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The numbers reading across and down in these squares are square: $\begin{array}{ccc} 1 & 4 & 4\\ 4 &8&4\\ 4&4&1 \end{array}$

$\begin{array}{ccc} 5&2&9\\ 2&5&6\\ 9&6&1 \end{array}$

$\begin{array}{cccc} 2&1&1&6\\ 1&2&2&5\\ 1&2&9&6\\ 6&5&6&1 \end{array}$

Are there any such square squares where the diagonals are also squares? If not in base 10, is it possible in other bases?

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    There are no 3x3 matrices that do what you want in base 10 :-)2011-11-09

2 Answers 2

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Those matrices are certainly hard to find, I tried it a lot and found that there are no $4\times4$ or $3\times3$ matrices that do what you want in base $10$ or below. However watch this one in base $11$:

$\begin{array}{ccc} 1 & 9 & 5\\ 9 & 6 & 1\\ 5 & 1 & 9 \end{array}$

You have

$169_{11}=196=14^2$

$195_{11}=225=15^2$

$519_{11}=625=25^2$

$565_{11}=676=26^2$

$961_{11}=1156=34^2$

All colums, rows, diagonals when read in any way (colums and rows only down and right) are squares, woha ;-)

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Uuuuh yes, take any n^2xn^2 matrix with the same number everywhere.

Wait, take any $n^2Xn^2$ matrix, and put any of your matrices in each spot.