On picture below is three-by-three magic square in which seven of the entries are squared integers, found by Andrew Bremner of Arizona State University (and independently by Lee Sallows of the University of Nijmegen):
$$%![enter image description here][1] \boxed{ \begin{array} {ccc} 373^2 & 289^2 & 565^2 \\ 360721 & 425^2 & 23^2 \\ 205^2 & 527^2& 222121 \end{array}}$$
What would be an efficient algorithm for finding a new example of a three-by-three magic square with seven squared entries that differs from the one already known ?
I know that general formula for $e_{ij}$ entry of an odd magic square is given by :
$$e_{ij}= n\cdot\left(\left(i+j-1+\left \lfloor \frac{n}{2} \right \rfloor \right) \bmod n \right)+\left((i+2j-2\right) \bmod n)+1$$
P.S.
Rotations, symmetries, and multiples of this known square don't count as new solutions.
EDIT :
I have found this one with six squared entries :
$$%![enter image description here][2] \boxed{\begin{array} {ccc} 17^2 & 35^2 & 19^2 \\ 697 & 25^2 & 553 \\ 889 & 5^2 & 31^2 \end{array}}$$