**Theorem 2, by Lee Morgenstern, December 2007.In a 6x6 pandiagonal
multiplicative
magic square, the
magic product is a 6th power.**

The proof is in 2 parts.

Part 1 proves that the magic product must be a square.

Part 2 proves that the magic product must be a cube.

Together, they prove that the magic product must be a 6th power.

Part 1 proves that the magic product must be a square.

Part 2 proves that the magic product must be a cube.

Together, they prove that the magic product must be a 6th power.

**Part 1**

a c a c a c

- b - b - b

a c a c a c

- b - b -
b

a c a c a c

- b - b - b

Let P be the magic product.

Let A, B, and C be the products of
the a-squares,

the b-squares, and the c-squares,
respectively.

The product of the 3 rows of a-squares and c-squares

A x C = P^3.

The product of the 3
columns of b-squares and c-squares

B x C =
P^3.

Thus we have

A = B.

The product of the 3 pan-diagonals of a-squares and b-squares

A x B = P^3,

but since A =
B, this is the same as

A^2 =
P^3,

which is true only if P is a
square.

**Part 2**

a a d a a d

c b - c b -

b c - b c -

a a d a a
d

c b - c b -

b c - b c -

Let P be the magic product.

Let A, B, C, and D be the products
of the a-squares,

the b-squares, the c-squares, and
the d-squares, respectively.

The product of the 2 rows of a-squares and d-squares,

A x D =
P^2.

The product of the 2 pan-diagonals of
b-squares and d-squares

B x D =
P^2.

The product of the 2 pan-diagonals of
c-squares and d-squares

C x D =
P^2.

Thus we have

A = B = C.

The product of the 4 columns of a-squares, b-squares, and c-squares

A x B x C =
P^4.

but since A = B = C, this is the same
as

A^3 = P^4,

which is true only if P is a cube.

**QED**

You can also use the diagrams above for a 6x6 additive pan-magic
square,
proving that the magic sum, S, must be a multiple of 6, using addition in
place of multiplication.

In Part 1, you would get 2A =
3S, proving that S is a multiple of 2.

In Part 2, you would get 3A = 4S, proving that S is a multiple of 3.

BTW, the reason that there is no normal 6x6 pan-magic square is that the normal magic sum of 1...36 divided by 6 is 111, an odd number, but an even number is required to be pan-magic.

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