A community for students.
Here's the question you clicked on:
 0 viewing
2bornot2b
 3 years ago
If \(d=gcd(a,b)\), then show that \(gcd(a^2,b^2)=d^2\)
2bornot2b
 3 years ago
If \(d=gcd(a,b)\), then show that \(gcd(a^2,b^2)=d^2\)

This Question is Closed

Rohangrr
 3 years ago
Best ResponseYou've already chosen the best response.0Given gcd(a, b) = d By linear combination of two numbers as + bt = d [where s and t are integers] Divide both sides by d as/d + bt/d = 1 Rearrange the equation s(a/d) + t(b/d) = 1 ==> gcd(a/d, b/d) = 1

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0@Rohangrr that was not my question. You proved something different from my question.

Rohangrr
 3 years ago
Best ResponseYou've already chosen the best response.0no its your answer @2bornot2b

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0You need to show that the gcd of \(a^2, b^2\) is equal to \(d^2\), and what have you shown? You have shown " gcd(a/d, b/d) = 1 "

JuanitaM
 3 years ago
Best ResponseYou've already chosen the best response.1Let a and b be integers, not both zero. Assume that gcd(a, b) = 1. Set g = gcd(a2, b2). I need to prove that g = 1. By Theorem 2.1.3 there exist integers x and y such that ax + by = 1. Now do some algebra 1 = 1^3 = (ax + by)^3 = a^3x^3 + 3a^2x^2by + 3axb^2y^2 + b^3y^3 = a^2(ax^3 + 3x^2by) + b^2(3axb^2y^2 + by^3). Set u = ax^3 +3x^2by and v = 3axb^2y^2 +by^3. Thus a^2u+b^2v = 1. Since g = gcd(a^2, b^2), there exists, t ∈ Z such that a^2 = gs and b^2 = gt. Hence 1 = a^2u + b^2v = gsu + gtv = g(su + tv), that is g1. Since g > 0 we conclude g = 1. Now assume that d = gcd(a, b) > 1. Then there exist j, k ∈ Z such that a = dj and b = dk. By Proposition 2.2.5 it follows that gcd(j, k) = 1. By the first part of this proof it follows that gcd(j^2, k^2) = 1. Since a^2 = d^2j^2 and b^2 = d^2k^2 and since j^2 and k^2 are relatively prime, Lemma 1 implies that gcd(a^2, b^2) = gcd(d^2j^2, d^2k^2) = d2.

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1@JuanitaM: Plagiarism is not a good idea.

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1@2bornot2b: HINT: Prime factorization.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0@nikvist you say that gcd(p, q) = 1, why do you say that? I mean did you assume it? Can you make such an assumption?

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1@2bornot2b: That's not an assumption. It is an (easy) inference.

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Please explain @FoolForMath

2bornot2b
 3 years ago
Best ResponseYou've already chosen the best response.0Let d be the gcd of \(a,b\) then there exist x and y such that \(ax+by=d\implies \frac{a}{d}x+\frac{b}{d}y=1\). So \(\frac{a}{d}\) and \(\frac{b}{d}\) are prime to each other. Is this the reason behind that step of nikvist?
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.