3x3 "Magic Square" of Prime Numbers

Question

During the thinking and analysis of some mathematical problems, I came up with this puzzle:

Just like any magic square, one has to fill in $9$ different numbers $P_1, P_2, \dots P_9$ to a $3 \times 3$ grid. But this time, all the numbers must be different prime numbers. In addition, the $8$ sums ($3$ horizontal, $3$ vertical and $2$ diagonal) must not only be different prime numbers among themselves, but also be different from the $9$ numbers in the grid. In other words, $P_1, P_2, \dots, P_9, S_1, S_2, \dots, S_8$ must be all different prime numbers.

I suppose there are infinitely many solutions, so the challenge is to minimize the sum $S_1 + S_2 + \dots + S_8$. Here is one answer I found:

The total of the $8$ sums is $480$. I believe there are very likely solutions that can beat this total. You are welcome to have a try.

Update (plus Spoiler Alert): It was verified (using computer program) that one of the answers here (the accepted answer) is the optimal solution that cannot be beaten. There are $8$ optimal solutions, but actually they are the same because if you rotate one of the solutions by $90$, $180$ and $270$ degrees, and also horizontally flip each of the resulting grid, you will get all $8$ answers. Hence the "open-ended" tag has been removed.

Answer 1

With a brute force program solver written in C#, I found a solution with sum 366:

    3 11  5 | 19
   37 17 13 | 67
    7 31 23 | 61
   ---------+---
29 47 59 41 | 43

I let the program run until the top left corner was 61, so I'm pretty sure that there are no better solutions, but feel free to look for yourself:

Source code at PasteBin (you might want to change the initial value of recordSum to something bigger than 366 to actually get some results).

Answer 2

Here's an example that beats the OP's:

  |11  41   7|  59
  |47   3  17|  67
  |13  83   5| 101
  ------------
23 71 127  29   19

I got this by putting the smallest possible number (3) in the most central position (which is involved in four of the eight sums) and then the next smallest numbers (5, 7, 11, 13) in the positions involved in three of the eight sums, so that $S_1+\dots+S_8$ is minimised.

Answer 3

I just came up with a solution summing up a total of 396:

   19 11 17 | 47
    5 29 37 | 71
    7  3 13 | 23
   ---------+---
53 31 43 67 | 61

In fact there are exactly 8 matrices summing up 396 using primes {3, 5, 7, 11, 13, 17, 19, 29, 37}.

I will be working around on this problem till I can find a better solution but I think the best possible is the one stated by our friend @schnaader

PS: In fact, there are exactly 8 matrices summing up to 366 using primes {3, 5, 7, 11, 13, 17, 23, 31, 37}, just like the one @schnaader showed us!

Answer 4

Here's my attempt to solve it:

    5   17  7   |  29
    61  3   19  |  83
    13  47  11  |  71
----------------+----
23  79  67  37  |  19

That sums up to 408 which is better than 480

I kept 3(smallest one) in the middle, because the number in the middle affects 4 different sums(vertical middle, horizontal middle and both diagonals), so increasing the middle number by 1 would increase total sum by 4.
Then, because each number on the diagonal also affects 3 other sums(vertical, horizontal and diagonal), I tried to keep those numbers as small as possible.
And as last, I added 4 last numbers(17, 19, 47 and 61) so that all the sums would be prime too.

Answer 5

Here's another solution that reaches 480(!)

   |17   79  13|  109
   | 5    3  59|  67
   | 7   19  11|  37
  ------------
23  29  101  83   31

Answer 6

This is more of a comment as opposed to an answer

---

    5   29  7   |  41
    23  3   17  |  43
    13  11  59  |  83
----------------+----
23  41  43  83  |  67

Hello! I am new to this, but above is what I found. I do not have a computer to find solutions, so I came up with this by hand.

$$\sum_{n = 1}^8 S_n = \boxed{424}$$

The occurrence of $41, 43$ and $83$ is extraordinary in my opinion, which is why I mentioned this result, but perhaps the best result nonetheless would have to be $366$.