Ran into this phenomenon recently.
Just have to share it.
Construct a 3-by-3 Magic Square.
. . \(\displaystyle \begin{array}{|c|c|c|}\hline6 & 1 & 8 \\ \hline7&5&3 \\ \hline2&9&4 \\ \hline \end{array}\)
Replace each number by its corresponding Fibonacci number:
. . \(\displaystyle \begin{array}{|c|c|c|}\hline F_6 & F_1 & F_8 \\ \hline F_7&F_5&F_3 \\ \hline F_2&F_9&F_4 \\ \hline \end{array} \quad\Rightarrow\quad \begin{array}{|c|c|c|}\hline8 & 1 & 21 \\ \hline13&5&2 \\ \hline1&34&3 \\ \hline \end{array}\)
Consider the products of the row numbers
. . and the products of the column numbers.
. . \(\displaystyle \begin{array}{|ccc|c|} \hline8&1&21 & 168 \\13&5&2 & 130 \\1&34&3 & 102 \\ \hline104&170&126 & \color{red}{400} \\ \hline \end{array}\)
The sum of the row-products and the sum
. . of the column-products are equal.
Just have to share it.
Construct a 3-by-3 Magic Square.
. . \(\displaystyle \begin{array}{|c|c|c|}\hline6 & 1 & 8 \\ \hline7&5&3 \\ \hline2&9&4 \\ \hline \end{array}\)
Replace each number by its corresponding Fibonacci number:
. . \(\displaystyle \begin{array}{|c|c|c|}\hline F_6 & F_1 & F_8 \\ \hline F_7&F_5&F_3 \\ \hline F_2&F_9&F_4 \\ \hline \end{array} \quad\Rightarrow\quad \begin{array}{|c|c|c|}\hline8 & 1 & 21 \\ \hline13&5&2 \\ \hline1&34&3 \\ \hline \end{array}\)
Consider the products of the row numbers
. . and the products of the column numbers.
. . \(\displaystyle \begin{array}{|ccc|c|} \hline8&1&21 & 168 \\13&5&2 & 130 \\1&34&3 & 102 \\ \hline104&170&126 & \color{red}{400} \\ \hline \end{array}\)
The sum of the row-products and the sum
. . of the column-products are equal.
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