The Magic Encyclopedia ™ DataBase

Luo-Shu Squares
(by Aale de Winkel)

discussions with <Robert Dickter> led me to investigate the Luo-Shu square once more, the generalisation Robert investigated on his site www.luo-shu.com inspired me to go one step further to define Luo-Shu(m,a) a relatively prime to m as a generalisation of the Luo-Shu square to other orders besides 3. Below this generalisation is discussed from the mathematical point of view leaving the fuzzy side of things to Robert to discuss.

Luo-Shu Square
www.research.att.com/~njas/sequences attributes the famous order 3 Luo-Shu square to the Chinese mathematician Fuh-Hi (2858-2738 BC)
the story of the turtle on the Luo-Shu river is widely known and described elsewhere
Luo-Shu square
order 3
square
3 8 1
2 4 6
7 0 5
high
1 2 0
0 1 2
2 0 1
low
0 2 1
2 1 0
1 0 2
the above shows the Luo-Shu square in analitic numberrange, the splitup in high and low component shows
that these squares are each others mirror images in the central vertical
Luo-Shu(m,a)
Luo-Shu(m,a) = m [ { a ( y + (m - 1 - x) ) + { a - 1 - (m - 1) / 2 } % m ] + [ { a ( y + x ) + { a - 1 - (m - 1) / 2 } } % m ] =
m [ { a (y - x) + (m - 1) / 2 } % m ] + [ { a (y + x + 1) + (m - 1) / 2 } % m ]
(m - 1) / 2 < a < m ; a relatively prime to m { GCD(a,m) = 1 }
the definition above { a - 1 - (m - 1) / 2 } asures the main diagonal a constant (m - 1) / 2 and thus summing to the magic constant
the monagonal and the subdiagonal hold all the digits and are summing to the magic constant that way
the resulting squares are therefore {magic}
with a relatively prime to m ensures all digits are present in a column
Luo-Shu squares
order 5
Luo-Shu(5,3)
10 23 06 19 02
03 11 24 07 15
16 04 12 20 08
09 17 00 13 21
22 05 18 01 14
Luo-Shu(5,4)
11 15 24 03 07
05 14 18 22 01
04 08 12 16 20
23 02 06 10 19
17 21 00 09 13
Luo-Shu squares
order 7
Luo-Shu(7,4)
21 46 15 40 09 34 03
04 22 47 16 41 10 28
29 05 23 48 17 35 11
12 30 06 24 42 18 36
37 13 31 00 25 43 19
20 38 07 32 01 26 44
45 14 39 08 33 02 27
Luo-Shu(7,5)
22 41 04 16 28 47 10
13 25 37 00 19 31 43
46 09 21 40 03 15 34
30 42 12 24 36 06 18
14 33 45 08 27 39 02
05 17 29 48 11 23 35
38 01 20 32 44 07 26
Luo-Shu(7,6)
23 29 35 48 05 11 17
15 21 34 40 46 03 09
07 20 26 32 38 44 01
06 12 18 24 30 36 42
47 04 10 16 22 28 41
39 45 02 08 14 27 33
31 37 43 00 13 19 25
<Robert Dickter> noted a couple of pythagorean triangles in the regular ranged Luo-Shu(m,(m+1)/2)
below a proof of those PT's; PT([(m-1)/2,0]) in all odd order squares; PT([(m+1)/2,0]) exist for m > 3.
PT([(m-1)/2,0]) at Luo-Shu(m,(m+1)/2)regular { [(m-1)/2 - 1, (m-1)/2 - 1] , [(m-1)/2 - 1, (m-1)/2] , [(m-1)/2, (m-1)/2] }
proof:
Luo-Shu(m,a) [x,x] = m (m-1)/2 + [ { a (2x+1) + (m-1)/2 } % m ]
Luo-Shu(m,a) [x,x+1] = m [ { a + (m-1)/2 } % m ] + [ { a (2x+2) + (m-1)/2 } % m ]

Luo-Shu(m,(m+1)/2) [x,x] = m (m-1)/2 + [ { (m+1)/2 (2x+1) + (m-1)/2 } % m ] = m (m-1)/2 + [{(m+1)x % m}]
Luo-Shu(m,(m+1)/2) [x,x+1] = m [ { (m+1)/2 + (m-1)/2 } % m ] + [ { (m+1)/2 (2x+2) + (m-1)/2 } % m = [{x + (m+1)/2} % m

Luo-Shu(m,(m+1)/2) [(m-1)/2,(m-1)/2] = m (m-1)/2 + [{(m+1)(m-1)/2 % m}] = (m+1)(m-1)/2 = (m2-1)/2
Luo-Shu(m,(m+1)/2) [(m-1)/2 - 1,(m-1)/2 - 1] = m (m-1)/2 + [{(m+1)((m-1)/2-1) % m}] = (m+1)(m-1)/2 - 1 = (m2-1)/2 - 1
Luo-Shu(m,(m+1)/2) [(m-1)/2 - 1,(m-1)/2] = m [ { (m+1)/2 + (m-1)/2 } % m ] + [ { (m+1)/2 (m-1) + (m-1)/2 } % m = m - 1

adding 1 to change to regular numberrange the triplet [ (m2-1)/2 , m , (m2+1)/2 ] coincides with PT([(m-1)/2,0])
Q.E.D.
PT([(m+1)/2,0]) at Luo-Shu(m,(m+1)/2)regular { [(m-1)/2 - 1, (m-1)/2 + 3] , [(m-1)/2, (m-1)/2 + 3] , [(m-1)/2, (m-1)/2 + 4] }
proof:
Luo-Shu(m,a) [x,x+3] = m [ { 3a + (m-1)/2 } % m ] + [ { a (2x+4) + (m-1)/2 } % m ]
Luo-Shu(m,a) [x,x+4] = m [ { 4a + (m-1)/2 } % m ] + [ { a (2x+5) + (m-1)/2 } % m ]

Luo-Shu(m,(m+1)/2) [x,x+3] = m [ { 3(m+1)/2 + (m-1)/2 } % m ] + [ { (m+1)/2 (2x+4) + (m-1)/2 } % m ] = m + [{(m+1)x + (m+1)/2} % m}]
Luo-Shu(m,(m+1)/2) [x,x+4] = m [ { 4(m+1)/2 + (m-1)/2 } % m ] + [ { (m+1)/2 (2x+5) + (m-1)/2 } % m = m(m+3)/2 + [{(m+1)x + 3(m+1) - 1} % m ]

Luo-Shu(m,(m+1)/2) [(m-1)/2 - 1,(m-1)/2 + 3] = m(m+3)/2 + [{(m+1){(m-1)/2-1} + 3(m+1) - 1 } % m] = m(m+3)/2 + (m+1)/2 = (m+1)(m+3)/2 - 1
Luo-Shu(m,(m+1)/2) [(m-1)/2,(m-1)/2 + 3] = m + [{(m+1)(m-1)/2 + (m+1)/2 } % m] = m + 1
Luo-Shu(m,(m+1)/2) [(m-1)/2,(m-1)/2 + 4] = m(m+3)/2 + [ { (m+1)(m-1)/2 + 3(m+1) - 1 } % m] = m(m+3)/2 + (m+3)/2 = (m+1)(m+3)/2

adding 1 to change to regular numberrange the triplet [ (m+1)(m+3)/2 , m+2 , (m+1)(m+3)/2+1 ] coincides with PT([(m+1)/2,0])
Q.E.D.
further <Robert Dickter> noted that the number above the squares center squared the number to the centers left
Luo-Shu(m,(m+1)/2)regular [(m-1)/2 , (m-1)/2 - 1] = { Luo-Shu(m,(m+1)/2)regular [(m-1)/2 - 1, (m-1)/2] }2
proof:
Luo-Shu(m,a) [x,x-1] = m [ { -a + (m-1)/2 } % m ] + [ { a (2x) + (m-1)/2 } % m ]

Luo-Shu(m,(m+1)/2) [x,x-1] = m [ { -(m+1)/2 + (m-1)/2 } % m ] + [ { (m+1)x + (m-1)/2 } % m ] = m(m-1) + [{(m+1)x + (m-1)/2} % m}]

Luo-Shu(m,(m+1)/2) [(m-1)/2,(m-1)/2 - 1] = m(m-1) + [{(m+1)(m-1)/2 + (m-1)/2 } % m] = m(m-1) + (m-1) = (m+1)(m-1) = m2-1

adding 1 to change to regular numberrange confirms the relation
Q.E.D.
as a last point <Robert Dickter> noted that the squares bottom right corner the sum of 1..m
Luo-Shu(m,(m+1)/2)regular [(m-1, (m-1)/2] = k=1m k = m(m+1)/2
proof:
Luo-Shu(m,(m+1)/2) [x,x] = m (m-1)/2 + [ { (m+1)/2 (2x+1) + (m-1)/2 } % m ] = m (m-1)/2 + [{(m+1)x % m}]

Luo-Shu(m,(m+1)/2) [(m-1),(m-1)] = m(m-1)/2 + [{(m+1)(m-1)/2 + (m-1)/2 } % m] = m(m-1)/2 + (m-1) = m(m+1)/2 - 1

adding 1 to change to regular numberrange confirms the relation
Q.E.D.
furthermore it seems Luo-Shu(m,a) = Luo-Shu(m,(m+1)/2)_[perm(??)]
illudes me at the moment but possibly in some future upload.