Please answer in exercise 7.2 number 2 Only al xb in Z/nZ are as many as are the

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Please answer in exercise 7.2 number 2 Only

al xb in Z/nZ are as many as are the elements of Z/dZ. This finishes the proof of item (2. As usual, the case when pis a prime number is very simple: if a [0] in Z/pZ, then gcd(a, p) = 1, so the solution to laj.X-?is unique. Of course, if [al is the reciprocal of [a], which exists since [a] [0], then, that (unique) solution is given by Exercise 7.2 (1) Solve the congruences (i)AX 2 mod 6 () 4x1 mod 6 (iii) 4X 4 mod 6 (e) 256X 179 mod 337 iv) 3X800 mod 11 (vi) 1215X560 mod 2755. (2) Let p be a prime number, and a, b be integers such that 1 S a Sp-1. Prove that the congruence ax b mod p has a solution a-1mod p. Equations of Higher Degree in Z/nZ 7.2 7.2.1 General Remarks and Notations

Explanation / Answer

1.a) For solving linear congruences

ax?b (mod m) , where x is an unknown integer

so here a=4,b=2,m=6

Using Euclid's Algorithm

6(-l)+4x=2

applying gcd(6,4)=2

6(-l)+4*2=2

-6l=2-8

-6l=-6

l=1

so, x=2

b)4X?1 mod 6

here a=4,b=1,m=6

gcd(6,4)=2

4X?1 mod 6 has no solution because 2 does not divide 1.

c)4X?4 mod 6

here a=4,b=4,m=6

gcd(6,4)=2

4X?4 mod 6 has no solution because 2 does not divide 4.

d)3X?800 mod 11

here a=3,b=800,m=11

gcd(3,11)=1

3X?800 mod 11 has no solution because 1 does not divide 800.

e)256X?179 mod 337

here a=256,b=179,m=337

gcd(256,337)=1

256X?179 mod 337 has no solution because 1 does not divide 179.

f)1215X?560 mod 2755

gcd(1215,2755)=5

5 divides 560.So,X=5.

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