**Bimagic squares of order 4, using distinct
numbers, are impossible**

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

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