panmagic squares of squared order (orders: m = p^{2})  

Latin component square generating formula LSC(a,b) factor: p(p1) 
Latin component squares obtained by formula  
LSC(a,b): LC_{i,j} = [a (j % p) + b (j div p) + i] % p ; i,j ε [0,..,m1]; a = 0 .. p  1 ; b = 1 .. p  1 

Latin square factor: [p(p1)]^{2} 
Latin squares obtained by combining two component squares  
LS(a,b,c,d): L_{i,j} = p LSC(a,b)^{t} + LSC(c,d)  
The latin components combine into latin squares, the transposition is necessairy to avoid double number on each line. These latin squares inherit the panmagicness from the components, digit changing however need to be done on the components in order to keep the panmagic feature on these latin squares. 

most basic panmagic squares factor: [p(p1)]^{2} * 2 * [(p(p1)1)(p(p1)2)1] preliminairy 
S = m LS(a,b,c,d) + LS(e,f,g,h) or S = m LS(a,b,c,d) + LS^{t}(e,f,g,h) 

Experimentations with an order 9 spreadsheet suggest 19 low components LS's with each high component LS suggesting [p(p1)]^{2}[(p1)(p2)1] panmagic squares obtainable by this theory. (note: formula based on order 9 only, so might be faulty) 

In order to derive the panmagic squares in normalized position from the above the same priciples of course apply as in the prime order case, complicated by the fact that the digit changing now apply on each of the four concidered components, since we want the upperleft corner to remain 0 each component contributes a possible (p1) due to digit changing, the "[p,q,r,s]condition" (see prime orders) give rise to a factor (p2) in stead of (p1) in one of the 3 higher components (??), so thus argued (p1)^{3}(p2) panmagic squares are thus obtained. 

panmagic squares factor: 2 * [p(p1)]^{2} * [(p(p1)1)(p(p1)2)1](p!)^{4} status: preliminairy note the given formulae need to be verified (generalising from one orders samples is tricky, prone to be faulty) 
S = m [p LSC(a,b)^{t}_{p1} + LSC(c,d)_{p2}] +
[p LSC(e,f)^{t}_{p3} + LSC(g,h)_{p4}] or S = m [p LSC(a,b)^{t}_{p1} + LSC(c,d)_{p2}] + [p LSC(e,f)^{t}_{p3} + LSC(g,h)_{p4}]^{t} 

The various factors here denoted as _{p1} .. _{p4} are allowed here to change full range, hence the factor is (p!)^{4} allowing this full range digit changing one also obtains reflectional variants, also all m^{2} panrelocations are obtained this way, thus an extra deviding factor of 4m^{2} is needed to obtain the amount of different panmagic squares. 

2 * [p(p1)]^{2}[(p(p1)1)(p(p1)2)1](p!)^{4}  
9 (p=3) 25 (p=5) 
2 * 886.464 (2 * 2.736) 2 * 28.283.904.000.000 (2 * 11.313.561.600) 

The number of LSC's is currently under investigation the Latin Component Hypercube article suggest LSC's are formed by adding "order 4 addonns" to the trivial hypercube. This scheme also produces more irregularly shaped LSC's than form by the discussed 2parameter method 

LSC's by means of augmented patterns  
Pattern  a p by p square with digits radix p  
The pattern form the basic content of each p by p subsquare  
Augmentator  a p by p square with digits radix p  
Each augmentator cell represents a p by p block of cells with the digit as element to each block a pattern is added (mod p) to obtain a LSC. 

Colorisor  a p by p square with in each cell a permutation of the digits [0..p]  
Each colorisor cell contains a digit changing permutation, to be applied to the pattern which is pasted onto the p by p block each cell represents 

for order 9 the the agmentation and colorisation scheme are equivalent, currently I know of 8 augmentators, each have at least 12 working patterns as patterns which form LSC components of order 9 magic squares. This is under investigation since recenty I found 19 working patterns for one of the augmentators. The DataBase holds a full listing of the 124 panmagic LSC's found. A spreadsheet is provided to combine LSC's and test the resulting squares for regularity. 

Order p^{2} panmagic squares  
Given the set of panmagic LSC's a panmagic square is obtained by combining those LSC's LSC_{1} * p^{3} + LSC_{2} * p^{2} + LSC_{3} * p + LSC_{4} Currently it is mere trial and error to find LSC quadruplets which combine into regular squares Thus for now it isn't possible to predict the total amount of squares, nor do I see possibility to predict how many LSC's there are for other orders. The reader is invited to investigate! 