proofs 3

Prove that any three sets of three numbers will form a 3 by 3 magic square providing:

• each set of three must be in a linear sequence with the same difference 'a' (e.g. 1, 7, 13 & 10, 16, 22 & 19, 25, 31 all with a common difference of 6)
• the first terms in the three sequences must also be in a linear sequence, difference 'b' (1 , 10 and 19 in the above example)

this is why any 3 by 3 block from a calendar (with a = 1 and b = 7) will form a magic square

for any magic square

• use the fact that the corners are the average of the two outer middles not in line with the corner (proved elsewhere) to establish that the lowest and highest numbers must be in a middle edge cell position
• use the line total property to establish that the four corner numbers sum to a multiple of 4
• what happens when (using the terminology above) b = 2a? try to explain why this happens