Nonexistence proof of {panmagic} squares of doubly odd order  

Suppose ^{2q}S is a pandiagonal square of order m = 2q and define: A_{i,j} = ^{2q}S_{2i,2j}, B_{i,j} = ^{2q}S_{2i+1,2j}, C_{i,j} = ^{2q}S_{2i,2j+1} and D_{i,j} = ^{2q}S_{2i+1,2j+1} i,j = 0 .. q1 The magic sum S is reach by all rows, columns and diagonals thus: S = _{i=0}∑^{q1} (A_{i,j} + B_{i,j}) = _{i=0}∑^{q1} (A_{j,i} + C_{j,i}) = _{i=0}∑^{q1} (C_{i,j} + D_{i,j}) = _{i=0}∑^{q1} (B_{j,i} + D_{j,i}) = _{i=0}∑^{q1} (A_{i,j+i} + D_{i,j+i}) = _{i=0}∑^{q1} (C_{i,j+i} + B_{i,j+i}) j = 0 .. q1 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}∑^{q1} 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(4q^{2}+1)/2 = q(4q^{2}+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 ndimensional hypercube, only for the fact that the sum is different in this case of course S = 2q(4q^{n}+1)/2 = q(4q^{n}+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 panragonal (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 

^{n}H_{q}[_{j}i] = _{l=0}∑^{n1}_{k=0}∑^{m/q1} ^{n}H_{m}[_{j}i + k _{l}q]; j = 0..n1  
If ^{n}H_{m} is panragonal then so is ^{n}H_{q}
so either ^{n}H_{q} is a constant cube (single number on every cell) or ^{n}H_{q} is a real panragonal magic hypercube, thus if it is impossible for ^{n}H_{q} to be panragonal so it is impossible for ^{n}H_{m} to be panragonal, unless ^{n}H_{q} is constant. 

Hypothesis  ^{n}H_{(2n1)} 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,)2,5agonal (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, = 23w if w >= 8. verified by computer that this hypercube is indeed normal and pan1,2,5agonal. n = 4 (not yet verified) 

theorem  Let q be an even divisor of m and m/q odd then ^{n}H_{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(s1)(2k+1))} can't be {pandiagonal monagonal} 

theorem  ^{n}H_{(2n)} {pandiagonal monagonal} then ^{n}H_{(2n1)} constant  
since ^{n}H_{(2n1)} not {pandiagonal monagonal} 