There is another 3x3 square of numbers, with some entries missing, such that the first two in each row or column add up to give the last entry in that row or column. It so happens that we can write four numbers in the square in almost any way so that the rest of the square can be completed.
\[ \begin{array}{ccc} 2 & 3 & ? \\ 4 & 6 & ? \\ ? & ? & ? \end{array} \]
The square is completed for the first two rows with
\[2+3=5\]
\[4+6=10\]
We have then
\[ \begin{array}{ccc} 2 & 3 & 5 \\ 4 & 6 & 10 \\ ? & ? & ? \end{array} \]
Now sum the columns to give
\[ \begin{array}{ccc} 2 & 3 & 5 \\ 4 & 6 & 10 \\ 6 & 9 & 15 \end{array} \]
.Now notice for the last row
\[5+10=15\]
.This always happens.
If we fill out the square differently, say
\[ \begin{array}{ccc} 1 & ? & 7 \\ 1 & ? & 8 \\ ? & ? & ? \end{array} \]
.Completing the columns first gives
\[ \begin{array}{ccc} 1 & ? & 7 \\ 1 & ? & 8 \\ 2 & ? & 15 \end{array}\]
.Completing the rows gives
\[ \begin{array}{ccc} 1 & 6 & 7 \\ 1 & 7 & 8 \\ 2 & 13 & 15 \end{array}\]
.Again the columns and rows add up to the number at the bottom or right respectively.