**Impossibility of normal tetramagic squares 12x12 ... 23x23**

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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