The Magic Encyclopedia ™

Nonexistence proof of {panmagic} squares of doubly odd order
Nonexistence proof of {pandiagonal magic} hypercubes of doubly odd order
{note: investigative article}
(by Aale de Winkel)

For quite some time the nonexistence theorem of panmagic squares for doubly odd order illuded me, until <George Chen> sent me for order 6 the crux for the following proof which shows that the magic sum needs to be an even number
<Günter Sterten> alerted me that the proof is valid for all hypercubes of any dimension, so there are no ⁿH_2(2k+1){pandiagonal magic} for any n >= 2.

Nonexistence proof of {panmagic} squares of doubly odd order
Suppose ^2qS is a pandiagonal square of order m = 2q and define:
A_i,j = ^2qS_2i,2j, B_i,j = ^2qS_2i+1,2j, C_i,j = ^2qS_2i,2j+1 and D_i,j = ^2qS_2i+1,2j+1 i,j = 0 .. q-1
The magic sum S is reach by all rows, columns and diagonals thus:
S = _i=0∑^q-1 (A_i,j + B_i,j) = _i=0∑^q-1 (A_j,i + C_j,i) = _i=0∑^q-1 (C_i,j + D_i,j) = _i=0∑^q-1 (B_j,i + D_j,i) =
_i=0∑^q-1 (A_i,j+i + D_i,j+i) = _i=0∑^q-1 (C_i,j+i + B_i,j+i) j = 0 .. q-1
summing these q seperate lines in each direction together gives:
qS = A + B = A + C = A + D = B + C = B + D = C + D with A = _i,j=0∑^q-1 A_i,j etc.
which gives A = B = C = D all independent terms hence qS needs to be a multiple of 4
when q is odd S = 2q(4q²+1)/2 = q(4q²+1) which is on odd number
and thus qS is odd, which shows that a {pandiagonal} square of doubly odd order can't
exists using the regular number range, when q is even qS is of course a multiple of 4.
Q.E.D.

Nonexistence proof of {pandiagonal} magic hypercubes of doubly odd order
The reasoning doesn't change when we consider the pandiagonals of the orthogonal
plane of an n-dimensional hypercube, only for the fact that the sum is different
in this case of course S = 2q(4qⁿ+1)/2 = q(4qⁿ+1) which is of still an odd
number when q = odd, so {pandiagonal} hypercubes of doubly odd order are impossible.

Note the nonexistence of {pandiagonal} magic hypercubes of doubly odd order means also that
there are no {perfect} hypercubes od doubly odd order.

As Abe's pantriagonal cubes show further conclusions can't be drawn, Abe's cube shows in
oblique planes off sums in only one direction, which translates into A+B = C+D = A+D = B+C
which results in A=C and B=D, so the doubly odd order hypercubes can be pan-r-agonal (r != 2)

The nonexistence of doubly odd order {panmagic} squares is well known, one of the references George mentioned is:
C. K. Planck in the following book: Benson and Jacoby: New Recreation with Magic Squares, Dover, 1976
(for order 6 as it was reported to me),
The more general statement that doubly odd order {pandiagonal} hypercubes in any dimension was new to me until <Günter Sterten> opted the generalisation of the proof mentioned above.
the reader is invited to send me other references to be added

The summation done above can be generalized to all dimensions this decomposes in a way the hypercubes of composite order into hypercubes of a divisor, the current theory is that the quality of the hypercube is reflected in the quality of the resulting hypercube, thus:
NOTE: The reader be aware: there still is much work to be done in the following

Hypercube decomposition theorem
the elements of a composite order hypercubes can be summed over

divisor order
summing cube Let q be a divisor of m then the following hypercube of order q can
be formed from the given hypercube of order m
ⁿH_q[_ji] = _l=0∑^n-1_k=0∑^m/q-1 ⁿH_m[_ji + k _lq]; j = 0..n-1

If ⁿH_m is pan-r-agonal then so is ⁿH_q so either ⁿH_q is a constant cube (single number on every cell)
or ⁿH_q is a real pan-r-agonal magic hypercube, thus if it is impossible for ⁿH_q to be pan-r-agonal so it is
impossible for ⁿH_m to be pan-r-agonal, unless ⁿH_q is constant.

Hypothesis ⁿH_(2^n-1) can't be {pandiagonal monagonal}
n = 2 by exhaustive placing of 1,2,3,4 in square
n = 3 by computeranalysis of 144 equations involving 64 variables (<Günter Sterten>)
n == 4 yet unproven, n == 5 falsified
<Mitsutoshi Nakamura> sent me the following falsification for n = 5
suppose H(x,y,z,u,v), where x,y,z,u,v vary from 0 to 15,
is a pan-(1,)2,5-agonal (normal) magic hypercube of n = 5 and m = 16:

H(x,y,z,u,v) = (16^4)f(a) + (16^3)f(b) + (16^2)f(c) + 16f(d) + f(e) + 1, where
a = ( x + 2y + 3z + 4u + 5v) mod 16,
b = (5x + y + 2z + 3u + 4v) mod 16,
c = (4x + 5y + z + 2u + 3v) mod 16,
d = (3x + 4y + 5z + u + 2v) mod 16,
e = (2x + 3y + 4z + 5u + v) mod 16,
f(w) = w if w < 8, = 23-w if w >= 8.

verified by computer that this hypercube is indeed normal and pan-1,2,5-agonal.
n = 4 (not yet verified)

theorem Let q be an even divisor of m and m/q odd then ⁿH_q can't be constant.
UNPROVEN

corollaries impossibility statements
^(n>=2)H_2(2k+1) can't be {pandiagonal monagonal}
^(n>=3)H_4(2k+1) can't be {pandiagonal monagonal}
given the hypothesis: (for s = 2, 3, 4?)
^(n>=s)H_{(2^(s-1)(2k+1))} can't be {pandiagonal monagonal}

theorem ⁿH_(2ⁿ) {pandiagonal monagonal} then ⁿH_(2^n-1) constant
since ⁿH_(2^n-1) not {pandiagonal monagonal}

Nonexistence proof of {panmagic} squares of doubly odd order
Suppose ^2qS is a pandiagonal square of order m = 2q and define: A_i,j = ^2qS_2i,2j, B_i,j = ^2qS_2i+1,2j, C_i,j = ^2qS_2i,2j+1 and D_i,j = ^2qS_2i+1,2j+1 i,j = 0 .. q-1 The magic sum S is reach by all rows, columns and diagonals thus: S = _i=0∑^q-1 (A_i,j + B_i,j) = _i=0∑^q-1 (A_j,i + C_j,i) = _i=0∑^q-1 (C_i,j + D_i,j) = _i=0∑^q-1 (B_j,i + D_j,i) = _i=0∑^q-1 (A_i,j+i + D_i,j+i) = _i=0∑^q-1 (C_i,j+i + B_i,j+i) j = 0 .. q-1 summing these q seperate lines in each direction together gives: qS = A + B = A + C = A + D = B + C = B + D = C + D with A = _i,j=0∑^q-1 A_i,j etc. which gives A = B = C = D all independent terms hence qS needs to be a multiple of 4 when q is odd S = 2q(4q²+1)/2 = q(4q²+1) which is on odd number and thus qS is odd, which shows that a {pandiagonal} square of doubly odd order can't exists using the regular number range, when q is even qS is of course a multiple of 4. Q.E.D.
Nonexistence proof of {pandiagonal} magic hypercubes of doubly odd order
The reasoning doesn't change when we consider the pandiagonals of the orthogonal plane of an n-dimensional hypercube, only for the fact that the sum is different in this case of course S = 2q(4qⁿ+1)/2 = q(4qⁿ+1) which is of still an odd number when q = odd, so {pandiagonal} hypercubes of doubly odd order are impossible.
Note the nonexistence of {pandiagonal} magic hypercubes of doubly odd order means also that there are no {perfect} hypercubes od doubly odd order.
As Abe's pantriagonal cubes show further conclusions can't be drawn, Abe's cube shows in oblique planes off sums in only one direction, which translates into A+B = C+D = A+D = B+C which results in A=C and B=D, so the doubly odd order hypercubes can be pan-r-agonal (r != 2)

Hypercube decomposition theorem
the elements of a composite order hypercubes can be summed over
divisor order summing cube	Let q be a divisor of m then the following hypercube of order q can be formed from the given hypercube of order m
divisor order summing cube	ⁿH_q[_ji] = _l=0∑^n-1_k=0∑^m/q-1 ⁿH_m[_ji + k _lq]; j = 0..n-1
If ⁿH_m is pan-r-agonal then so is ⁿH_q so either ⁿH_q is a constant cube (single number on every cell) or ⁿH_q is a real pan-r-agonal magic hypercube, thus if it is impossible for ⁿH_q to be pan-r-agonal so it is impossible for ⁿH_m to be pan-r-agonal, unless ⁿH_q is constant.
Hypothesis	ⁿH_(2^n-1) can't be {pandiagonal monagonal}
	n = 2 by exhaustive placing of 1,2,3,4 in square n = 3 by computeranalysis of 144 equations involving 64 variables (<Günter Sterten>) n == 4 yet unproven, n == 5 falsified
	<Mitsutoshi Nakamura> sent me the following falsification for n = 5 suppose H(x,y,z,u,v), where x,y,z,u,v vary from 0 to 15, is a pan-(1,)2,5-agonal (normal) magic hypercube of n = 5 and m = 16: H(x,y,z,u,v) = (16^4)f(a) + (16^3)f(b) + (16^2)f(c) + 16f(d) + f(e) + 1, where a = ( x + 2y + 3z + 4u + 5v) mod 16, b = (5x + y + 2z + 3u + 4v) mod 16, c = (4x + 5y + z + 2u + 3v) mod 16, d = (3x + 4y + 5z + u + 2v) mod 16, e = (2x + 3y + 4z + 5u + v) mod 16, f(w) = w if w < 8, = 23-w if w >= 8. verified by computer that this hypercube is indeed normal and pan-1,2,5-agonal. n = 4 (not yet verified)
theorem	Let q be an even divisor of m and m/q odd then ⁿH_q can't be constant.
	UNPROVEN
	corollaries	impossibility statements
	corollaries	^(n>=2)H_2(2k+1) can't be {pandiagonal monagonal} ^(n>=3)H_4(2k+1) can't be {pandiagonal monagonal} given the hypothesis: (for s = 2, 3, 4?) ^(n>=s)H_{(2^(s-1)(2k+1))} can't be {pandiagonal monagonal}
theorem	ⁿH_(2ⁿ) {pandiagonal monagonal} then ⁿH_(2^n-1) constant
theorem	since ⁿH_(2^n-1) not {pandiagonal monagonal}