How do you prove 12? Prove Exercise 4.6.12 in the course notes. Then, assuming that ged(a, n)|b, determine how many solutions(in Z„) the congruenceax = b (mod n)has. Explain your reasoning and provide examples to support your claim.(Hint: Look at Example 4.3.2.)12. Prove that if gcd(a, n) † b, then ax = b (mod n) has no solutions. Example 4.3.2. Solve for z. Solve the congruence2z = 6 (mod 10).and express your answer using congruence classes of the original modulus.SolutionWe can apply Proposition 4.2.14 here to divide through by 2, and we getProposition 4.2.14. Dividing both sides of a Congruence. Letne Nand a, b,d e Z lif ad = bd (mod n). thenab (modged(d, n)If d and n are relatively prime, then ad = bd (mod n) impliesa =b (mod n).Proof.Suppose that ad= bd (mod n). This means ad = bd + kn for some keZ.Since god(d, n) divides both dand n, we can divide through to getd= b-+k.god(d, n)ged(d, n)god(d, n)This is equivalent todgod(d, n)|(a - b)god(d, n)Since god1. we can apply the result ofgod(d, n)' god(d, 71)Checkpoint 1.3.13 to obtainCheckpoint 1.3.13. Another Divisibility Property. Letm, a, beN. Using Theorem 1.3.7. prove that if m| ab andgod(a, m) = 1, then m|b.Theorem 1.3.7. Bezout's Identity. Let a, be Z not both zero.Then there exist m,neZ such that am + bn = gcd(a, b).in-contextJknowinhm-beout.htmin-contextSknowles-review-botuclhemged(d, n)(a - b).or that a= b (modas desired.gad(d, n)).If god(d, n) =1, the statement reduces to a =b (mod n).!!knowliprool-6-hiddenin-contextknowliprop-cong-division.himi3 (mod 5) -Modulo 10, this is the same asz= 3 (mod 10) and 1 = 8 (mod 10).So the original congruence has two solutions in {[0], [1],..., [9]}, namely (3]and (8).

Name: How do you prove 12? Prove Exercise 4.6.12 in the course notes. Then,
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Description: How do you prove 12? Prove Exercise 4.6.12 in the course notes. Then, assuming that ged(a, n)|b, determine how many solutions(in Z„) the congruenceax = b (mod n)has. Explain your reasoning and provide examples to support your claim.(Hint: Look at Exa