Order 16 Hexamagic Series Impossibility
Theorem. There are no order 16 hexamagic series.
Proof.
S1 = 2056 ≡ 1 (mod 5)
S2 = 351576 ≡ 1 (mod 5)
S3 = 67634176 ≡ 1 (mod 5)
S4 = 13878462600 ≡ 0 (mod 5)
S5 = 2966502593536
S6 = 652201827924456
From the Modulo 5 Tetramagic Series Lemma,
this hexamagic series must have
5a+1 entries of 0 (mod 5),
5b+2 entries of 1 (mod 5),
5c+1 entries of 2 (mod 5),
5d+1 entries of 3 (mod 5),
5e+1 entries of 4 (mod 5),
where a+b+c+d+e = 2.
Let 5Aj+1 be the square of an entry of 1 or 4 (mod 5), j=1..5(b+e)+3.
Let 5Bj+4 be the square of an entry of 2 or 3 (mod 5), j=1..5(c+d)+2.
Let 25Cj be the square of an entry of 0 (mod 5), j=1..5a+1.
Let Tn = sum (Aj)n, j=1..5(b+e)+3, n=1,2,3.
Let Un = sum (Bj)n, j=1..5(c+d)+2, n=1,2,3.
Let Vn = sum (Cj)n, j=1..5a+1, n=1,2,3.
S2 = 5T1 + 5(b+e)+3 + 5U1 + 20(c+d)+8 + 25V1
S6 = 125T3 + 75T2 + 15T1 + 5(b+e)+3 +
125U3 + 300U2 + 240U1 + 320(c+d)+128 + 15625V3
(S2 - 11)/5 =
T1 + (b+e) + U1 + 4(c+d) + 5V1 = 70313
thus
T1 + (b+e) + U1 + 4(c+d) ≡ 3 (mod 5)
(S6 - 131)/5 =
25T3 + 25T2 + 3T1 + (b+e) +
25U3 + 60U2 + 48U1 + 64(c+d) + 3125V3 = 130440365584865
thus
3T1 + (b+e) + 3U1 + 4(c+d) ≡ 0 (mod 5)
Eliminating T1 and U1 yields
(b+e) + 4(c+d) ≡ 2 (mod 5)
Since b+c+d+e <= 2, we must have
(b+e) = 2, c = d = 0
and since a+b+c+d+e = 2,
a = 0.
Starting over with b+e = 2, a=c=d=0
Let 5A be the entry of 0 (mod 5).
Let 5Bj+1 be an entry of 1 (mod 5), j=1..5b+2.
Let 5C+2 be the entry of 2 (mod 5).
Let 5D+3 be the entry of 3 (mod 5).
Let 5Ej+4 be an entry of 4 (mod 5), j=1..5e+1.
where b+e = 2
Let Tn = sum (Bj)n, j=1..5b+2, n=1..6.
Let Un = sum (Ej)n, j=1..5e+1, n=1..6.
S5 = 3125A5 +
3125T5 + 3125T4 + 1250T3 + 250T2 + 25T1 + 5b+2 +
3125C5 + 6250C4 + 5000C3 + 2000C2 + 400C + 32 +
3125D5 + 9375D4 + 11250D3 + 6750D2 + 2025D + 243 +
3125U5 + 12500U4 + 20000U3 + 16000U2 + 6400U1 + 1024(5e+1)
(S5 - 1301)/5 =
625A5 +
625T5 + 625T4 + 250T3 + 50T2 + 5T1 + b +
625C5 + 1250C4 + 1000C3 + 400C2 + 80C +
625D5 + 1875D4 + 2250D3 + 1350D2 + 405D +
625U5 + 2500U4 + 4000U3 + 3200U2 + 1280U1 + 1024e = 593300518447
thus
b + 4e ≡ 2 (mod 5)
Since b+e = 2, we must have
b=2,e=0.
Starting over with b=2,a=c=d=e=0
Let 5A be the entry of 0 (mod 5).
Let 5Bj+1 be an entry of 1 (mod 5), j=1..12.
Let 5C+2 be the entry of 2 (mod 5).
Let 5D+3 be the entry of 3 (mod 5).
Let 5E+4 be the entry of 4 (mod 5).
Let Tn = sum (Bj)n, j=1..12, n=1..6.
S1 = 5A +
5T1 + 12 +
5C + 2 +
5D + 3 +
5E + 4
S2 = 25A2 +
25T2 + 10T1 + 12 +
25C2 + 20C + 4 +
25D2 + 30D + 9 +
25E2 + 40E + 16
S3 = 125A3 +
125T3 + 75T2 + 15T1 + 12 +
125C3 + 150C2 + 60C + 8 +
125D3 + 225D2 + 135D + 27 +
125E3 + 300E2 + 240E + 64
S4 = 625A4 +
625T4 + 500T3 + 150T2 + 20T1 + 12 +
625C4 + 1000C3 + 600C2 + 160C + 16 +
625D4 + 1500D3 + 1350D2 + 540D + 81 +
625E4 + 2000E3 + 2400E2 + 1280E + 256
S5 = 3125A5 +
3125T5 + 3125T4 + 1250T3 + 250T2 + 25T1 + 12 +
3125C5 + 6250C4 + 5000C3 + 2000C2 + 400C + 32 +
3125D5 + 9375D4 + 11250D3 + 6750D2 + 2025D + 243 +
3125E5 + 12500E4 + 20000E3 + 16000E2 + 6400E + 1024
S6 = 15625A6 +
15625T6 + 18750T5 + 9375T4 + 2500T3 + 375T2 + 30T1 + 12 +
15625C6 + 37500C5 + 37500C4 + 20000C3 + 6000C2 + 960C + 64 +
15625D6 + 56250D5 + 84375D4 + 67500D3 + 30375D2 + 7290D + 729 +
15625E6 + 75000E5 + 150000E4 + 160000E3 + 96000E2 + 30720E + 4096
(S1 - 21)/5 =
A + T1 + C + D + E = 407
(S2 - 41)/5 =
5(A2 + T2 + C2 + D2 + E2) +
2T1 + 4C + 6D + 8E = 70307
(S3 - 111)/5 =
25(A3 + T3 + C3 + D3 + E3) +
15(T2 + 2C2 + 3D2 + 4E2) +
3T1 + 12C + 27D + 48E = 13526813
(S4 - 365)/5 =
125(A4 + T4 + C4 + D4 + E4) +
100(T3 + 2C3 + 3D3 + 4E3) +
30(T2 + 4C2 + 9D2 + 16E2) +
4T1 + 32C + 108D + 256E = 2775692447
(S5 - 1311)/25 =
125(A5 + T5 + C5 + D5 + E5) +
125(T4 + 2C4 + 3D4 + 4E4) +
50(T3 + 4C3 + 9D3 + 16E3) +
25(T2 + 8C2 + 27D2 + 64E2) +
T1 + 16C + 81D + 256E = 118660103689
(S6 - 4901)/5 =
3125(A6 + T6 + C6 + D6 + E6) +
3750(T5 + 2C5 + 3D5 + 4E5) +
1875(T4 + 4C4 + 9D4 + 16E4) +
500(T3 + 8C3 + 27D3 + 64E3) +
75(T2 + 16C2 + 81D2 + 256E2) +
6T1 + 192C + 1458D + 6144E = 130440365583911
This produces the (mod 5) system
A + T1 + C + D + E ≡ 2 (mod 5)
2T1 + 4C + D + 3E ≡ 2 (mod 5)
3T1 + 2C + 2D + 3E ≡ 3 (mod 5)
4T1 + 2C + 3D + E ≡ 2 (mod 5)
T1 + C + D + E ≡ 4 (mod 5)
T1 + 2C + 3D + 4E ≡ 1 (mod 5)
Its unique solution is
A ≡ 3 (mod 5)
C ≡ 3 (mod 5)
D ≡ 1 (mod 5)
E ≡ 4 (mod 5)
T1 ≡ 1 (mod 5)
Let A = 5u+3
Let C = 5y+3
Let D = 5w+1
Let E = 5v+4
Let T1 = 5z+1
u + z + y + w + v = 79
25( u2 + y2 + w2 + v2 ) +
T2 + 2z + 34y + 16w + 48v + 30u = 14016
5( (5u+3)3 + T3 + (5y+3)3 + (5w+1)3 + (5v+4)3 ) +
25( 6y2 + 9w2 + 12v2 ) +
3T2 + 3z + 192y + 117w + 528v = 2705056
25( (5u+3)4 + T4 + (5y+3)4 + (5w+1)4 + (5v+4)4 ) +
20( T3 + 2(5y+3)3 + 3(5w+1)3 + 4(5v+4)3 ) +
25( 24y2 + 54w2 + 96v2 ) +
6T2 + 4z + 752y + 648w + 4096v = 555136437
25( (5u+3)5 + T5 + (5y+3)5 + (5w+1)5 + (5v+4)5 ) +
25( T4 + 2(5y+3)4 + 3(5w+1)4 + 4(5v+4)4 ) +
10( T3 + 4(5y+3)3 + 9(5w+1)3 + 16(5v+4)3 ) +
5( T2 + 8(5y+3)2 + 27(5w+1)2 + 64(5v+4)2 ) +
z + 16y + 81w + 256v = 23732020507
625( (5u+3)6 + T6 + (5y+3)6 + (5w+1)6 + (5v+4)6 ) +
750( T5 + 2(5y+3)5 + 3(5w+1)5 + 4(5v+4)5 ) +
375( T4 + 4(5y+3)4 + 9(5w+1)4 + 16(5v+4)4 ) +
100( T3 + 8(5y+3)3 + 27(5w+1)3 + 64(5v+4)3 ) +
15( T2 + 16(5y+3)2 + 81(5w+1)2 + 256(5v+4)2 ) +
6z + 192y + 1458w + 6144v = 26088073111459
This produces the (mod 5) system
u + z + y + w + v ≡ 4 (mod 5)
T2 + 2z + 4y + w + 3v ≡ 1 (mod 5)
3T2 + 3z + 2y + 2w + 3v ≡ 1 (mod 5)
T2 + 4z + 2y + 3w + v ≡ 2 (mod 5)
z + y + w + v ≡ 2 (mod 5)
z + 2y + 3w + 4v ≡ 4 (mod 5)
Its unique solution is
z ≡ 4 (mod 5)
y ≡ 3 (mod 5)
w ≡ 1 (mod 5)
v ≡ 4 (mod 5)
u ≡ 2 (mod 5)
T2 ≡ 3 (mod 5)
Let z = 5t+4
Let y = 5s+3 --> C = 25s+18 --> 5C+2 = 125s+92 = { 92, 217 }
Let w = 5r+1 --> D = 25r+6 --> 5D+3 = 125r+33 = { 33, 158 }
Let v = 5q+4 --> E = 25q+24 --> 5E+4 = 125q+124 = { 124, 249 }
Let u = 5p+2 --> A = 25p+13 --> 5A = 125p+65 = { 65, 190 }
Let T2 = 5n+3
+---------------------------------------------+
| Because an order 16 normal hexamagic series |
| must use entries in the range 1 ... 256, |
| the values of p,q,r,s must each be 0 or 1. |
+---------------------------------------------+
p + t + s + r + q = 13
5( (5p+2)2 + (5s+3)2 + (5r+1)2 + (5q+4)2 ) +
n + 2t + 34s + 16r + 48q + 30p = 2727
25( 625p3 + 975p2 + 507p ) +
25( 625s3 + 1350s2 + 972s ) +
25( 625r3 + 450r2 + 108r ) +
25( 625q3 + 1800q2 + 1728q ) +
5( 6(5s+3)2 + 9(5r+1)2 + 12(5q+4)2 ) +
T3 + 3n + 3t + 192s + 117r + 528q = 518377
5( (25p+13)4 + T4 + (25s+18)4 + (25r+6)4 + (25q+24)4 ) +
25( 5000s3 + 10800s2 + 7776s ) +
25( 7500r3 + 5400r2 + 1296r ) +
25( 10000q3 + 28800q2 + 27648q ) +
5( 24(5s+3)2 + 54(5r+1)2 + 96(5q+4)2 ) +
4T3 + 6n + 4t + 752s + 648r + 4096q = 110752991
5( (25p+13)5 + T5 + (25s+18)5 + (25r+6)5 + (25q+24)5 ) +
5( T4 + 2(25s+18)4 + 3(25r+6)4 + 4(25q+24)4 ) +
25( 5000s3 + 10800s2 + 7776s ) +
25( 11250r3 + 8100r2 + 1944r ) +
25( 20000q3 + 57600q2 + 55296q ) +
25( 200s2 + 288s + 675r2 + 324r + 1600q2 + 3072q ) +
2T3 + 5n + t + 16s + 81r + 256q = 4745870527
125( (25p+13)6 + T6 + (25s+18)6 + (25r+6)6 + (25q+24)6 ) +
150( T5 + 2(25s+18)5 + 3(25r+6)5 + 4(25q+24)5 ) +
75( T4 + 4(25s+18)4 + 9(25r+6)4 + 16(25q+24)4 ) +
20( T3 + 8(25s+18)3 + 27(25r+6)3 + 64(25q+24)3 ) +
25( 1200s2 + 1728s + 6075r2 + 2916r + 19200q2 + 36864q ) +
15n + 6t + 192s + 1458r + 6144q = 5217614150288
This produces the (mod 5) system
p + t + s + r + q ≡ 3 (mod 5)
n + 2t + 4s + r + 3q ≡ 2 (mod 5)
T3 + 3n + 3t + 2s + 2r + 3q ≡ 2 (mod 5)
4T3 + n + 4t + 2s + 3r + q ≡ 1 (mod 5)
2T3 + t + s + r + q ≡ 2 (mod 5)
t + 2s + 3r + 4q ≡ 3 (mod 5)
Its unique solution in terms of r is
q ≡ 3 + r (mod 5)
s ≡ 2 + 4r (mod 5)
t ≡ 2 (mod 5)
p ≡ 1 + 4r (mod 5)
n ≡ 1 (mod 5)
T3 ≡ 2r (mod 5)
+------------------------------------+
| r and q must each be 0 or 1, |
| but if r = 0, then q is at least 3 |
| and if r = 1, then q is at least 4.|
+------------------------------------+
An order 16 hexamagic series must use entries greater than 256,
therefore there is no order 16 normal hexamagic series.