12. Prove that if gcd(a, n) { b, then ax = b (mod n) has no solutions.

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Chapter2: Second-order Linear Odes
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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 congruence
ax = 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.
Transcribed Image Text:Prove Exercise 4.6.12 in the course notes. Then, assuming that ged(a, n)|b, determine how many solutions (in Z„) the congruence ax = 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 congruence
2z = 6 (mod 10).
and express your answer using congruence classes of the original modulus.
Solution
We can apply Proposition 4.2.14 here to divide through by 2, and we get
Proposition 4.2.14. Dividing both sides of a Congruence. Let
ne Nand a, b,d e Z lif ad = bd (mod n). then
ab (mod
ged(d, n)
If d and n are relatively prime, then ad = bd (mod n) implies
a =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 get
d
= b-
+k.
god(d, n)
ged(d, n)
god(d, n)
This is equivalent to
d
god(d, n)
|(a - b)
god(d, n)
Since god
1. we can apply the result of
god(d, n)' god(d, 71)
Checkpoint 1.3.13 to obtain
Checkpoint 1.3.13. Another Divisibility Property. Let
m, a, beN. Using Theorem 1.3.7. prove that if m| ab and
god(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-context
Jknowinhm-beout.htm
in-context
Sknowles-review-botuclhem
ged(d, n)
(a - b).
or that a= b (mod
as desired.
gad(d, n)).
If god(d, n) =1, the statement reduces to a =b (mod n).
!!
knowliprool-6-hidden
in-context
knowliprop-cong-division.himi
3 (mod 5) -
Modulo 10, this is the same as
z= 3 (mod 10) and 1 = 8 (mod 10).
So the original congruence has two solutions in {[0], [1],..., [9]}, namely (3]
and (8).
Transcribed Image Text:Example 4.3.2. Solve for z. Solve the congruence 2z = 6 (mod 10). and express your answer using congruence classes of the original modulus. Solution We can apply Proposition 4.2.14 here to divide through by 2, and we get Proposition 4.2.14. Dividing both sides of a Congruence. Let ne Nand a, b,d e Z lif ad = bd (mod n). then ab (mod ged(d, n) If d and n are relatively prime, then ad = bd (mod n) implies a =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 get d = b- +k. god(d, n) ged(d, n) god(d, n) This is equivalent to d god(d, n) |(a - b) god(d, n) Since god 1. we can apply the result of god(d, n)' god(d, 71) Checkpoint 1.3.13 to obtain Checkpoint 1.3.13. Another Divisibility Property. Let m, a, beN. Using Theorem 1.3.7. prove that if m| ab and god(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-context Jknowinhm-beout.htm in-context Sknowles-review-botuclhem ged(d, n) (a - b). or that a= b (mod as desired. gad(d, n)). If god(d, n) =1, the statement reduces to a =b (mod n). !! knowliprool-6-hidden in-context knowliprop-cong-division.himi 3 (mod 5) - Modulo 10, this is the same as z= 3 (mod 10) and 1 = 8 (mod 10). So the original congruence has two solutions in {[0], [1],..., [9]}, namely (3] and (8).
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