Impossibility of normal tetramagic squares 12x12 ... 23x23
by Lee Morgenstern, February-March 2013.
(16x16 proof corrected, August 2013)

Proofs are short and simple and show that either there is no magic series or that it is impossible to make disjoint rows. No reference to columns is needed.

Magic sum formulas for NxN tetramagic squares

S1 = N(N^2 + 1) / 2
S2 = N(N^2 + 1)(2n^2 + 1) / 6
S3 = (N^3)(N^2 + 1)^2 / 4
S4 = N(N^2 + 1)(2N^2 + 1)(3N^4 + 3N^2 - 1) / 30

Modulo 5 Tetramagic Lemma

   x   x^2 x^3 x^4  (mod 5)
-----------------
A: 1   1   1   1
B: 2   4   3   1
C: 3   4   2   1
D: 4   1   4   1

A, B, C, D are the number of 1's, 2's, 3's, 4's (mod 5) respectively, in a tetramagic series.

A + 2B + 3C + 4D = S1 (mod 5)
A + 4B + 4C +  D = S2 (mod 5)
A + 3B + 2C + 4D = S3 (mod 5)
A +  B +  C +  D = S4 (mod 5)

The unique solution is

A = 4(S3 + S4) - (S1 + S2) (mod 5)
B = 3(S3 - S1) + (S2 - S4) (mod 5)
C = 2(S3 - S1) + (S2 - S4) (mod 5)
D = 4(S2 - S3) + (S1 - S4) (mod 5)

12x12 Impossibility

S1 = 870        = 0 (mod 5)
S2 = 83810      = 0 (mod 5)
S3 = 9082800    = 0 (mod 5)
S4 = 1049954918 = 3 (mod 5)

Using Modulo 5 Tetramagic Lemma yields

A = B = C = D = 2 (mod 5).

Thus there is 1 tetramagic series (mod 5), { 0,0,0,0,1,1,2,2,3,3,4,4 }.

12 rows of these use 48 0's, but there are only 28 0's in the range 1 ... 12^2.

13x13 Impossibility

S1 = 1105
S2 = 124865
S3 = 15873325
S4 = 2152397897 = 9 (mod 16)

There must be 9 odd entries in every magic series. 13 rows use 117 odd entries, but there are only 85 odd entries in the range 1 .. 13^2.

14x14 Impossibility

S1 = 1379
S2 = 180649
S3 = 26622974
S4 = 4185095383

S3 is even, thus a magic series requires an even number of odd entries. S1, S2, S4 are odd, thus a magic series requires an odd number of odd entries.

Both can't be true, therefore there are no magic series.

15x15 Impossibility

S1 = 1695
S2 = 254815
S3 = 43095375
S4 = 7774354687 = 15 (mod 16)

All 15 entries in all rows must be odd, therefore no even numbers can be used.

16x16 Impossibility

S1 = 2056        = 4 (mod 9)
S2 = 351576      = 0 (mod 9)
S3 = 67634176    = 4 (mod 9)
S4 = 13878462600 = 0 (mod 9)

The modulo 3/9 Tetramagic Lemma yields

(D+E+F) = 7 (mod 9)
(A+B+C) = 2 (mod 9)

Since (A+B+C) >= 2, (D+E+F) < 14, thus (D+E+F) = 7.

16 rows use 112 2 (mod 3) entries, but there are only 85 2 (mod 3) entries in the range 1 ... 16^2.

17x17 Impossibility

S1 = 2465        = 0 (mod 5)
S2 = 475745      = 0 (mod 5)
S3 = 103295825   = 0 (mod 5)
S4 = 23923217921 = 1 (mod 5)

Using Modulo 5 Tetramagic Lemma yields

A = B = C = D = 4 (mod 5).

Thus there is 1 tetramagic series (mod 5), { 0,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4 }.

17 rows of these use only 17 0's, but there are 57 0's in the range 1 ... 17^2 that need to be used.

18x18 Impossibility

S1 = 2925
S2 = 632775
S3 = 154001250
S4 = 39978597945

S3 is even, thus a magic series requires an even number of odd entries. S1, S2, S4 are odd, thus a magic series requires an odd number of odd entries.

Both can't be true, therefore there are no magic series.

19x19 Impossibility

S1 = 3439
S2 = 828799
S3 = 224707699
S4 = 64985300791 = 7 (mod 16)

There must be 7 odd entries in a magic series. 19 rows use 133 odd entries, but there are 181 odd entries in the range 1 .. 19^2 that need to be used.

20x20 Impossibility

S1 = 4010         = 5 (mod 9)
S2 = 1070670      = 3 (mod 9)
S3 = 321602000    = 5 (mod 9)
S4 = 103041066666 = 3 (mod 9)

Let A, D, B, E, C, F be the number of 1, 2, 4, 5, 7, 8 (mod 9) entries, respectively, in a tetramagic series and thus, (A+B+C) is the number of 1 (mod 3) entries and (D+E+F) is the number of 2 (mod 3) entries.

   x  x^2 x^3 x^4  (mod 9)
----------------
A: 1   1   1   1
D: 2   4   8   7
   3   0   0   0
B: 4   7   1   4
E: 5   7   8   4
   6   0   0   0
C: 7   4   1   7
F: 8   1   8   1

[1] A + 7B + 4C + 4D + 7E +  F = S2 = 3 (mod 9)
[2] A +  B +  C + 8D + 8E + 8F = S3 = 5 (mod 9)
[3] A + 4B + 7C + 7D + 4E +  F = S4 = 3 (mod 9)

Subtract [2] from [1].

[4]     6B + 3C + 5D + 8E + 2F = 7 (mod 9)

Subtract [2] from [3].

[5]     3B + 6C + 8D + 5E + 2F = 7 (mod 9)

Multiply [5] by 2.

[6]     6B + 3C + 7D +  E + 4F = 5 (mod 9)

Subtract [4] from [6].

[7] 2D + 2E + 2F = 7 (mod 9)

Multiply [7] by 5.

[8] (D+E+F) = 8 (mod 9)

Substitute [8] into [2].

[9] (A+B+C) = 4 (mod 9)

Since (A+B+C) + (D+E+F) <= 20, we have

from [9], (A+B+C) = 4 or 13, and
from [8], (D+E+F) = 8 or 17,

but

13 + 17 > 20,
13 + 8 > 20,
4 + 17 > 20,

therefore a tetramagic series must contain

8 2 (mod 3) entries,
4 1 (mod 3) entries, and
8 0 (mod 3) entries.

20 rows of these use 160 0 (mod 3) entries but there are only 134 0 (mod 3) entries in the range 1 ... 20^2.

21x21 Impossibility

S1 = 4641
S2 = 1366001
S3 = 452316501
S4 = 159757914953 = 9 (mod 16)

There must be 9 odd entries in a magic series. 21 rows use 189 odd entries, but there are 221 odd entries in the range 1 .. 21^2 that need to be used.

22x22 Impossibility

S1 = 5335
S2 = 1723205
S3 = 626168950
S4 = 242702740379

S3 is even, thus a magic series requires an even number of odd entries. S1, S2, S4 are odd, thus a magic series requires an odd number of odd entries. Both can't be true, therefore there are no magic series.

23x23 Impossibility

S1 = 6095
S2 = 2151535
S3 = 854427575
S4 = 361935090463 = 15 (mod 16)

There must be 15 odd entries in a magic series. 23 rows use 345 odd entries, but there are only 265 odd entries in the range 1 .. 23^2.

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