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Name Ihsan ul ghafoor Roll #08030609-001 SubjectNumber theory SectionA Department Methamatics Semester4 th. Topic. Different methods of Solving Congruence's. Congruence. What is congruence? Equivalence of congruence:- If a Ξ b(mod m) - PowerPoint PPT Presentation
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Name Ihsan ul ghafoor
Roll # 08030609-001
Subject Number theory
Section A
Department Methamatics
Semester 4th
Topic
Different methods of Solving
Congruence's
CongruenceWhat is congruence?
Equivalence of congruence:-If a Ξ b(mod m)Then we can use ‘a’ as ‘b’ and ‘b’ as ‘a’ in
other congruence's under same modulus m.
CongruenceSolution of congruence:-If ax Ξ b(mod m)Then solution will belong to following set of
numbersX={0,1,2,3,………,m-1}
If solution is greater than that set we can convert it in to the above set number by using equivalence.
Solution of CongruenceFirst we find either solution exists or not.If ax Ξ b(mod m)
Then solution exists if and only ifd|b where d=gcd(a,m)
The congruence will have ‘d’ solutions.
Solution of CongruenceSome methods of solving congruence are
followingTrial method:-Simple method
Put the value one by one from the solution set and check until the congruence is true.
Solution of CongruenceExample:-2x Ξ 3(mod5)As (2,5)=1 and 1|3 solution exists. As d=1 so congruence
has unique solutionX={0,1,2,3,4}
Now put the value of x one by one2(0) Ξ 3(mod 5) false2(1) Ξ 3(mod 5) false2(2) Ξ 3(mod 5) false2(3) Ξ 3(mod 5) false2(4) Ξ 3(mod 5) trueSo x Ξ 4 (mod 5) is the required solution.
Solution of CongruenceDiophantine eq method:-Every congruence relation can be written in
the form of linear Diophantine eqs.
If ax Ξ b(mod m)m|ax-bax-b=my ; yЄ Zax-my=b
Solution of CongruenceThen find the value of x .And that will be the solution.
Further solutions can be findUsing formulax’=x-(bt/d) ; d=(a,m)
Solution of CongruenceExample:-2x Ξ 3(mod5)As (2,5)=1 and 1|3 solution exists. As d=1 so
congruence has unique solution2x – 5y = 3 (in Diophantine form)
now we will find the value of x in diophanitne eq.
Solution of CongruenceAs (2,5)=1=d1=5(1)+2(-2) (gcd as a linear combination)3=-5(-3)+2(-6) (multiply by 3)3=-2(-6) - 5(-3)So value of x = -6-6(mod5) = 4 5|-6 -6 Ξ 4 (mod 5) -6=5(-1)+(-1) but r>=0
so -6=5(-2)+4;
So solution of congruence x Ξ 4(mod 5)
Solution of CongruenceSymbolic fraction method :-If ax Ξ b(mod m)Then solution will bex Ξ {b+mh (mod m)} / a ; h Є ZIff a| (b + mh)
Solution of CongruenceExample:-2x Ξ 3(mod 5)As (2,5)=1 and 1|3 solution exists. As d=1 so
congruence has unique solution
x Ξ {3 + 5h (mod 5)} /2 [eq 1]2| (2+5h) ( a | b + mh)3+5h Ξ 0 (mod 2)1+ h Ξ 0 (mod 2) 5h Ξ h (mod 2), 3 Ξ 1(mod 2), h Ξ -1 (mod 2) mean h= -1
Solution of CongruencePut h = -1 in [eq 1]x Ξ {3 + 5(-1) (mod 5)} /2x Ξ {-2 (mod 5)} /2x Ξ -1 (mod 5)x Ξ 4 (mod 5) { -1 Ξ 4 (mod 5) }
as { -1 = 5(-1) + 4 }
Solution of CongruenceBy Fermate theorem :-
If ax Ξ b(mod m)
Then solution will be
x Ξ b aФ(m)-1 (mod m)
Ф(n) is Euler phi function.
Solution of CongruenceExample:-2x Ξ 3(mod 5)As (2,5)=1 and 1|3 solution exists. As d=1 so
congruence has unique solutionx Ξ 3.2Ф(5)-1
(mod 5)x Ξ 3.24-1
(mod 5)x Ξ 3.23
(mod 5)x Ξ 24 (mod 5)x Ξ 4 (mod 5) { 24 Ξ 4 (mod 5) }
Solution of CongruenceLet take another example and then find the
solution by methods explain above.
If 4 x Ξ 1 (mod 7)
As (4,7)=1 and 1|1 so unique solution exists.
Solution of Congruence 4 x Ξ 1 (mod 7)
By trial method:-
Put value of x one by one from the setX = {0,1,2,3,4,5,6}
x Ξ 2 (mod 7)
Solution of Congruence 4 x Ξ 1 (mod 7)By Diophantine eq method :-
4 x - 7 y = 1 gcd linear combination
7=4(1)+3 1=4(1)-3(1)
4=3(1)+1 1=4(1)-{7(1)-4(1)}(1)
So x = 2 1=4(1)-7(1)+4(1)x Ξ 2 (mod 7) is solution 1=4(2)-7(1)
Solution of Congruence 4 x Ξ 1 (mod 7)
By symbolic fraction method:-x Ξ {1 + 7 h(mod 7)} / 4 eq 1As 4 | 1 + 7 h1 + 7 h Ξ 0 (mod 4)1 + 3 h Ξ 0 (mod 4) 7 h Ξ 3h (mod 4)3h Ξ {-1 (mod 4) } h Ξ {-1 + 4 k(mod 4) } / 3 eq 23 | -1 + 4k-1 + 4k Ξ 0 (mod 3)-1 + k Ξ 0 (mod 3) 4k Ξ k (mod 3)k Ξ 1 (mod 3)
Solution of CongruencePut k =1 in eq 2h Ξ {-1 + 4(1) (mod 4) } / 3h Ξ {3 (mod 4) } / 3h Ξ 1 (mod 4)Put h =1 in eq 2x Ξ {1 + 7 (1)(mod 7)} / 4x Ξ {8 (mod 7)} / 4x Ξ 2 (mod 7)Required solution
Solution of Congruence4 x Ξ 1 (mod 7)
By Fermate theorem :- x Ξ 1.4Ф(7)-1 (mod 7)x Ξ 46-1 (mod 7)x Ξ 45 (mod 7)x Ξ 1024 (mod 7)x Ξ 2 (mod 7) { 1024 Ξ 2 (mod 7)}
ENDAny Question ?