Is it possible complete a $3\times3$ magic square such that

Question

Is it possible complete a $3\times3$ magic square such that the constant is equal to $(-22)$ and the initial form is : $$ \left[ \begin{array} {ccc} b& (+2) & a \\ d & c & (-34) \\ e & f & (+8) \end{array} \right]$$ I found different possibilities for $a,\ b, \ c, \ d, \ e, \ f$ so it seems that this magic square is not solvable.

What would be interesting to change ?

Thanks in advance !

Answer 1

\begin{bmatrix}4 &9 & 2\\3 & 5& 7\\8 & 1& 6 \end{bmatrix}is a magic matrix, add unknoun to the matrix to find what you want. \begin{bmatrix}4+x &9+x & 2+x\\3+x & 5+x& 7+x\\8+x & 1+x & 6+x \end{bmatrix} so $$(4+x)+(5+x)+(6+x)=-22\\15+3x=-22 \to 3x=-15-22\\3x=-37\\x=-\frac{37}{3}$$so initial value is $1+x=1-\frac{37}{3}$ there is not an integer solution for the table.
but if $-22$ was for example $-21$ then $$3x=-15-21=-36\\\to 1+x=-11 $$

Answer 2

Your initial values give an unique solution for a magic square.

I've calculated $ Q = \small \begin{bmatrix} -28 & 2 & -4 \\ 14 & -10 & -34 \\ -16 & -22 & 8 \end{bmatrix}$ - but it has the "magic sum" $m=-30$.

Remark: Since I see a $-22$ in the last row: did you perhaps mess your initial structure?

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Background:
Re-naming the entries in your matrix $Q$ in the following way: $ Q = \small \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ gives with a simple matrix-multiplication by a coefficients matrix $C$ , matrix of initial values $A=[b,f,i] = [2,-34,8] $ the resulting values for $R=[a,b,c,d,e,f,g,h,i]$ by

$$ C\cdot A= R \qquad : \qquad \small \begin{array} {} & * & \begin{bmatrix} b \\ f \\ i \end{bmatrix} \\ \begin{bmatrix} -1 & 1 & 1 \\ 1 & 0 & 0 \\ -3/2 & 1/2 & 2 \\ -1 & 0 & 2 \\ -1/2 & 1/2 & 1 \\ 0 & 1 & 0 \\ 1/2 & 1/2 & 0 \\ -2 & 1 & 2 \\ 0 & 0 & 1 \end{bmatrix} & = & \begin{bmatrix} a\\b\\c\\d\\e\\f\\g\\h\\i \end{bmatrix} \end{array}$$ Having $R$, the matrix $Q$ can then be formed by simple quadratic $3\times3$ - arrangement of the computed values in $R$

You may like to read about this calculation-ansatz in this text where I discuss the magic square with squared entries; however the initial ansatz is general and not confined to squared entries.