Four 3x3 semimagic squares in a 5x5 grid

Question

Can you place every number from 1 to 25 in a 5x5 grid such that it contains four 3x3 semimagic squares? A semimagic square is a square whose rows and columns all sum to the same number. This is intended to be solved with a computer.

Answer 1

Yes, $4$ is the most possible (take the subsquares with $12$, $3$, $18$, and $13$ in their top-left corners as the semi-magic squares). \begin{matrix} 12& 24& 3& 25& 11 \\ 9& 7& 23& 10& 6 \\18& 8& 13& 4& 22 \\20& 14& 5& 19& 15 \\1& 17& 21& 16& 2 \\\end{matrix}

Answer 2

Here's a different one, obtained via integer linear programming:

 18 19  1 12 25
 17  5 16 20  2
  3 14 21  6 11
 13 15 10 24  4
 22  9  7  8 23

Answer 3

Logic underpinning the other answers:

If two subsquares share a full row and/or column, their semimagic sums are equal. Consequently, no two subsquares share two full rows and/or columns, because their third rows/columns would be forced to be equal. We are left with three possible configurations, up to rotation: Zero Corners (top), Three Corners (left), and Four Corners (right), where subsquares of the same color have the same sum.

Four Corners is the only valid configuration.
In the Zero Corners configuration, R1C2-4 must have the same sum as R4C2-4, but R1Cn differs from R4Cn by the difference in sum between the green and red squares. Consequently, the green and red sums are equal, and identical numbers abound.
In the Three Corners configuration, R2C5 and R5C2 are equal*, being both different from R2C2 by the difference between the red and green sums. Likewise for R1C4 and R4C1 (via R4C4).
*(R1C1 is the top left.)

The central cell is congruent to 5 modulo 8 (i.e. 5, 13, or 21).
In the Four Corners configuration, the values of the entire 5×5 square are determined by the central 3×3 as follows, where $S$ is the magic sum:
\begin{array}{ccccc} a+b+d+e-S & S-a-d & S-b-e & S-c-f & b+c+e+f-S\\ S-a-b & a & b & c & S-b-c\\ S-d-e & d & e & f & S-e-f\\ S-g-h & g & h & i & S-h-i\\ d+e+g+h-S & S-d-g & S-e-h & S-f-i & e+f+h+i-S \end{array}
Summing the entries yields $325 = 8S+e$ and reducing modulo 8 yields the claim.

The number of solutions is divisible by 128.
The Four Corners configuration has five validity-preserving symmetries due to its structure:
- Rotation and reflection (order 8)
- Interchanging rows 1 and 2 (order 2)
- Interchanging rows 4 and 5 (order 2)
- Interchanging columns 1 and 2 (order 2)
- Interchanging columns 4 and 5 (order 2)
In fact, there are exactly four solutions up to these symmetries (i.e. 512 in total) - user1502040's, RobPratt's, and the squares formed by replacing each x with 26-x in either of those squares.