Name Ihsan ul ghafoor Roll # 08030609-001 Subject Number theory Section A

1. Name Ihsan ulghafoor • Roll # 08030609-001 • Subject Number theory • Section A • Department Methamatics • Semester 4th

2. Topic Different methods of Solving Congruence's

3. Congruence • What 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.

4. Congruence • Solution of congruence:- If ax Ξ b(mod m) • Then solution will belong to following set of numbers X={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.

5. Solution of Congruence • First we find either solution exists or not. If ax Ξ b(mod m) • Then solution exists if and only if d|b where d=gcd(a,m) • The congruence will have ‘d’ solutions.

6. Solution of Congruence • Some methods of solving congruence are following • Trial method:- • Simple method • Put the value one by one from the solution set and check until the congruence is true.

7. Solution of Congruence Example:- 2xΞ 3(mod5) As (2,5)=1 and 1|3 solution exists. As d=1 so congruence has unique solution X={0,1,2,3,4} Now put the value of x one by one 2(0) Ξ 3(mod 5) false 2(1) Ξ 3(mod 5) false 2(2) Ξ 3(mod 5) false 2(3) Ξ 3(mod 5) false 2(4) Ξ 3(mod 5) true So x Ξ 4 (mod 5) is the required solution.

8. Solution of Congruence • Diophantine eq method:- Every congruence relation can be written in the form of linear Diophantine eqs. If ax Ξ b(mod m) m|ax-b ax-b=my ; yЄ Z ax-my=b

9. Solution of Congruence Then find the value of x . And that will be the solution. Further solutions can be find Using formula x’=x-(bt/d) ; d=(a,m)

10. Solution of Congruence Example:- 2xΞ 3(mod5) As (2,5)=1 and 1|3 solution exists. As d=1 so congruence has unique solution 2x – 5y = 3 (in Diophantine form) now we will find the value of x in diophanitne eq.

11. Solution of Congruence As (2,5)=1=d 1=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)

12. Solution of Congruence • Symbolic fraction method :- If ax Ξ b(mod m) Then solution will be xΞ {b+mh (mod m)} / a ; h Є Z Iff a| (b + mh)

13. Solution of Congruence Example:- 2xΞ 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

14. Solution of Congruence Put h = -1 in [eq 1] x Ξ {3 + 5(-1) (mod 5)} /2 x Ξ {-2 (mod 5)} /2 x Ξ -1 (mod 5) x Ξ 4 (mod 5) { -1 Ξ 4 (mod 5) } as { -1 = 5(-1) + 4 }

15. Solution of Congruence • By Fermate theorem :- If ax Ξ b(mod m) Then solution will be xΞ b aФ(m)-1 (mod m) Ф(n) is Euler phi function.

16. Solution of Congruence Example:- 2xΞ 3(mod 5) As (2,5)=1 and 1|3 solution exists. As d=1 so congruence has unique solution xΞ 3.2Ф(5)-1 (mod 5) xΞ 3.24-1 (mod 5) xΞ 3.23 (mod 5) xΞ 24 (mod 5) x Ξ 4 (mod 5) { 24 Ξ 4 (mod 5) }

17. Solution of Congruence • Let 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.

18. Solution of Congruence 4 x Ξ 1 (mod 7) • By trial method:- Put value of x one by one from the set X = {0,1,2,3,4,5,6} x Ξ 2 (mod 7)

19. Solution of Congruence 4 x Ξ 1 (mod 7) • By Diophantine eq method :- 4 x - 7 y = 1 gcdlinear combination 7=4(1)+31=4(1)-3(1) 4=3(1)+11=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)

20. Solution of Congruence 4 x Ξ 1 (mod 7) • By symbolic fraction method:- x Ξ {1 + 7 h(mod 7)} / 4 eq 1 As 4 | 1 + 7 h 1 + 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 2 3 | -1 + 4k -1 + 4k Ξ 0 (mod 3) -1 + k Ξ 0 (mod 3) 4k Ξ k (mod 3) k Ξ 1 (mod 3)

21. Solution of Congruence • Put k =1 in eq 2 • h Ξ {-1 + 4(1) (mod 4) } / 3 • h Ξ {3 (mod 4) } / 3 • h Ξ 1 (mod 4) • Put h =1 in eq 2 • x Ξ {1 + 7 (1)(mod 7)} / 4 • x Ξ {8 (mod 7)} / 4 • x Ξ 2 (mod 7) • Required solution

22. Solution of Congruence 4 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) { 1024Ξ 2 (mod 7)}

23. END • Any Question ?