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

A	B	C	D
E	F	G	H
I	J	K	L
M	N	O	P

be a magic square.

Then:

• A+B+C+D = S1
• M+N+O+P = S1
• A+F+K+P = S1

• D+G+J+M = S1
• B+F+J+N = S1
• C+G+K+O = S1

 Adding the top 3, and subtracting the bottom 3:

• 2(A+P) = 2(G+J)

Therefore A+P = G+J.

Similarly, by the bimagic property, A²+P² = G²+J².

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

QED

Return to the home page http://www.multimagie.com