Diskmat kapittel 4 del 2

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What are the solutions of the linear congruence 3x≡4(mod7)?

1. We find the inverse of 3 mod 7 using the extended euclidian algorithm 7 = 3*2 + 1 3 = 1*3 + 0 1 = 7-2*3 The inverse is -2. 2. We multiply both sides of the congruence by the inverse -2*3x≡-2*4(mod7) -6x≡-8(mod7) -6 ≡ 1≡(mod7) Therefore, we get x≡-8(mod7) x = 6,13,20,27... and so on. x can also be negative

Find 7²²²(mod 11)

11 is prime 7 is not divisible by 11 Therefore, we can use Fermat's little theorem. We know that 7¹⁰≡1(mod 11) So (7¹⁰)ⁿ≡1(mod 11) for every positive integer k. 222=22*10+2 7²²²=(7²²)¹⁰7²≡(1)²²49≡5(mod 11) 7²²²(mod 11)=5.

What is an inverse of a (mod) m?

An integer such that (image)

What is a linear congruence?

It is a congruence of the form ax≡b(mod m) Where m is a positive integer, a and b are integers, and x is a variable.

Use the chinese remainder theorem to find x when you know that x≡2(mod3) x≡3(mod5) x≡2(mod7)

Since the numbers 3, 5 and 7 are pairwise relatively prime, we can use the chinese remainder theorem to solve the congruence equation set. If all the mods (3, 5 and 7) are all pairwise relatively prime, we can use the chinese rest theorem. This means that gcd(3,5)=gcd(5,7)=gcd(3,7)=1 1. We set a "global modulo" m = 3*5*7 = 105 2. We set M₁=m/3=35, M₂=m/5=21 and M₃=m/7=15. 3. To each of these Ms, we find an inverse. The inverse of 35(mod3) is 2=y₁ The inverse of 21(mod5) is 1=y₂ The inverse of 15(mod7) is 1=y₃ 4. x≡a₁M₁y₁+a₂M₂y₂+a₃M₃y₃ x≡2*35*2+3*21*1+2*15*1 = 233 233≡23(mod105) 23 is the smallest positive integer that is a solution for the equation set. So 23 is the smallest positive integer that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 5, and a remainder of 2 when divided by 7.

When does an inverse of a mod m exist?

Whenever a and m are relatively prime, and m>1.

Find an inverse of 3 modulo 7

gcd(3,7)=1, so an inverse of 3 modulo 7 must exist. Etended euclidian algorithm: 7 = 3*2 + 1 3 = 1*3 + 0 1 = 7-2*3 The inverse of a mod m is a number b so that ba≡1(modm) a = 3, so b = -2 So you always have to look for the factor of a when finding bezout coefficients, and that factor will be the inverse.


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