First notice that we need to get 10 numbers {10..19} from the sums & products of the 3 rows, 3 columns & 2 diagonals. That means we need
at least 2 products.
Any product without 1 is at least 2*3*4=24.
Any product without {23} is at least 1*4*5=20.
=> We can only use products with 1 and at least one of {23}.
Also note that the corresponding sum of a product must be from {10..19}.
The only possible triple with 1 & 3 is {136}, giving 1+3+6=10 & 1*3*6=18.
And then we need another triple, which must be with 1 & 2, to generate a sum and a product from {11..17,19}
The only possible triple is {128}, giving 1+2+8=11 & 1*2*8=16.
And with these no more products are available, the remaining 6 numbers must all be given by the sums.
Now we have the 8 sums as 10,11,12,13,14,15,17,19 (16 & 18 have been taken care by the products).
45-rule: all 3 rows sum to 45, so as all 3 columns.
So the total of the sums of the 3 rows & 3 columns = 45+45 = 90.
From the above list of 8 numbers they must be {12,13,14,15,17,19}, in other words 10 & 11 must be presented as sums of diagonals.
Therefore {128} & {136} are the diagonals, forcing 1 to be the centre cell.
Without loss of generality, the square must be of the following pattern (all other cases are reflection/rotation of it):
Code:
2 4579 3
4579 1 4579
6 4579 8
To give the sum of 19, r3c123 must be [658].
To give the sum of 17, r123c1 must be [296].
To give the sum of 15, r123c3 must be [348].
Thus r1c123=[273] to give the sum of 12.
r2c123=[914] to give the sum of 14.
r123c2=[715] to give the sum of 13.
Hence the square must be:
Code:
2 7 3
9 1 4
6 5 8