Impossibility proofs of multimagic series for cubes
by Lee Morgenstern, April-May 2015.


In this page, proofs:


Order 11, no tetramagic series,
S1 = 7326, S2 = 6503046, S3 = 6494103396, S4 = 6917517636810

S4 = 10 (mod 16), thus a tetramagic series must consist of 10 odd entries and 1 even entry.
Let 2Aj+1 be an odd entry, j=1..10.
Let 2B be the even entry.
Let Tn = sum (Aj)^n, j=1..10, n=1..4.

S1 = 2T1 + 10 + 2B
S2 = 4T2 + 4T1 + 10 + 4B^2
S3 = 8T3 + 12T2 + 6T1 + 10 + 8B^3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 10 + 16B^4
(S3 - 10)/2 =  4T3 + 6T2 + 3T1 + 4B^3 = 3247051693

thus  T1 is odd and T2,T3,T4 are odd.

(S1 - 10)/2 =  T1 + B = 3658

thus  B is odd.

- - - - - - - - - -

Let  B = 2z+1
Let T1 = 2y+1
Let T2 = 2x+1
Let T3 = 2w+1
Let T4 = 2v+1

((S3-10)/2 - 17)/2 = 4w + 6x + 3y + 2(8z^3 + 12z^2 + 6z) = 1623525838
thus  y is even.

((S1-10)/2 - 2)/2 = y + z = 1828
thus  z is even.

((S2-10)/4 - 3)/2 = x + y + 2(z^2 + z) = 812878
thus  x is even.

- - - - - - - - - -

Divide everything by 2.

((S3-10)/2 - 17)/4 = 2w + 3x + 3y/2 + (8z^3 + 12z^2 + 6z) = 811762919
thus  y/2 is odd.

((S1-10)/2 - 2)/4 = y/2 + z/2 = 914
thus  z/2 is odd.

((S2-10)/4 - 3)/4 = x/2 + y/2 + (z^2 + z) = 406439
thus  x/2 is even.

((S4-10)/8 - 12)/4 = v + 2w + 3x/2 + y/2 + (8z^4 + 16z^3 + 12z^2 + 4z^3) = 216172426147
thus  v is even.

- - - - - - - - - -

Let y/2 = 2t+1 so that T1 = 8t+5

(((S3-10)/2-17)/4 - 3)/2 = w + 3x/2 + 3t + (4z^3 + 6z^2 + 3z) = 405881458

thus  w + t is even.

(((S2-10)/4-3)/4 - 1)/2 = x/4 + t + z^2/2 + z/2 = 203219

thus  x/4 + t is even.

(((S4-10)/8-12)/4 - 1)/2 = v/2 + w + 3x/4 + t + (4z^4 + 8z^3 + 6z^2 + 2z^3) = 108086213073

thus  v/2 + t is odd.

- - - - - - - - - -

We now have

t = w = x/4 (mod 2),
t /= v/2 (mod 2), and
T1 = 1 (mod 4).

Case 1 of 2, t is even.

  x/4 is even, thus T2 = 1 (mod 8).
  w is even,   thus T3 = 1 (mod 4).
  v/2 is odd,  thus T4 = 5 (mod 8).

The sum of even entries = T1 - T3 = 0 (mod 4) implies an even number of singly even entries;
the sum of even squares = T2 - T4 = 4 (mod 8) implies an odd number of single even entries;
therefore T1 and T3 are inconsistent with T2 and T4.

Case 2 of 2, t is odd.

  x/4 is odd,  thus T2 = 1 (mod 8).
  w is odd,    thus T3 = 3 (mod 4).
  v/2 is even, thus T4 = 1 (mod 8).

The sum of even entries = T1 - T3 = 2 (mod 4) implies an odd number of singly even entries;
the sum of even squares = T2 - T4 = 0 (mod 8) implies an even number of single even entries;
therefore T1 and T3 are inconsistent with T2 and T4.


Order 12, no tetramagic series,
S1 = 10374, S2 = 11954306, S3 = 15497262144, S4 = 21429611701862

S1 = 4 (mod 5), S2 = 1 (mod 5), S3 = 4 (mod 5), S4 = 22 (mod 5)

From the Modulo 5 Tetramagic Series Lemma (in Zip file), an order 12 tetramagic series must contain

  0 entries of 0 (mod 5),
  4 entries of 1 (mod 5),
  4 entries of 2 (mod 5),
  4 entries of 3 (mod 5), and
  0 entries of 4 (mod 5).

Let 5Aj+1 be an entry of 1 (mod 5), j=1..4.
Let 5Bj+2 be an entry of 2 (mod 5), j=1..4.
Let 5Cj+3 be an entry of 3 (mod 5), j=1..4.
Let Tn = sum (Aj)^n, j=1..4, n=1..4.
Let Un = sum (Bj)^n, j=1..4, n=1..4.
Let Vn = sum (Cj)^n, j=1..4, n=1..4.

S1 = 5T1 + 4 +
     5U1 + 8 +
     5V1 + 12

S2 = 25T2 + 10T1 + 4 +
     25U2 + 20U1 + 16 +
     25V2 + 30V1 + 36

S3 = 125T3 + 75T2 + 15T1 + 4 +
     125U3 + 150U2 + 60U1 + 32 +
     125V3 + 225V2 + 135V1 + 108

S4 = 625T4 + 500T3 + 150T2 + 20T1 + 4 +
     625U4 + 1000U3 + 600U2 + 160U1 + 64 +
     625V4 + 1500V3 + 1350V2 + 540V1 + 324

(S1 - 24)/5 =
  T1 + U1 + V1 = 2070

(S2 - 56)/5 =
  5T2 + 2T1 + 5U2 + 4U1 + 5V2 + 6V1 = 2390850

(S3 - 144)/5 =
  25T3 + 15T2 + 3T1 +
  25U3 + 30U2 + 12U1 +
  25V3 + 45V2 + 27V1 = 3099452400

(S4 - 392)/5 =
  125T4 + 100T3 + 30T2 + 4T1 +
  125U4 + 200U3 + 120U2 + 32U1 +
  125V4 + 300V3 + 270V2 + 108V1 = 4285922340294

This produces the system

[1]  T1 +  U1 +  V1 = 0 (mod 5)
[2] 2T1 + 4U1 +  V1 = 0 (mod 5)
[3] 3T1 + 2U1 + 2V1 = 0 (mod 5)
[4] 4T1 + 2U1 + 3V1 = 4 (mod 5)

Subtract [1] from [2].
[2a] T1 + 3U1 = 0 (mod 5)

Multiply [1] by 2.
[1a] 2T1 + 2U1 + 2V1 = 0 (mod 5)

Subtract [1a] from [3].
[3a] T1 = 0 (mod 5)

Substitute [3a] into [2a].
[2b] U1 = 0 (mod 5).

Substitute [3a] and [2b] into [1].
[1b] V1 = 0 (mod 5).

[4] is inconsistent with T1 = U1 = V1 = 0 (mod 5).


Order 13, no pentamagic series,
S1 = 14287, S2 = 20930455, S3 = 34496004361, S4 = 60643971480547, S5 = 111054263191528657

S4 = 3 (mod 16), so there must be 3 odd entries and 10 even entries.

Let 2Aj+1 be an odd entry, j=1..3
Let 2Bj be an even entry, j=1..10
Let Tn = sum (Aj)^n, j=1..3, n=1..4
Let Un = sum (Bj)^n, j=1..10, n=1..4

S1 = 2T1 + 3 + 2U1
S2 = 4T2 + 4T1 + 3 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 3 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 3 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 3 + 32U5

(S3 - 3)/2 =  4T3 + 6T2 + 3T1 + 4U3 = 17248002179
thus  T1 is odd and T2,T3,T4,T5 are odd.

(S1 - 3)/2 =  T1 + U1 = 7142
thus  U1 is odd and U2,U3,U4,U5 are odd.

(S2 - 3)/4 =  T2 + T1 + U2 = 5232613

(S4 - 3)/8 =  2T4 + 4T3 + 3T2 + T1 + 2U4 = 7580496435068

(S5 - 3)/2 =  16T5 + 40T4 + 40T3 + 20T2 + 5T1 + 16U5 = 55527131595764327

-------------

Let T1 = 2z+1
Let U1 = 2y+1
Let T2 = 2x+1
Let U2 = 2w+1

((S3-3)/2 - 9)/2 =  2T3 + 6x + 3z + 2U3 = 8624001085
thus  z is odd.

((S1-3)/2 - 2)/2 =  z + y = 3570
thus  y is odd.

((S4-3)/8 - 4)/2 =  T4 + 2T3 + 3x + z + U4 = 3790248217532
thus  x is odd which also implies T2 = T4 = 3 (mod 4).

((S2-3)/4 - 3)/2 =  x + z + w = 2616305
thus  w is odd,

((S5-3)/2 - 25)/2 =  8T5 + 20T4 + 20T3 + 20x + 5z + 8U5 = 27763565797882151

------------

Let z = 2v+1
Let y = 2u+1
Let x = 2t+1
Let w = 2s+1
Let T4 = 4r+3

(((S3-3)/2-9)/2 - 9)/2 =  T3 + 6t + 3v + U3 = 4312000538
thus  v is even.

(((S5-3)/2-25)/2 - 85)/2 =  4T5 + 40r + 10T3 + 20t + 5v + 4U5 = 13881782898941033
thus  v is odd.

v can't be both even and odd.


Order 15, no pentamagic series,
S1 = 25320, S2 = 56978440, S3 = 144248040000, S5 = 1095708063757320000

S4 = 8 (mod 16), so there must be 8 odd entries and 7 even entries.

Let 2Aj+1 be an odd entry, j=1..8
Let 2Bj be an even entry, j=1..7
Let Tn = sum (Aj)^n, j=1..8, n=1..4
Let Un = sum (Bj)^n, j=1..7, n=1..4

S1 = 2T1 + 8 + 2U1
S2 = 4T2 + 4T1 + 8 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 8 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 8 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 8 + 32U5

(S3 - 8)/2 =  4T3 + 6T2 + 3T1 + 4U3 = 72124019996
thus  T1 is even and T2,T3,T4,T5 are even.

(S1 - 8)/2 =  T1 + U1 = 12656
thus  U1 is even and U2,U3,U4,U5 are even.

(S2 - 8)/4 =  T2 + T1 + U2 = 14244608

(S4 - 8)/8 =  2T4 + 4T3 + 3T2 + T1 + 2U4 = 48690924477538

(S5 - 8)/2 =  16T5 + 40T4 + 40T3 + 20T2 + 5T1 + 16U5 = 547854031878659996

-------------------------

Dividing everything by 2,

(S3 - 8)/4 =  2T3 + 3T2 + 3T1/2 + 2U3 = 36062009998
thus  T1/2 is even.

(S1 - 8)/4 =  T1/2 + U1/2 = 6328
thus  U1/2 is even.

(S4 - 8)/16 =  T4 + 2T3 + 3T2/2 + T1/2 + U4 = 24345462238769
thus  T2/2 is odd.

(S2 - 8)/8 =  T2/2 + T1/2 + U2/2 = 7122304
thus  U2/2 is odd.

(S5 - 8)/4 =  8T5 + 20T4 + 20T3 + 10T2 + 5T1/2 + 8U5 = 273927015939329998

---------------

Let T2/2 = 2z+1
Let U2/2 = 2y+1

((S3-8)/4 - 6)/2 =  T3 + 6z + 3T1/4 + U3 = 18031004996
thus  T1/4 is even.

((S5-8)/4 - 20)/2 =  4T5 + 10T4 + 10T3 + 20z + 5T1/4 + 4U5 = 13696350796966498
thus  T1/4 is odd.

T1/4 can't be both even and odd.


Order 16, no hexamagic series,
S1 = 32776, S2 = 89511256, S3 = 275012141056, S4 = 901269770766472, S5 = 3076709603561783296, S6 = 10803206156110098495976

S2 = 1 (mod 9), S3 = 7 (mod 9), S4 = 1 (mod 9)

Hexamagic series are also tetramagic series and from the Modulo 9/3 Tetramagic Series Lemma (in Zip file), an order 16 tetramagic series must consist of
  6 entries of 0 (mod 3),
  4 entries of 1 (mod 3), and
  6 entries of 2 (mod 3).

Let 3Aj+2 be an entry of 2 (mod 3),  j=1..6.
Let 3Bj+1 be an entry of 1 (mod 3),  j=1..4.
Let 3Cj   be an entry of 0 (mod 3),  j=1..6.

Let Tn = sum (Aj)^n, j=1..6, n=1..6.
Let Un = sum (Bj)^n, j=1..4, n=1..6.
Let Vn = sum (Cj)^n, j=1..6, n=1..6.

S2 = 9T2 + 12T1 + 24 +
     9U2 + 6U1 + 4 +
     9V2.

S6 = 729T6 + 6x243x2T5 + 15x81x4T4 + 20x27x8T3 + 15x9x16T2 + 6x3x32T1 + 384 +
     729U6 + 6x243U5 + 15x81U4 + 20x27U3 + 15x9U2 + 6x3U1 + 4 +
     729V6.

(S2 - 28)/3 = 3T2 + 4T1 + 3U2 + 2U1 + 3V2 = 29837076 = 0 (mod 3)
thus  T1 + 2U1 = 0 (mod 3).

(S6 - 388)/9 =
  81T6 + 6x27x2T5 + 15x9x4T4 + 20x3x8T3 + 15x16T2 + 64T1 +
  81U6 + 6x27U5 + 15x9U4 + 20x3U3 + 15U2 + 2U1 +
  81V6 = 1200356239567788721732 = 1 (mod 3)
thus  T1 + 2U1 = 1 (mod 3).

T1 + 2U1 can't be both 0 and 1 (mod 3),
therefore S6 is inconsistent with S2.


Order 17, no tetramagic series,
S1 = 41769, S2 = 136821321, S3 = 504203665329, S4 = 1981923740310969

From the Modulo 9/3 Tetramagic Series Lemma (in Zip file), an order 17 tetramagic series must contain
  9a+8 entries of 0 (mod 3)
  9b   entries of 1 (mod 3)
  9c   entries of 2 (mod 3)
where
  a+b+c = 1.

-------

case of all 17 entries being 0 (mod 3)

S4 would have to be divisible by 3^4, but it isn't.

-------

case of 8 entries of 0 (mod 3) and 9 entries of 1 (mod 3)

Let 3Aj+1 be an entry of 1 (mod 3), j=1..9
Let 3Bj   be an entry of 0 (mod 3), j=1..8

Let Tn = sum (Aj)^n, j=1..9, n=1..4
Let Un = sum (Bj)^n, j=1..8, n=1..4

S2 = 9T2 + 12T1 + 9 + 9U2
S3 = 27T3 + 3x9T2 + 3x3T1 + 9 + 27U3

(S2 - 9)/3 =  3T2 + 4T1 + 3U2 = 45607104 = 0 (mod 3)
thus  T1 = 0 (mod 3).

(S3 - 9)/9 =  3T3 + 3T2 + T1 + 3U3 = 56022629480 = 2 (mod 3)
thus  T1 = 2 (mod 3).

T1 can't be both 0 and 2 (mod 3),

-----

case of 8 entries of 0 (mod 3) and 9 entries of 2 (mod 3).

This is the complement of the previous case, which means they are either both possible or both impossible.

Since the previous case is impossible they both are impossible.


Return to the home page http://www.multimagie.com