Order 7 Trimagic Series Impossibility
Theorem. There are no order 7 normal trimagic series.
Proof.
Part 1 - Modulo 5 Pattern
Part 2 - Modulo 3 Pattern
Part 3 - Modulo 4 Pattern
Part 4 - Multiple Modulus Proof
Part 1 of 4 - Modulo 5 Pattern
S1 = 175
S2 = 5775
S3 = 214375
Let A,B,C,D be the number of entries
of 1,2,3,4 (mod 5), respectively, in a trimagic series.
[1] A + 2B + 3C + 4D ≡ 0 (mod 5)
[2] A + 4B + 4C + D ≡ 0 (mod 5)
[3] A + 3B + 2C + 4D ≡ 0 (mod 5)
From [2]
A+D ≡ B+C (mod 5)
From [3] minus [1]
B ≡ C (mod 5)
and thus
A ≡ D (mod 5)
Let Z be the number of entries of 0 (mod 5).
Since Z+A+B+C+D = 7, we have the following possibilities.
Z A+D B+C
- --- ---
3 2 2
7 0 0
Conclusion used in part 4:
Z ≥ 3
Part 2 of 4 - Modulo 3 Pattern
S1 = 175 ≡ 1 (mod 3)
S2 = 5775 ≡ 0 (mod 3)
S3 = 214375 ≡ 4 (mod 9) ≡ 1 (mod 3)
Let A,B be the number of entries
of 1,2 (mod 3), respectively, in a trimagic series.
[1] A + 2B ≡ 1 (mod 3)
[2] A + B ≡ 0 (mod 3)
From [1] minus [2],
A ≡ 2 (mod 3)
and thus
B ≡ 1 (mod 3)
Let Z be the number of entries of 0 (mod 3).
Since Z+A+B = 7, we have
A = 2 or 5 and
B = 1 or 4,
thus we have the following possibilities,
each checked for S3 ≡ 4 (mod 9).
Z A B
- - -
4 2 1 but 2x13 + 1x23 ≡ 1 (mod 9)
1 2 4 but 2x13 + 4x23 ≡ 7 (mod 9)
1 5 1 and 5x13 + 1x23 ≡ 4 (mod 9)
Conclusion used in part 4:
Z=1, A=5, B=1
Part 3 of 4 - Modulo 4 Pattern
S1 = 175
S2 = 5775
S3 = 214375
Since the sums are odd, there are an odd number
of odd entries in a trimagic series.
Let 2Aj+1 be an odd entry, j=1..2a+1
Let 2Bj be an even entry, j=1..7-(2a+1)
where a = {0,1,2,3}
Let Tn = sum (Aj)n, j=1..2a+1, n=1,2,3
Let Un = sum (Bj)n, j=1..7-(2a+1)
S1 = 2T1 + 2a+1 + 2U1
S2 = 4T2 + 4T1 + 2a+1 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 2a+1 + 8U3
(S2 - 1)/2 =
2T2 + 2T1 + a + 2U2 = 2887
thus
a is odd so a = {1,3}
(S3 - 1)/2 =
4T3 + 6T2 + 3T1 + a + 4U3 = 107187
thus
T1 is even and T2,T3 are even
(S1 - 1)/2 =
T1 + a + U1 = 87
thus
U1 is even and U2,U3 are even
---
Let a = 2z+1, z = {0,1}
T2 + T1 + z + U2 = 1443
thus
z is odd
and
a = 3
Therefore,
there are 7 odd entries and 0 even entries
in a trimagic series.
---
Starting over ...
Let 2Aj+1 be an odd entry, j=1..7
Let Tn = sum (Aj)n, j=1..7, n=1..3
S1 = 2T1 + 7 = 175
S2 = 4T2 + 4T1 + 7 = 5775
S3 = 8T3 + 12T2 + 6T1 + 7 = 214375
T1 = (175 - 7)/2 = 84
T2 = (5775 - 7 - 4T1)/4 =
(5768 - 4x84)/4 = 1358
T3 = (214375 - 7 - 6T1 - 12T2)/8 =
(214368 - 6x84 - 12x1358)/8 = 24696
The Aj contain an even number of odd entries.
Let 2Bj+1 be an odd entry, j=1..2a
Let 2Cj be an even entry, j=1..7-2a
where a = {0,1,2,3}
Let Vn = sum (Bj)n, j=1..2a, n=1,2,3
Let Wn = sum (Cj)n, j=1..7-2a, n=1,2,3
T1 = 2V1 + 2a + 2W1
T2 = 4V2 + 4V1 + 2a + 4W2
T3 = 8V3 + 12V2 + 6V1 + 2a + 8W3
2V2 + 2V1 + a + 2W2 = 679
thus
a is odd so a = {1,3}
4V3 + 6V2 + 3V1 + a + 4W3 = 12348
thus
V1 is odd and V2,V3 are odd
V1 + a + W1 = 42
thus
W1 is even and W2,W3 are even
---
Let a = 2z+1, z = {0,1}
Let V1 = 2y+1
V2 + 2y + z + W2 = 338
thus
z is odd
and
a = 3
Aj contains 6 odd entries and 1 even entry.
Conclusion used in part 4:
The trimagic series contains
6 entries of 3 (mod 4) and 1 entry of 1 (mod 4).
Part 4 of 4 - Multiple Modulus - Range Checking Proof
Let A and B be the number of entries of 1 (mod 4) and 3 (mod 4) that are also 0 (mod 3).
Let C and D be the number of entries of 1 (mod 4) and 3 (mod 4) that are also 1 (mod 3).
Let E and F be the number of entries of 1 (mod 4) and 3 (mod 4) that are also 2 (mod 3).
| | 1 (mod 4) | 3 (mod 4) | |
| 0 (mod 3) | A | B | 1 |
| 1 (mod 3) | C | D | 5 |
| 2 (mod 3) | E | F | 1 |
| | 1 | 6 | |
A + B = 1
C + D = 5
E + F = 1
A + C + E = 1
B + D + F = 6
A+C+E = 1 thus C = 0 or 1.
C+D = 5 thus D = 4 or 5,
but there are only 4 entries that are
both 1 (mod 3) and 3 (mod 4) in the range 1..49,
namely,
7, 19, 31, 43
thus D = 4 and A=E=0, B=C=F=1
and all 4 entries of D must be used.
We need 1 entry each from B,C,F and
we also need at least 3 entries of 0 (mod 5) thus
there is only one choice each for B,C,F
in the range 1..49.
B: 3 (mod 4), 0 (mod 3), 0 (mod 5) --> 15
C: 1 (mod 4), 1 (mod 3), 0 (mod 5) --> 25
F: 3 (mod 4), 2 (mod 3), 0 (mod 5) --> 35
Checking the sums of powers, we have
7 + 15 + 19 + 25 + 31 + 35 + 43 = 175
72 + 152 + 192 + 252 + 312 + 352 + 432 = 5295
73 + 153 + 193 + 253 + 313 + 353 + 433 = 178375
The only trimagic series that is compatible with
modulo 3,4,5 has incorrect sums for S2 and S3,
therefore there is no order 7 normal trimagic series.