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

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

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

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

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

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

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

 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

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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.

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