, 2021

Suppose that the integer in the bottom left corner is $n$. In this case, the sum of the integers in the first column is $64+70+n$ or $n+134$. Thus, the sum of the integers in each row, in each column, and on each diagonal also equals $n+134$. (This means that this square is in fact a magic square, since the sum of the numbers in each row, in each column, and on each diagonal is the same.) Using the top row, the top right integer equals $(n+134)-64-10$ or $n+60$. Using the northeast diagonal, the centre integer equals $(n+134) - n - (n+60)$ or $74-n$. Using the second row, the middle integer in the right column equals $(n+134)-70-(74-n)$ or $2n-10$. Using the southeast diagonal, the bottom right integer equals $(n+134)-64 - (74-n)$ or $2n-4$. Using the third row, the middle integer is $x = (n+134)-n-(2n-4) = 138 - 2n$. $$\begin{array}{|c|c|c|} \hline 64 & 10 & n+60 \\ \hline 70 & 74-n & 2n-10 \\ \hline n & 138-2n & 2n-4\\ \hline \end{array}$$ Using the third column, $$\begin{aligned} (n+60)+(2n-10)+(2n-4) & = n + 134 \\ 5n + 46 & = n + 134 \\ 4n & = 88 \\ n & = 22\end{aligned}$$ Therefore, $x= 130-2n = 130 - 44 = 94$ and the complete grid is $\begin{array}{|c|c|c|} \hline 64 & 10 & 82 \\ \hline 70 & 52 & 34 \\ \hline 22 & 94 & 40 \\ \hline \end{array}$ .