2bornot2b
  • 2bornot2b
If \(d=gcd(a,b)\), then show that \(gcd(a^2,b^2)=d^2\)
Mathematics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

2bornot2b
  • 2bornot2b
If \(d=gcd(a,b)\), then show that \(gcd(a^2,b^2)=d^2\)
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

2bornot2b
  • 2bornot2b
anonymous
  • anonymous
Given 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
  • 2bornot2b
@Rohangrr that was not my question. You proved something different from my question.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
no its your answer @2bornot2b
2bornot2b
  • 2bornot2b
You 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 "
anonymous
  • anonymous
Let 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 g|1. 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.
anonymous
  • anonymous
@JuanitaM: Plagiarism is not a good idea.
anonymous
  • anonymous
@2bornot2b: HINT: Prime factorization.
nikvist
  • nikvist
1 Attachment
2bornot2b
  • 2bornot2b
@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?
anonymous
  • anonymous
@2bornot2b: That's not an assumption. It is an (easy) inference.
2bornot2b
  • 2bornot2b
Please explain @FoolForMath
2bornot2b
  • 2bornot2b
Let 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.