Trimagic Square General Theorems

A trimagic square is also bimagic, so all the bimagic theorems apply.
The bimagic squares with special structures have an added benefit
when used for trimagic squares.

You might recognize the alternative bimagic structure used by
Walter Trump's 12x12 trimagic square.
The columns all come in complementary pairs and
the diagonals form a complementary pair in trade for using
a pair or more of self-complementary rows.

We can understand the complementary pairs of columns because we know
that if two complementary columns are magic, then they are also bimagic.

But in his square, every row is self-complementary.
We know that self-complementary rows all have the same sum,
but aren't necessarily bimagic.

So why so many self-complementary rows?


Theorem
If two self-complementary rows are bimagic,
then they are also trimagic.
If
     a +(P-a) +...+b +(P-b)  = c +(P-c) +...+d +(P-d)  = NP/2,
and
     a2+(P-a)2+...+b2+(P-b)2 = c2+(P-c)2+...+d2+(P-d)2,
then
     a3+(P-a)3+...+b3+(P-b)3 = c3+(P-c)3+...+d3+(P-d)3.


Proof
Multiplying out the sum of squares,
      NP2 + 2(a2+...+b2) - 2P(a+...+b) = NP2 + 2(c2+...+d2) - 2P(c+...+d).
Subtracting NP2 and dividing by 2, then multiplying by 3NP/2,
(1)   3NP(a2+...+b2)/2 - 3NP2(a+...+b)/2 = 3NP(c2+...+d2)/2 - 3NP2(c+...+d)/2.

Multiplying out the sum of cubes for the left hand side,
      NP3/2 + 3NP(a2+...+b2)/2 - 3NP2(a+...+b)/2.
Substituting (1), we get
      NP3/2 + 3NP(c2+...+d2)/2 - 3NP2(c+...+d)/2,
which equals the sum of cubes for the right hand side.
QED