The document provides definitions and theorems related to Latin squares and their connections to algebra, number theory, and cryptography. It begins with formal definitions of Latin squares and Latin functions. It then discusses various methods for constructing Latin squares using finite groups, finite fields, permutation polynomials, and properties of quadratic residues modulo prime numbers. Theorems are provided on Latin squares constructed from group operations, automorphisms, and using properties of finite fields and prime numbers.
3 سيپٽمبر 2015 تبليغ قرآن و سنت جي عالمگير غير سياسي تحريڪ دعوت اسلامي جي تحت هفته وار سنتن ڀريي اجتماع ۾ ٿيڻ وارو سنتن ڀريو بيان ”عاشقن جو حج“ توهان جي لاءِ هڪ تمام گهڻو مفيد ۽ اهم سنتن ڀريو بيان جنهن کي پڙهڻ سان توهان جي علم ۽ نيڪين ۾ اِن شآء الله عز و جل اضافو ٿيندو. توهان هن بيان کي ويب سائٽ تي موجود رهندي آن لائن پڙهڻ جي لاءِ Read جي بٽڻ ۽ ڊاؤن لوڊ ڪرڻ جي لاءِ Download جي بٽڻ تي ڪلڪ ڪريو. هن بيان جي باري ۾ پنهنجا تاثرات هيٺ ڏنل Comments Box ۾ ڏيو.
Jakarta Football Festival - GrabBike Rusun Cup 2015 Coaching Clinic Tahap ke-...Uni Papua Football
Jakarta Football Festival - GrabBike Rusun Cup 2015
Coaching Clinic Tahap ke-2 dihari ke-2
Lokasi berada di Lapangan Sepakbola Cakung, Jakarta Timur
#UniPapuaFootball #UniPapuaFc #Papua #Indonesia #Rusunawa #Cakung
#SepakbolaSosial #Sepakbola #FIFA #UniPapua #footballicious #Rusun
#coachesacrosscontinents #oneworldplayproject #avantis #JakTim
#ahok #dki #jakarta #pemprovdki #rusuncup #rusuncup2015 #RumahSusun
#jakartafootballfestival #grabbike #GrabBikeRusunCup2015 #SocialFootball
www.unipapua.net
www.unipapua.net/rusuncup2015
-Ys-
Annual leave is a right sanctioned by the Constitution of the Italian Republic. t was something workers fought for and obtained in the twentieth century with the advent of the Welfare State.
3 سيپٽمبر 2015 تبليغ قرآن و سنت جي عالمگير غير سياسي تحريڪ دعوت اسلامي جي تحت هفته وار سنتن ڀريي اجتماع ۾ ٿيڻ وارو سنتن ڀريو بيان ”عاشقن جو حج“ توهان جي لاءِ هڪ تمام گهڻو مفيد ۽ اهم سنتن ڀريو بيان جنهن کي پڙهڻ سان توهان جي علم ۽ نيڪين ۾ اِن شآء الله عز و جل اضافو ٿيندو. توهان هن بيان کي ويب سائٽ تي موجود رهندي آن لائن پڙهڻ جي لاءِ Read جي بٽڻ ۽ ڊاؤن لوڊ ڪرڻ جي لاءِ Download جي بٽڻ تي ڪلڪ ڪريو. هن بيان جي باري ۾ پنهنجا تاثرات هيٺ ڏنل Comments Box ۾ ڏيو.
Jakarta Football Festival - GrabBike Rusun Cup 2015 Coaching Clinic Tahap ke-...Uni Papua Football
Jakarta Football Festival - GrabBike Rusun Cup 2015
Coaching Clinic Tahap ke-2 dihari ke-2
Lokasi berada di Lapangan Sepakbola Cakung, Jakarta Timur
#UniPapuaFootball #UniPapuaFc #Papua #Indonesia #Rusunawa #Cakung
#SepakbolaSosial #Sepakbola #FIFA #UniPapua #footballicious #Rusun
#coachesacrosscontinents #oneworldplayproject #avantis #JakTim
#ahok #dki #jakarta #pemprovdki #rusuncup #rusuncup2015 #RumahSusun
#jakartafootballfestival #grabbike #GrabBikeRusunCup2015 #SocialFootball
www.unipapua.net
www.unipapua.net/rusuncup2015
-Ys-
Annual leave is a right sanctioned by the Constitution of the Italian Republic. t was something workers fought for and obtained in the twentieth century with the advent of the Welfare State.
La diversificación de espacios en la práctica, LC. Leticia Jiménezlettizya
En este archivo se describe, de forma breve, las características que deberán tener las áreas destinadas, así como las repercusiones que éstas tienen en la educación de los estudiantes.
When Should I Buy My Child A Smartphone (Or Tablet)?TeenSafe
The average child today starts accessing the internet at age three. But do we really know how this will affect our children? How young is too young to start using digital devices? When should you begin teaching your child responsible smartphone behavior? We've compiled advice from child development experts to create a full guideline of when to introduce your child to smartphones and tablets.
The simplest and most common form of mathematical induction infers that a statement involving a natural number n (that is, an integer n ≥ 0 or 1) holds for all values of n. The proof consists of two steps:
The base case (or initial case): prove that the statement holds for 0, or 1.
The induction step (or inductive step, or step case): prove that for every n, if the statement holds for n, then it holds for n + 1. In other words, assume that the statement holds for some arbitrary natural number n, and prove that the statement holds for n + 1
Best Approximation in Real Linear 2-Normed SpacesIOSR Journals
This pape r d e l i n e a t e s existence, characterizations and st rong unicity of best uniform
approximations in real linear 2-normed spaces.
AMS Su ject Classification: 41A50, 41A52, 41A99, 41A28.
Maksim Zhukovskii – Zero-one k-laws for G(n,n−α)Yandex
We study asymptotical behavior of the probabilities of first-order properties for Erdős-Rényi random graphs G(n,p(n)) with p(n)=n-α, α ∈ (0,1). The following zero-one law was proved in 1988 by S. Shelah and J.H. Spencer [1]: if α is irrational then for any first-order property L either the random graph satisfies the property L asymptotically almost surely or it doesn't satisfy (in such cases the random graph is said to obey zero-one law. When α ∈ (0,1) is rational the zero-one law for these graphs doesn't hold.
Let k be a positive integer. Denote by Lk the class of the first-order properties of graphs defined by formulae with quantifier depth bounded by the number k (the sentences are of a finite length). Let us say that the random graph obeys zero-one k-law, if for any first-order property L ∈ Lk either the random graph satisfies the property L almost surely or it doesn't satisfy. Since 2010 we prove several zero-one $k$-laws for rational α from Ik=(0, 1/(k-2)] ∪ [1-1/(2k-1), 1). For some points from Ik we disprove the law. In particular, for α ∈ (0, 1/(k-2)) ∪ (1-1/2k-2, 1) zero-one k-law holds. If α ∈ {1/(k-2), 1-1/(2k-2)}, then zero-one law does not hold (in such cases we call the number α k-critical).
We also disprove the law for some α ∈ [2/(k-1), k/(k+1)]. From our results it follows that zero-one 3-law holds for any α ∈ (0,1). Therefore, there are no 3-critical points in (0,1). Zero-one 4-law holds when α ∈ (0,1/2) ∪ (13/14,1). Numbers 1/2 and 13/14 are 4-critical. Moreover, we know some rational 4-critical and not 4-critical numbers in [7/8,13/14). The number 2/3 is 4-critical. Recently we obtain new results concerning zero-one 4-laws for the neighborhood of the number 2/3.
References
[1] S. Shelah, J.H. Spencer, Zero-one laws for sparse random graphs, J. Amer. Math. Soc.
1: 97–115, 1988.
In this note we, first, recall that the sets of all representatives of some special ordinary residue classes become (m, n)-rings. Second, we introduce a possible p-adic analog of the residue class modulo a p-adic integer. Then, we find the relations which determine, when the representatives form a (m, n)-ring. At the very short spacetime scales such rings could lead to new symmetries of modern particle models.
1. NCM LECTURE NOTES ON LATIN SQUARES.
NOW WE SHALL SEE THE MOST BEAUTIFUL RELATION S BETWEEN , ALGEBRA , NUMBER THEORY AND
COMBINATORICS , WHEN WE SOLVE PROBLEMS BASED ON “ LATIN SQUARES “. THEY TOO HAVE VERY
IMPORTANT CONNECTION IN “ CRYPTOGRAPHY “ ALSO.
NOW , WE WILL MOVE TO THE FORMAL DEFINITION OF “ LATIN SQUARES “.
LET “ n “ BE A GIVEN POSITIVE INTEGER AND LET “ S “ BE A GIVEN SET OF “ n “ DISTINCT
ELEMENTS .
S = { s1 , s2 , s3 , . . . , sn – 1 , sn } . A LATIN SQUARE OF ORDER “ n “ BASED ON “ S “ IS AN
“ n - by – n array “ , each of whose entries is an element of “ S “ SUCH THAT EACH OF THE “ n “
elements of “ S “ OCCURS EXACTLY ONCE IN EACH ROW AND EXACTLY ONCE IN EACH COLUMN .
THUS EACH OF THE ROWS AND EACH OF THE COLUMNS OF A LATIN SQURE IS A PERMUTATION OF
THE ELEMENTS OF “ S “. LET “ G “ BE A GIVEN FINITE GROUP OF “ n “ ELEMENTS.
SUPPOSE “ f “ IS A FUNCTION : S x S …… > S
SUCH THAT f ( s i , s j ) = si . sj LIES IN “ S “ WITH THE FOLLOWING PROPERTY
FOR A FIXED “ i “ , si . sj IS A PERMUTATION ON “ S “ FOR ALL “ j “ VARIES FROM 1 TO “ n “.
Similarly for a fixed “ j “ , si . sj IS A PERMUTATION ON “ S “ FOR ALL “ i “ varies from 1 TO “ n “.
SUCH A FUNCTION “ f “ IS CALLED A LATIN FUNCTION DEFINED ON S x S TO S .
G = { g1 , g2 , g3 , … , gi , … , gj , … , gn – 1 , gn } . WITHOUT LOSS OF GENERALITY , ASSUME g1 = e , the
identity element of “ G “. NOW WE SHALL DEFINE AN “ n – by – n “ MATRIX “ A “ AS FOLLOWS :
A = [ ai j ] = [ gi . gj ]. Now it is easy to verify “ A “ IS A LATIN SQUARE OF ORDER “ n “ BASED ON
“ G “
THE i th ROW OF “ A “ = { gi . g1 , gi . g2 , gi . g3 , … , gi . gi , … , gi . gj , … ,gi . gn }.
THE j th COLUMN OF “ A “ = { g1 . gj , g2 . gj , g3 . gj , … , gi . gj , … , gj . gj , … , gn . gj } .
THEOREM – 1 . LET “ G “ BE A GIVEN ( ANY ) FINITE GROUP OF ORDER | G | .
G = { g1 = e , g2 , g3 , … , gi , … , gj , … , g| G | } . LET “ m “ , “ n “ BE GIVEN POSITIVE INTEGERS SUCH
THAT ( m , | G | ) = ( n , | G | ) = 1 .
DEFINE | G | x | G | matrix “ A “ AS A = [ gi
m
. gj
n
] ; i , j belongs to { 1 , 2 , 3 , … , | G | } .
2. “ A “ IS A LATIN SQUARE OF ORDER | G | , BASED ON “ G “.
THEOREM – 2 : LET “ a “ , “ b “ BE ANY TWO FIXED ELEMENTS OF “ G “ , GIVEN FINITE GROUP.
DEFINE A = [ ( a . gi ) . ( gj . b ) ] , B = [ ( a . gi ) . ( b . gj ) ] , C = [ ( gi . a ) . ( gj . b ) ] ,
D = [ ( gi . a ) . ( b. gj ) ] . A , B , C , D ARE ALL “ LATIN SQUARES “ OF ORDER | G | , BASED ON “ G “ .
THEOREM - 3
LET T1 , T2 BE ANY TWO AUTOMORPHISMS OF A FINITE GROUP “ G “ .
LET “ X “ BE A MATRIX OF ORDER | G | x | G | , DEFINED BY
X = [ T1 ( g i ) . T 2 ( g j ) ] . THEN “ X “ IS A LATIN SQUARE OF ORDER | G | , BASED ON
“ G “.
NOW LET US DO SOME GOOD “ NUMBER THEORY “ .
LET “ N “ BE A GIVEN POSITIVE INTEGER.
Z N = { 0 , 1 , 2 , 3 , … , N – 2 , N – 1 } BE THE RING OF INTEGERS MOD ( N ) .
+ N , x N IS THE ADDITION MOD N AND MULTIPLICATION MOD N DEFINED ON “ Z N “ .
DEFINE Z N
* = Z N - { 0 } WITH “ a “ LIES IN Z N
* IF AND ONLY IF ( a , N ) = 1 .
SELECT FIXED a , b LIES IN Z N
* [ ( a , N ) = ( b , N ) = 1 ].
SELECT “ r “ , “ s “ [ FIXED , LIES IN Z N ] .
NOW DEFINE A BIJECTIVE MAP f a , r : Z N ……..> Z N
By f a ,r ( x ) = a . x + r ( MOD N ) . SIMILARLY
We can define g b , s ( x ) = b . x + s ( MOD N ) .
NOW DEFINE A = [ f a , r ( i ) + g b , s ( j ) ( MOD N ) ] .
A IS A LATIN SQUARE OF ORDER “ N “ BASED ON Z N .
AS A SIMPLE CORLLARY A = [ ( i + j ) ( mod N ) ] , ALWAYS DEFINES A LATIN SQUARE
OF ORDER “ N “ BASED ON Z N
3. ALSO DEFINE A MATRIX B = [ ( a . i + j ) ( MOD N ) ] , ALWAYS DEFINE A LATIN
SQUARE OF ORDER “ N “ BASED ON Z N .
DEFINE “ M = 2 . N + 1 “ . CONSTRUCT AN M by M MATRIX “ C “ AS FOLLOWS :
C = [ ( N + 1 ) . ( i + j ) ( mod M ) ] IS A LATIN SQUARE OF ORDER “ M “ BASED ON ZM
THIS IS THE BEAUTIFUL EXAMPLE OF IDEMPOTENT SYMMECTRIC LATIN SQUARE OF ODD
ORDER.
DEFINE FOR ANY FIXED NON NEGATIVE INTEGER “ k “ , M = 2 k + 1 + 5
CONSTRUCT AN “ M by M “ matrix D = [ ai j ] = [ ( 2k + 3 ) ( i + j ) ( MOD M ) ]
FOR i , j belongs TO { 0 , 1 , 2 , 3 , 4 , … , 2k + 1 + 4 }.
THIS “ D “ ALSO DEFINES AN IDEMPOTENT SYMMECTRIC LATIN SQUARE OF ORDER “ M “
BASED ON “ ZM “
NOW LET US CONSTRUCT MORE GENERALLY IDEMPOTENT SYMMECTRIC LATIN SQUARE
AS FOLLOWS : DEFINE M = 2 k + 1 + p , where “ p “ is ODD PRIME.
DEFINE A MATRIX OF ORDER “ M “ AS X = [ ai j ] =
[ [ 2k + ( p + 1 / 2 ) ] ( i + j ) ( MOD M ) ]
Here i , j BELONGS TO { 0 , 1 , 2 , 3 , . . . , 2k + 1 + p - 1 }. X IS IDEMPOTENT AND SYMMECTRIC
LATIN SQUARE OF ORDER “ M “ BASED ON ZM .
LET “ p “ BE ANY GIVEN ODD PRIME. LET US SELECT A POSITIVE INTEGER “ e “ SUCH THAT
G.C.D ( e , p - 1 ) = 1. NOW DEFINE A MATRIX “ X “ OF ORDER “ p x p “ as follows.
X = [ ai j ] = [ ( p + 1 / 2 ) . ( ie + je ) ( MOD p ) ] , i , j LIES IN { 0 , 1 , 2 , 3 , . .. , p – 1 } is
a symmectric LATIN SQUARE.
4. NOW LET US CONSTRUCT R.S.A ( PUBLIC KEY CRYPTOGRAPHY ) LATIN SQUARE.
SELECT TWO DISTINCT VERY LARGE ODD PRIMES “ p , q “ .
Select two fixed positive integers e , d such that ( e , ( p - 1 ) . ( q - 1 ) ) = 1
And ( d , ( p - 1 ) . ( q - 1 ) ) = 1
Now define a matrix A = [ ie + jd ( MOD N ) ] , N = p . q .
A is a LATIN SQUARE OF ORDER “ N “ BASED ON Z N .
Y = [ ( a . ie + j ) ( MOD N ) ] , ( a , p . q ) = 1 IS A LATIN SQUARE OF ORDER
“ N = p. q “ BASED ON ZN ; i , j BELONGS TO { 0 , 1 , 2 , 3 , … ,N – 1 }. HERE a , e ARE FIXED.
NOW WE ARE GOING TO USE TWO BEAUTIFUL RESULTS , NAMELY PERMUTATION
POLYNOMIALS OVER FINITE FIELDS.
LET US OBSERVE THE FOLLOWING FACTS :
FACT – 1 : LET “ K “ BE A FINITE FIELD WITH “ q = pn “ elements . THEN xq = x FOR ALL
“ x “ LIES IN “ K “ .
FACT – 2 : IF “ f [ t ] , g [ t ] “ are two polynomials over “ K “ OF DEGREES < q , and
If f ( a ) = g ( a ) , for all “ a “ LIES IN “ K “, THEN f = g .
FACT – 3 : LET “ f “ be a given polynomial over “ K “ . f [ t ] LIES IN K [ t ]. THEN THERE
EXISTS A POLYNOMIAL f* LIES IN K [ t ] OF DEGREE < q , SUCH THAT f * ( a ) = f ( a ) , for all
“ a “ LIES IN “ K “ .
FACT - 4 : LET “ T “ BE ANY FUNCTION FROM “ K “ TO ITSELF . T : K ……> K. THEN THERE
EXISTS A POLYNOMIAL f [ t ] LIES IN K [ t ] SUCH THAT T ( a ) = f ( a ) for all “ a “ LIES
IN “ K “ .
5. DEFINITION : A POLYNOMIAL “ f “ OVER A FINITE FIELD “ K “ IS CALLED A PERMUTATION
POLYNOMIAL IF THE MAPPING f : K ……. > K , DEFINED BY “ a …….> f ( a ), for “ a “ LIES
IN “ K “ IS ONE - TO – ONE.
NOW OUR AIM IS TO CONSTRUCT A LATIN SQUARE , USING PERMUTATION POLYNOMIALS
OVER A FINITE FIELD “ K “ with “ q “ elements.
THEOREM – 1 ( R.A . MOLLIN AND C. SMALL ) . LET K = GF ( q ) , GIVEN FINITE FIELD
WITH “ q “ elements. ASSUME THE CHARACTERISTIC OF K IS DIFFERENT FROM “ 3 “.
THAT IS CHAR ( K ) NOT EQUAL TO 3. THEN f ( x ) = a x3 + b x2 + c x + d ( a LIES IN K* )
THAT IS “ a “ is non –zero element of “ K “ is a PERMUTATION POLYNOMIAL ON “ K “
I F AND ONLY IF “ b 2 = 3 . a .c “ and q IS CONGRUENT TO 2 ( MOD 3 ) [ q = 2 ( MOD 3 ) ]
THEOREM – 2 ( RAJESH PRATAP SINGH AND SOUMEN MAITY ) . LET “ p “ = ODD PRIME.
LET “ K = Zp = FIELD OF INTEGERS MOD “ p “ .
DEFINE f [ t ] LIES IN Zp [ t ] such that f [ t ] = ta ( t ( p – 1 / 2 ) + b ) with ( a , p – 1 ) = 1
And “ b LIES IN Zp* “ ( non – zero element in Zp ). Then f [ t ] is a permutation polynomial
ON Zp if and only if ( b2 - 1 )( p – 1 / 2 ) = 1 ( MOD p ) .
IF WE WANT TO USE THESE TWO THEOREMS EFFECTIVELY FOR THE CONSTRUCTION OF LATIN
SQUARE MOD “ p = ODD PRIME “ , WE NEED TO UNDERSTAND THE BASIC CONCEPTS
ABOUT “ QUADRATIC RESIDUES “
THEREFORE , NOW WE ARE CONCERNED WITH QUADRATIC CONGRUENCES OF THE FORM
“ x 2 = n ( MOD p ) “ , where p = ODD PRIME AND “ n “ is NOT CONGRUENT TO
0 ( MOD p ) . IF “ x “ is a solution so is “ – x “ . HENCE THE NUMBER OF SOLUTIONS IS
EITHER “ 0 “ OR “ 2 “.
6. DEFINITION : SUPPOSE “ p “ is an ODD PRIME and “ a “ is an integer. “ a “ is defined to
be a quadratic residue MOD p if
1. a is NOT congruent to 0 ( MOD p )
2. THE CONGRUENCE x 2 = a ( MOD p ) has a solution “ x “ LIES IN Zp*
“ a “ is said to be a quadratic non – residue MOD p if “ a “ IS NOT CONGRUENT
TO 0 ( MOD p ) and “ a “ is NOT a quadratic residue MOD p .
FACT - 1 : LET “ p “ be an ODD PRIME. THEN “ EVERY REDUCED SYSTEM MOD p “
Contains exactly ( p - 1 / 2 ) quadratic residues and exactly ( p - 1 / 2 )
quadratic non residues MOD p.
THE QUADRATIC RESIDUES BELONG TO THE RESIDUE CLASSES CONTAING THE NUMBERS
12 ( MOD p ) , 22 ( MOD p ) , 32 ( MOD p ) , . . . , ( ( p - 1 ) / 2 ) 2 ( MOD p )
Let us have the following example. TAKE p = 1 1 . THEN QUADRATIC RESIDUES MOD 1 1
ARE { 12 = 1 , 22 = 4 , 32 =9 , 42 = 5 , 52 = 3 } AND QUADRATIC NON – RESIDUES ARE
{ 2 , 6 , 7 , 8 , 10 }. NOW LET US PROVE THIS FACT – 1 BY GROUP THEORY.
PROOF : Z p
* = { 1 , 2 , 3 , . . . , p - 2 , p – 1 } forms an ABELIAN GROUP ( IN FACT CYCLIC
GROUP ) WITH RESPECT TO THE BINARY OPERATION “ MULTIPLICATION MOD p “ ( = x p )
| Zp
* | = p - 1 . CONSIDER A MAP T : Z p
* ………..> Zp
* , defined by
T ( x ) = x 2. THIS MAP “ T “ IS A GROUP HOMORPHISM FROM Z p
* to ITSELF.
KERNEL ( T ) = { x LIES IN Z p
* | x2 = 1 }. SINCE | KERNEL ( T ) | = 2 , “ T “ IS NOT
ONE – TO – ONE ( THEREFORE “ T “ IS NOT “ ONTO “ )
LET “ H = T ( Z p
* ) “ , proper subgroup of Z p
*
H = { x 2 | x LIES IN Zp
* } = set of all quadratic residues MOD p AND
| H | = ( p – 1 ) / 2 . MORE OVER H IS THE NORMAL SUBGROUP OF INDEX “ TWO “.
THEREFORE Zp
* = H U H .y ( y is a quadratic non – residue ( MOD p ) ). THIS IS THE
IMPORTANT OBSERVATION , HERE.
7. IN THIS CONTEXT , WE WOULD LIKE TO REGISTER TWO CELEBRATED RESULTS, NAMELY
EULER’S CRITERIAN AND THE FAMOUS “ THE QUADRATIC RECIPROCITY LAW
NOW WE SHALL DEFINE Legendre ‘ s symbol . LET “ a “ be any given integer
( a / p ) = 0 if n = 0 ( MOD p )
( a / p ) = 1 if n is a quadratic residue MOD p
( a / p ) = - 1 if n is a quadratic non residue MOD p
EULER ‘ S CRITERIAN : LET “ p “ be given ODD PRIME. THEN FOR ANY INTEGER “ a “,
We have ( a / p ) = a ( p – 1 ) / 2 ( MOD p )
THE QUADRATIC RECIPROCITY LAW
IF “ p , q “ are distinct odd primes , then ( p / q ) . ( q / p ) = ( - 1 )( p – 1 ) . ( q – 1 ) / 4
THE Legendre ‘ s symbol ( a / p ) is a completely multiplicative function of “ a “.
LET “ p “ be a given fixed ODD PRIME. LET a 1 , a2 be any given positive integers .
THEN ( a1 . a2 / p ) = ( a 1 / p ) . ( a 2 / p ) .
NOW ,LET US BEGIN OUR MAIN TASK OF CONSTRUCTING A LATIN SQUARE BASED ON Zp , p is
ODD PRIME. AGAIN , LET US RECALL ,THE BEAUTIFUL RESULT OF R.A. MOLLIN AND C.SMALL.
TAKE GF ( q ) = Z p , where “ p “ is ODD PRIME.
HAVE A CLOSE LOOK AT THE CUBIC POLYNOMIAL f ( x ) = a x3 + bx2 + cx + d LIES IN Z p [ x ].
Here 1. We observe b 2 = 3 . a . c and p = 2 ( MOD 3 )
THEREFORE THE Legendre ‘s symbol ( 3.a.c / p ) = 1
This ,further reduces to ( 3 / p ) . ( a / p ) . ( c / p ) = 1 AND p = 2 ( MOD 3 )
CASE - 1 : ( 3 / p ) = 1 . THIS IMPLIES p = 1 ( MOD 12 ) OR p = 11 ( MOD 12 ).
SINCE p = 2 ( MOD 3 ) , HERE p = 1 1 ( MOD 12 ).
8. THEREFORE IN THIS CASE OUR ODD PRIME p = 1 2 .k + 1 1, for some suitable integer “ k “.
Now ( a / p ) . ( c / p ) = 1 implies ( a / p ) = ( c / p ) = 1 ( OR ) ( a / p ) = ( c / p ) = - 1.
Again let us recall the beautiful decomposition “ Zp = H U H. y “.
THIS FORCES US TO SELECT a , c LIES IN H ( OR ) SELECT a , c LIES IN H.y.
THEN , b 2 = 3a c IN Z p , where p = 11 ( MOD 12 ).
AT THIS TIME , LET US CONSTRUCT TWO POLYNOMIALS f( x ) , g ( x ) LIES IN Z p [ x ] .
That is f ( x ) = a x3 + bx2 + cx + d with a , c non – zero and a , c LIES IN “ H “.
THEN WE CAN SOLVE b , such that b 2 = 3 ac.
Define g ( x ) = a’ x3 + b’ x2 + c’ x + d’, here a’ , c’ are non – zero LIES IN H. y.
Then b’2 = 3 a’ c’ is SOLVABLE IN Z p.
Now define a matrix “ M “ OF ORDER “ p x p “ , by
M = [ f (i ) + g ( j ) ], i , j LIES IN Z p , p = 12 k + 1 1 .
Then “ M “ IS A LATIN SQUARE OF ORDER “ p “ .
NOW CONSIDER THE CASE , WHERE ( 3 / p ) = - 1. USING
“ QUADRATIC RECIPROCITY LAW “ , WE MUST CONCLUDE p = 5 ( MOD 12 ) ( OR )
p = 7 ( MOD 12 ) . SINCE p = 2 ( MOD 3 ) , WE MUST HAVE p = 5 ( MOD 12 )
CASE – 2 : ( 3 / p ) = - 1 AND p = 5 ( MOD 1 2 ). p = 12 k + 5.
Then ( a . c / p ) = - 1. THIS IMPLIES ( a / p ) . ( c / p ) = - 1 .
Here also , we have Zp
* = H U H . y.
If “ a “ LIES IN H , THEN “ c “ lies in H .y
If “ a “ LIES IN H .y , THEN “ c “ LIES IN H.
THEN ONLY b 2 = 3.ac , is solvable , for “ b “ LIES IN Z p
* , here p = 12 k + 5 .
IF WE CONSTRUCT f [ x ] , g [ x ] with the above conditions,
9. .
THEN M = [ f [i ] + g [ j ] ] defines a LATIN SQUARE OF ORDER “ p = 12 k + 5 “.
Now , again , let us have a close look at theorem – 2 of R. P.Singh and Soumen Maity
HERE “ p “ is any ODD PRIME . Select “ a “ such that ( a , p – 1 ) = 1 . b LIES IN Zp
*
With ( ( b 2 - 1 ) / p ) = 1 . here f [ t ] = ta ( t( p – 1 ) / 2 + b ) .
Now ( ( b 2 - 1 ) / p ) = 1 IMPLIES ( ( b – 1 ) / p ) . ( ( b + 1 ) / p ) = 1.
There SELECT “ b “ such that , b – 1 AND b + 1 LIES IN “ H “ ( OR ) b – 1 and b + 1 LIES IN
H .y. IN THIS CASE ALSO , WE CAN CONSTRUST M = [ f[ i ] + g [ j ] ] , LATIN SQUARE OF ORDER
“ p “
NOW LET US LEARN TWO IMPORTANT CONCEPTS FROM FINITE GROUP THEORY.
DEFINITION – 1 : COMPLETE MAPPING OF “ G “
LET “ G “ BE A GIVEN FINITE GROUP. LET f : G …….> G BE A PERMUTATION.
WE CALL “ f “ A COMPLETE MAPPING OF “ G “ IF THE MAPPING “ T “ : g …… > g. f ( g ) is
also a permutation on “ G “.
DEFINITION - 2 : WE CALL THE MAPPING “ f “ ( which is a permutation on “ G “ ) , AN
ORTHOMORPHISM OF “ G “ IF THE MAPPING “ U “ : g ……. > g – 1 . f ( g ) is a permutation
on “ G “
WE CALL A MAPPING “ f “ OF “ G “ , A STRONG COMPLETE MAPPING OF “ G “ IF IT IS
BOTH A COMPLETE MAPPING AND AN ORTHOMORPHISM OF “ G “
THEOREM – 1 : “ f “ IS A COMPLETE MAPPING OF “ G “ I f and only if the mapping “ T “
DEFINIED BY T ( g ) = g. f ( g ) is an orthomorphism of “ G “.
A mapping “ f “ is an orthomorphism of “ G “ I f and only if THE MAPPING “ U “ ,
DEFINED BY U ( g ) = g – 1 . f ( g ) IS A COMPLETE MAPPING OF “ G “ .
NOTE : LET G = { g 1 , g2 , g3 , …. gn – 1 , gn } be a given group. LET “ f “ be a given
permutation on “ G “ .
10. DEFINE A MATRIX A = [ ai , j ] = [ f ( gi ) . gj ] IS A LATIN SQUARE OF ORDER | G | , BASED
ON “ G “ .
LEMMA - 1 : SUPPOSE “ f “ IS A COMPLETE MAPPING OF “ G “.
DEFINE A = [ f ( gi ) . gj ] and B = [ gi . f ( gi ) . gj ] . THEN “ A “ IS ORTHOGONAL
TO “ B “ .
LEMMA – 2 : LET “ G “ BE A GIVEN GROUP OF ORDER “ n “ .
Define “ M “ BE THE “ n x n “ CAYLEY TABLE WITH “ i j “ th ENTRY = gi . gj
M = [ gi . gj ] . LET “ f “ BE A GIVEN PERMUTATION ON “ G “.
Mf = [ gi . f ( gj ) ]. IT IS EASY TO SEE THAT M IS A LATIN SQUARE , AND THAT Mf IS OBTAINED
FROM “ M “ BY PERMUTING ITS COLUMNS.
“Mf “ IS ORTHOGONAL TO “ M “ I f and only if “ f “ IS AN “ORTHOMORPHISM “ OF “ G .
LEMMA - 3 : LET “ f , g “ BE ANY TWO PERMUTATIONS OF “ G “.
NOW CONSTRUCT “ Mf AND Mg “ . THEN “ Mf IS ORTHOGONAL TO Mg “
I F AND ONLY I F THE MAPPING “ T “ DEFINED BY T ( x ) = ( g ( x ) )– 1 . f ( x ) IS A
PERMUTATION OF “ G “ . [ x is an element of “ G “ ].
AT THIS POINT OF TIME I WOULD LIKE TO INTRODUCE A VERY IMPORTANT DEFINITION
DEFINITION : LET “ f , g “ BE ANY TWO MAPPINGS OF A GROUP “ G “ .
WE SAY “ f AND g “ ARE ORTHOGONAL
IF THE MAPPING “ T “ , DEFINED BY T ( x ) = ( g ( x ) ) – 1 . f ( x ) IS A PERMUTATION
ON “ G “ .
FACT – 1 : A MAPPING f : G ---- G IS A “ COMPLETE MAPPING “ OF “ G “ IF IT IS
ORTHOGONAL TO THE MAPPINGS “ k* “ AND “ l* “
Where k* ( x ) = 1 ( the identity element of “ G “ ) FOR ALL “ x “ LIES IN “ G “
AND l* ( x ) = x – 1 FOR ALL “ x “ LIES IN “ G “ .
FACT – 2 : A MAPPING “ f “ IS AN ORTHOMORPHISM IF IT IS ORTHOGONAL TO “ k* “ AND
IDENTITY MAPPING “ IG “ ( IG ( x ) = x FOR ALL “ x “ LIES IN “ G “ ) .