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I consider this more a "recreational math" problem, possibly lacking a solution (because it stems from the question of "magic square of squares") or simply intractable with reasonable effort.

Consider the following matrix-equation in integers with otherwise arbitrary choosable values for $e,h,i$ only with the additional condition, that they must all be different, all must be odd, and more precisely, $e^2,h^2,i^2$ must be $1 \pmod {12}$ $$ \begin{array} {rcrrrr} \begin{bmatrix} a^2\\b^2\\c^2\\d^2\\f^2\\g^2 \end{bmatrix} & =& \left[ \begin{array} {rrr} 2 & 0 & -1\\ 2 & -1 & 0 \\ -1 & 1 & 1 \\ -2 & 1 & 2 \\ 4 & -1 &-2 \\ 3 & -1 & -1 \\ \end{array} \right] & \cdot \begin{bmatrix} e^2\\ h^2 \\ i^2 \end{bmatrix} \end{array} $$ Is it possible to find some $e,h,i$ with the given conditions, such that the lhs contains only squares? I thought, it might by possible by analysis of the coefficients matrix, but after some fiddling I didn't see a promising path... On the other hand, there might be some obvious(?) argument, that such a combination of squares is impossible. Or - that the given coefficients-matrix cannot -for some good reason- be of help for this problem.

[update] One could extend the conditions, such that we describe the unknowns $d^2,f^2,g^2$ as dependent on $a^2,b^2,c^2$ by the following further matrix equation $$ \begin{array} {} 9 \cdot \begin{bmatrix} d^2\\f^2\\g^2\end{bmatrix} &=& \left[ \begin{array} {rrr} -6 & 3 & 12 \\ 12 & 3 & -6 \\ 6 & 6 & -3 \end{array} \right] \cdot \begin{bmatrix} a^2 \\ b^2 \\ c^2 \end{bmatrix} \end{array} $$

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does $g$ have a scale factor of $3$ (or an odd number of them)? If not a simple parity of factors of $3$ would prove there is not a solution since each equation has a square on the left and a factor of $3$ on the right. – adam W Jan 22 '13 at 15:31
@adamW Well, at least the solution $a=\ldots=i=1$ is possible (except that the "must all be different" rules it out). Thus I guess that modulo observations cannot rule out the existence of a solution. – Hagen von Eitzen Jan 22 '13 at 16:05
@adamW: I do not know more about the dependent variables than that they must be odd and congruent 1 (mod 12). However, it came to my mind in the afternoon that it is a field of old and intense research, how sums of squares can be equal ... hmmm... – Gottfried Helms Jan 22 '13 at 16:36
by exhaustive search mod $12$ ( looking at squares it was only $64$ different cases) I see there is not a solution. I will type up my method for you now. – adam W Jan 22 '13 at 16:37
I must apologize as I calculated with an incorrect digit. Also, I have derived the exact same data as is in the question and thus verify its correctness. I am erasing my answer! – adam W Jan 23 '13 at 2:20

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