If the order M square and the order N square are both associative, the order MN composite square is also associative.
Consider squares A, B and the A-B composite C:
(The range of square B is given as 0 to N2-1 to avoid having to subtract
1 from the B numbers at each reference).
A B C
----- ----- ------
order: M N Q = MN
cells: ai,j bi,j ci,j
numbers: 1..M2 0..N2-1 1..Q2
then ci,j = air,jr + M2biq,jq
where i, j are 0 .. Q-1
ir,jr are the remainders of the division of i,j by M
iq,jq are the quotients of the division of i,j by M
let k = Q-1-i, l = Q-1-j,
TM be the order M complementary pair sum = M2+1
TN be the order N complementary pair sum = N2-1
TQ be the order Q complementary pair sum = Q2+1
then the center symmetric pair:
ci,j + ck,l =
air,jr + M2biq,jq + akr,lr + M2bkq,lq =
(air,jr + akr,lr) + M2(biq,jq + bkq,lq) =
TM + M2TN = (M2+1) + M2(N2-1) = M2N2+1 = Q2+1 = TQ
For example, a 4x4 associative square A, a 3x3 associative square B, and the 12x12 A-B composite associative square C: .txt
A B C
--- --- -----
1 8 12 13 7 0 5 113 120 124 125 1 8 12 13 81 88 92 93
14 11 7 2 2 4 6 126 123 119 114 14 11 7 2 94 91 87 82
15 10 6 3 3 8 1 127 122 118 115 15 10 6 3 95 90 86 83
4 5 9 16 116 117 121 128 4 5 9 16 84 85 89 96
33 40 44 45 65 72 76 77 97 104 108 109
46 43 39 34 78 75 71 66 110 107 103 98
47 42 38 35 79 74 70 67 111 106 102 99
36 37 41 48 68 69 73 80 100 101 105 112
49 56 60 61 129 136 140 141 17 24 28 29
62 59 55 50 142 139 135 130 30 27 23 18
63 58 54 51 143 138 134 131 31 26 22 19
52 53 57 64 132 133 137 144 20 21 25 32
order: M = 4 N = 3 Q = MN = 12
1 .. 16 0 .. 8 1 .. 144
the center symmetric pair:
c0,0 + c11,11 = (a0,0 + M2b0,0) + (a3,3 + M2b2,2)
c0,0 + c11,11 = (a0,0 + a3,3) + M2(b0,0 + b2,2)
113 + 32 = (1 + 16) + 42(7 + 1)
145 = 17 + 128
If the order M square and the order N square are both pandiagonal, the order MN composite square is also pandiagonal.
Consider the squares A, B and the A-B composite C:
(The range of square B is given as 0 to N2-1 to avoid having to subtract
1 from the B numbers at each reference).
A B C
----- ----- ------
order: M N Q = MN
cells: ai,j bi,j ci,j
numbers: 1..M2 0..N2-1 1..Q2
then ci,j = air,jr + M2biq,jq
where i, j are 0 .. Q-1
ir,jr are the remainders of the division of i,j by M
iq,jq are the quotients of the division of i,j by M
let pr be the remainder of the division of Q-1 by M
pq be the quotient of the division of Q-1 by M
SM be the order M magic sum = M(M2+1)/2
SN be the order N magic sum = N(N2-1)/2
SQ be the order Q magic sum = Q(Q2+1)/2
then the main \diagonal sum is:
c0,0 + c1,1 + ... + cQ-1,Q-1 =
(a0,0 + M2b0,0) + (a1,1 + M2b0,0) + ... + (apr,pr + M2b0,0) +
(a0,0 + M2b1,1) + (a1,1 + M2b1,1) + ... + (apr,pr + M2b1,1) +
... ... ...
(a0,0 + M2bpq,pq) + (a1,1 + M2bpq,pq) + ... + (apr,pr + M2bpq,pq) =
Na0,0 + Na1,1 + ... + Napr,pr +
M(M2b0,0) + M(M2b1,1) + ... + M(M2bpq,pq) =
N(a0,0 + a1,1 + ... + apr,pr) +
M3(b0,0 + b1,1 + ... + bpq,pq) =
N(SM) + M3(SN) = N(M(M2+1)/2) + M3(N(N2-1)/2) =
(M3N + MN + M3N3 - M3N)/2 = MN((MN)2+1)/2 = Q(Q2+1)/2 = SQ
and the second \diagonal sum is:
c1,0 + c2,1 + ... + c0,Q-1 =
(a1,0 + M2b0,0) + (a2,1 + M2b0,0) + ... + (apr,pr-1 + M2b0,0) +
(a0,pr + M2b1,0) +
(a1,0 + M2b1,1) + (a2,1 + M2b1,1) + ... + (apr,pr-1 + M2b1,1) +
(a0,pr + M2b2,1) +
... ... ...
(a1,0 + M2bpq,pq) + (a2,1 + M2bpq,pq) + ... + (apr,pr-1 + M2bpq,pq) +
(a0,pr + M2b0,pq) =
Na1,0 + Na2,1 + ... + Napr,pr-1 +
Na0,pr +
M2b1,0 + M2b2,1 + ... + M2b0,pq +
(M-1)(M2b0,0) + (M-1)(M2b1,1) + ... + (M-1)(M2bpq,pq) =
N(a1,0 + a2,1 + ... + apr,pr-1 + a0,pr) +
M2(b1,0 + b2,1 + ... + b0,pq) +
(M-1)M2(b0,0 + b1,1 + ... + bpq,pq) =
N(SM) + M2(SN) + (M-1)M2(SN) =
N(SM) + M3(SN) = N(M(M2+1)/2) + M3(N(N2-1)/2) =
(M3N + MN + M3N3 - M3N)/2 = MN((MN)2+1)/2 = Q(Q2+1)/2 = SQ
and so on for the other \diagonals and similarly for the /diagonals.
For example, a 4x4 pandiagonal square A, a 5x5 pandiagonal square B, and the 20x20 A-B composite pandiagonal square C: .txt
A B
--- -----
6 3 16 9 0 6 12 18 24
15 10 5 4 13 19 20 1 7
1 8 11 14 21 2 8 14 15
12 13 2 7 9 10 16 22 3
17 23 4 5 11
order: M = 4 N = 5
numbers: 1 .. 16 0 .. 24
C
---------------------
6 3 16 9 102 99 112 105 198 195 208 201 294 291 304 297 390 387 400 393
15 10 5 4 111 106 101 100 207 202 197 196 303 298 293 292 399 394 389 388
1 8 11 14 97 104 107 110 193 200 203 206 289 296 299 302 385 392 395 398
12 13 2 7 108 109 98 103 204 205 194 199 300 301 290 295 396 397 386 391
214 211 224 217 310 307 320 313 326 323 336 329 22 19 32 25 118 115 128 121
223 218 213 212 319 314 309 308 335 330 325 324 31 26 21 20 127 122 117 116
209 216 219 222 305 312 315 318 321 328 331 334 17 24 27 30 113 120 123 126
220 221 210 215 316 317 306 311 332 333 322 327 28 29 18 23 124 125 114 119
342 339 352 345 38 35 48 41 134 131 144 137 230 227 240 233 246 243 256 249
351 346 341 340 47 42 37 36 143 138 133 132 239 234 229 228 255 250 245 244
337 344 347 350 33 40 43 46 129 136 139 142 225 232 235 238 241 248 251 254
348 349 338 343 44 45 34 39 140 141 130 135 236 237 226 231 252 253 242 247
150 147 160 153 166 163 176 169 262 259 272 265 358 355 368 361 54 51 64 57
159 154 149 148 175 170 165 164 271 266 261 260 367 362 357 356 63 58 53 52
145 152 155 158 161 168 171 174 257 264 267 270 353 360 363 366 49 56 59 62
156 157 146 151 172 173 162 167 268 269 258 263 364 365 354 359 60 61 50 55
278 275 288 281 374 371 384 377 70 67 80 73 86 83 96 89 182 179 192 185
287 282 277 276 383 378 373 372 79 74 69 68 95 90 85 84 191 186 181 180
273 280 283 286 369 376 379 382 65 72 75 78 81 88 91 94 177 184 187 190
284 285 274 279 380 381 370 375 76 77 66 71 92 93 82 87 188 189 178 183
order: Q = MN = 20
numbers: 1 .. 400
the main \diagonal sum is:
c0,0 + c1,1 + ... + c19,19 =
6 + 10 + 11 + 7 + 310 + 314 + 315 + 311 + 134 + 138
+ 139 + 135 + 358 + 362 + 363 + 359 + 182 + 186 + 187 + 183 =
(a0,0 + 42b0,0) + (a1,1 + 42b0,0) + ... + (a3,3 + 42b0,0) +
(a0,0 + 42b1,1) + (a1,1 + 42b1,1) + ... + (a3,3 + 42b1,1) +
... ... ...
(a0,0 + 42b4,4) + (a1,1 + 42b4,4) + ... + (a3,3 + 42b4,4) =
5a0,0 + 5a1,1 + ... + 5a3,3 +
4(42b0,0) + 4(42b1,1) + ... + 4(42b4,4) =
5(a0,0 + a1,1 + ... + a3,3) +
43(b0,0 + b1,1 + ... + b4,4) =
5(6 + 10 + 11 + 7) + 43(0 + 19 + 8 + 22 + 11) =
5(34) + 64(60) =
4010 = SQ
and the second \diagonal sum is:
c1,0 + c2,1 + ... + c0,19 =
15 + 8 + 2 + 217 + 319 + 312 + 306 + 41 + 143 + 136
+ 130 + 265 + 367 + 360 + 354 + 89 + 191 + 184 + 178 + 393 =
(a1,0 + 42b0,0) + (a2,1 + 42b0,0) + (a3,2 + 42b0,0) +
(a0,3 + 42b1,0) +
(a1,0 + 42b1,1) + (a2,1 + 42b1,1) + (a3,2 + 42b1,1) +
(a0,3 + 42b2,1) +
... ... ...
(a1,0 + 42b4,4) + (a2,1 + 42b4,4) + (a3,2 + 42b4,4) =
(a0,3 + 42b0,4) =
5a1,0 + 5a2,1 + 5a3,2 +
5a0,3 +
42b1,0 + 42b2,1 + ... + 42b0,4 +
3(42b0,0) + 3(42b1,1) + ... + 3(42b4,4) =
5(a1,0 + a2,1 + a3,2 + a0,3) +
42(b1,0 + b2,1 + b3,2 + b4,3 + b0,4) +
3(42(b0,0 + b1,1 + b2,2 + b3,3 + b4,4) =
5(15 + 8 + 2 + 9) + 42(13 + 2 + 16 + 5 + 24) + 3(42(0 + 19 + 8 + 22 + 11)) =
5(34) + 16(60) + 48(60) =
5(34) + 64(60) =
4010 = SQ
and so on for the other \diagonals and similarly for the /diagonals.
If the order M square and the order N square are both 2-multimagic, the order MN composite square is also 2-multimagic.
Consider squares A, B and the A-B composite C:
(The range of square B is given as 0 to N2-1 to avoid having to subtract
1 from the B numbers at each reference).
A B C
----- ----- ------
order: M N Q = MN
cells: ai,j bi,j ci,j
numbers: 1..M2 0..N2-1 1..Q2
then ci,j = air,jr + M2biq,jq
where i, j are 0 .. Q-1
ir,jr are the remainders of the division of i,j by M
iq,jq are the quotients of the division of i,j by M
let pr be the remainder of the division of Q-1 by M
pq be the quotient of the division of Q-1 by M
SM be the order M magic sum = M(M2+1)/2
SN be the order N magic sum = N(N2-1)/2
SQ be the order Q magic sum = Q(Q2+1)/2
S2M be the order M bimagic sum = M(M2+1)(2M2+1)/6
S2N be the order N bimagic sum sum = N(N2-1)(2N2-1)/6
S2Q be the order Q bimagic sum sum = Q(Q2+1)(2Q2+1)/6
then the first row bimagic sum is:
c0,02 + c0,12 + ... + c0,Q-12 =
(a0,0 + M2b0,0)2 + (a0,1 + M2b0,0)2 + ... + (a0,pr + M2b0,0)2 +
(a0,0 + M2b0,1)2 + (a0,1 + M2b0,1)2 + ... + (a0,pr + M2b0,1)2 +
... ... ...
(a0,0 + M2b0,pq)2 + (a0,1 + M2b0,pq)2 + ... + (a0,pr + M2b0,pq)2 =
a0,02 + 2a0,0(M2b0,0) + (M2b0,0)2 + ...
a0,02 + 2a0,0(M2b0,1) + (M2b0,1)2 + ...
...
a0,02 + 2a0,0(M2b0,pq) + (M2b0,pq)2 + ... =
N(a0,02 + a0,12 + ... + a0,pr2) +
2a0,0M2(b0,0 + b0,1 + ... + b0,pq) +
2a0,1M2(b0,0 + b0,1 + ... + b0,pq) +
...
2a0,prM2(b0,0 + b0,1 + ... + b0,pq) +
M(M2)2(b0,02 + b0,12 + ... + b0,pq2) =
N(S2M) + 2M2(SM)(SN) + M5(S2N) =
NM(M2+1)(2M2+1)/6 + 2M3N((M2+1)/2)((N2-1)/2 + M5N((N2-1)(2N2-1)/6 =
MN(2M4N4+3M2N2+1)/6 = MN(M2N2+1)(2M2N2+1)/6 =
Q(Q2+1)(2Q2+1)/6 = S2Q
and so on for the other row bimagic sums,
and similarly for the column and main diagonal bimagic sums.
Not another attempt at squaring the circle! 
The verse, attributed to Adam C. Orr, is taken from Martin Gardner's The SCIENTIFIC AMERICAN book of Mathematical Puzzles & Diversions, 1959, Simon and Schuster, New York. The Adam C. Orr submission is in The Literary Digest, January 20, 1906, page 84. "The great immortal Syracusan" was Archimedes.
The word lengths give the digits of π to 30 decimal places.
Note: The image is a portion of the square coded in file Text.html by program CompositeCalligraphy. The file was edited to change the title to π r squared.