{bmatrix} 1& 2 &7 &3 \\ 6 & 11 &9 &5 \\ 8 &12 &16 &13 \\ 14 & 10 & 15 &4 \end{bmatrix}$

3:

$\begin{bmatrix} 2& 5 &1 &7 \\ 3 & 6 &9 &11 \\ 8 &12 &16 &13 \\ 10 & 14 & 15 &4 \end{bmatrix}$

4:

$\begin{bmatrix} 2& 5 &7 &1 \\ 3 & 6 &9 &11 \\ 8 &12 &16 &13 \\ 4 & 14 & 15 &10 \end{bmatrix}$

5:

$\begin{bmatrix} 3& 1 &2 &7 \\ 4 & 5 &9 &11 \\ 6 &12 &8 &15 \\ 10 & 13 & 16 &14 \end{bmatrix}$

6:

$\begin{bmatrix} 3& 1 &2 &7 \\ 4 & 9 &5 &11 \\ 6 &12 &8 &15 \\ 10 & 13 & 16 &14 \end{bmatrix}$

7:

$\begin{bmatrix} 1& 2 &4 &6 \\ 3 & 5 &7 &8 \\ 10 &9 &12 &16 \\ 11 & 15 & 14 &13 \end{bmatrix}$

8:

$\begin{bmatrix} 1& 2 &4 &6 \\ 3 & 5 &7 &8 \\ 12 &9 &10 &16 \\ 13 & 15 & 14 &11 \end{bmatrix}$

9:

$\begin{bmatrix} 1& 2 &4 &6 \\ 3 & 5 &7 &8 \\ 12 &9 &10 &16 \\ 13 & 15 & 14 &11 \end{bmatrix}$

10:

$\begin{bmatrix} 1& 2 &4 &8 \\ 3 & 6 &11 &5 \\ 9 &15 &7 &12 \\ 10 & 14 & 13 &16 \end{bmatrix}$

10:

$\begin{bmatrix} 1& 2 &5 &3 \\ 4 & 8 &7 &6 \\ 11 &12 &14 &10 \\ 13 & 9 & 15 &16 \end{bmatrix}$

I think there are too many solutions, so I stopped :)

• Hmm. The solution I found is not among these, as none of them have a row or column with 1+2+3+5=11. Sep 4 '17 at 15:17
• @JaapScherphuis still updating :)
– Oray
Sep 4 '17 at 15:19
• The last two of your solutions have a row sum of 25, which is not a prime. I get only 28 solutions, excluding row/column permutations. Sep 4 '17 at 22:13
• Oray is there a solution where diagonals also add up to Prime numbers? That could be unique
– DrD
Sep 4 '17 at 22:55
• @DeepakMahulikar According to my computer program there are 44 solutions where the rows, columns, and diagonals add up to distinct prime numbers. One of them is shown in my answer. Sep 5 '17 at 12:42