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2bornot2b

If \(d=gcd(a,b)\), then show that \(gcd(a^2,b^2)=d^2\)

  • 2 years ago
  • 2 years ago

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  1. 2bornot2b
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    @FoolForMath

    • 2 years ago
  2. Rohangrr
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    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   

    • 2 years ago
  3. 2bornot2b
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    @Rohangrr that was not my question. You proved something different from my question.

    • 2 years ago
  4. Rohangrr
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    no its your answer @2bornot2b

    • 2 years ago
  5. 2bornot2b
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    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 "

    • 2 years ago
  6. JuanitaM
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    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.

    • 2 years ago
  7. FoolForMath
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    @JuanitaM: Plagiarism is not a good idea.

    • 2 years ago
  8. FoolForMath
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    @2bornot2b: HINT: Prime factorization.

    • 2 years ago
  9. nikvist
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    • 2 years ago
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  10. 2bornot2b
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    @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?

    • 2 years ago
  11. FoolForMath
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    @2bornot2b: That's not an assumption. It is an (easy) inference.

    • 2 years ago
  12. 2bornot2b
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    Please explain @FoolForMath

    • 2 years ago
  13. 2bornot2b
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    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?

    • 2 years ago
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