How to fill this diagram in order to obtain a magic square?

Question

How to fill the following diagram in order to obtain a magic square?

Answer 1

To do it mechanically, let the magic constant be $z$. Then the center left is $z+3x-8y$. Anywhere you have two cells in a row filled in you can fill in the third. When you finish you can add the cells in rows or columns you didn't use. The sum must equal $z$. You will get equations that give you $z$ in terms of $x,y$ and can replace $z$ everywhere.

Added: you can also prove that the magic constant is three times the central square. If you add the three columns you get all the numbers in the grid for three times the magic constant. If you add the four lines through the center you get all the numbers in the grid plus three times the center square for four times the magic constant. Subtracting these give the magic constant is three times the central square, so $z=12x-3y$. You can now fill in the rest of the grid.

Answer 2

Let

$\begin{matrix} -2x+3y & -- &--\\ ax+by & 4x-y &-- \\-x+5y &-- &--\end{matrix}$

Addition of first column gives $(-3+a)x+(8+b)y$. Using this for second row (taking hints from Thomas Andrews) gives

$\begin{matrix} -2x+3y & -- &--\\ ax+by & 4x-y &-7x+9y \\-x+5y &-- &--\end{matrix}$

Again considering

$\begin{matrix} -2x+3y & -- &--\\ -- & 4x-y &-7x+9y \\-x+5y &-- &ax+by\end{matrix}$

Summing up main the main diagonal gives $(2+a)x+(2+b)y$, using this for last row gives

$\begin{matrix} -2x+3y & -- &--\\ -- & 4x-y &-7x+9y \\-x+5y &3x-3y &--\end{matrix}$

Again considering

$\begin{matrix} -2x+3y & -- &ax+by\\ -- & 4x-y &-7x+9y \\-x+5y &3x-3y &--\end{matrix}$

summing up the other diagonal gives $(3+a)x+(4+b)y$, using this for first row gives

$\begin{matrix} -2x+3y & 5x+y &--\\ -- & 4x-y &-7x+9y \\-x+5y &3x-3y &--\end{matrix}$

and we are done as second column sums up to $12x-3y$