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=1∑m 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.