Office Applications and Entertaiment, Magic Squares  
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5.4 Further Analysis, Symmetric Magic Squares
In a Symmetric Magic Square, the sum of each pair of elements, which can be connected with a straight line through the centre and which are equidistant to the centre, is 1 + n x n. For 5^{th} order (Pan) Magic Squares these pairs sum to 26.
This results in following additional equations:
which can be added to the equations describing either a Magic or a Pan Magic Square of the fifth order.
5.4.1 Associated Magic Squares
When the symmetry conditions are added to the equations describing a Magic Square of the fifth order (Section 3.2), the resulting square  also referred to as Associated Magic Square  is described by following set of linear equations: a(21) = s1  a(22)  a(23)  a(24)  a(25) a(16) = s1  a(17)  a(18)  a(19)  a(20) a(15) = s1 / 5 + a(16)  a(20) + a(21)  a(25) a(14) = s1 / 5 + a(17)  a(19) + a(22)  a(24) a(13) = s1 / 5 a(12) = 2 * s1 / 5  a(14) a(11) = 2 * s1 / 5  a(15) a(10) = 2 * s1 / 5  a(16) a( 9) = 2 * s1 / 5  a(17) a( 8) = 2 * s1 / 5  a(18) a( 7) = 2 * s1 / 5  a(19) a( 6) = 2 * s1 / 5  a(20) a( 5) = 2 * s1 / 5  a(21) a( 4) = 2 * s1 / 5  a(22) a( 3) = 2 * s1 / 5  a(23) a( 2) = 2 * s1 / 5  a(24) a( 1) = 2 * s1 / 5  a(25)
An optimized guessing routine (MgcSqr5c2), counted the 388352 (= 8 * 48544) possible Associated Magic Squares within half an hour, of which the first 744 are shown in Attachment 5.4.2 .
When the symmetry conditions are added to the equations describing a Pan Magic Square of the fifth order (Section 3.1), the resulting square  also referred to as Ultramagic Square  is described by following set of linear equations: a(21) = s1  a(22)  a(23)  a(24)  a(25) a(20) = 0.6 * s1  a(24)  a(25) a(19) = 0.6 * s1 + a(22)  a(23)  a(24)  a(25) a(18) = 0.6 * s1  a(22)  a(24) a(17) = 0.4 * s1 + 2 * a(24) + a(25) a(16) = 0.4 * s1 + a(23) + a(24) + a(25) a(15) = 0.2 * s1  a(22) + a(24) a(14) = 0.8 * s1 + a(23) + 2 * a(24) + 2 * a(25) a(13) = s1 / 5 a(12) = 2 * s1 / 5  a(14) a(11) = 2 * s1 / 5  a(15) a(10) = 2 * s1 / 5  a(16) a( 9) = 2 * s1 / 5  a(17) a( 8) = 2 * s1 / 5  a(18) a( 7) = 2 * s1 / 5  a(19) a( 6) = 2 * s1 / 5  a(20) a( 5) = 2 * s1 / 5  a(21) a( 4) = 2 * s1 / 5  a(22) a( 3) = 2 * s1 / 5  a(23) a( 2) = 2 * s1 / 5  a(24) a( 1) = 2 * s1 / 5  a(25)
An optimized guessing routine (MgcSqr5c), produced the 128 possible Ultramagic Squares within 18.8 seconds, which are shown in Attachment 5.4.1.

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