Bimagic squares of order 4, using distinct numbers, are impossible
by Dr. Luke Pebody, October 18th 2004


There are no bimagic squares of distinct integers for 4x4.

I do not use the lines through E, H, I or L in the diagram below, but I do use both bimagic diagonals.

Proof:

Let

be a magic square.

Then:

 Adding the top 3, and subtracting the bottom 3:

Therefore A+P = G+J.

Similarly, by the bimagic property, A+P = G+J.

Therefore {A,P}={G,J}.
A=G or A=J: impossible, distinct integers in the square.

QED


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