Sums of Squares in Lines of Order 3 Magic Square

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Theorem

Consider the order 3 magic square:

$\begin{array}{|c|c|c|}

\hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end{array}$


The sums of the squares of the top and bottom rows are equal, and differ by $18$ from the sums of the squares of the middle row
The sums of the squares of the left and right columns are equal , and differ by $18$ from the sums of the squares of the middle column.


Proof

For the rows:

\(\ds 2^2 + 7^2 + 6^2\) \(=\) \(\ds 4 + 49 + 36\)
\(\ds \) \(=\) \(\ds 89\)
\(\ds 4^2 + 3^2 + 8^2\) \(=\) \(\ds 16 + 9 + 64\)
\(\ds \) \(=\) \(\ds 89\)
\(\ds 9^2 + 5^2 + 1^2\) \(=\) \(\ds 81 + 25 + 1\)
\(\ds \) \(=\) \(\ds 107\)
\(\ds \) \(=\) \(\ds 89 + 18\)


For the colums:

\(\ds 2^2 + 9^2 + 4^2\) \(=\) \(\ds 4 + 81 + 16\)
\(\ds \) \(=\) \(\ds 101\)
\(\ds 6^2 + 1^2 + 8^2\) \(=\) \(\ds 36 + 1 + 64\)
\(\ds \) \(=\) \(\ds 101\)
\(\ds 7^2 + 5^2 + 3^2\) \(=\) \(\ds 49 + 25 + 9\)
\(\ds \) \(=\) \(\ds 83\)
\(\ds \) \(=\) \(\ds 101 - 18\)

$\blacksquare$


Sources

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