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Math Help - Abundant numbres proof

  1. #1
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    Abundant numbres proof

    Prove that there exist infinitely many odd abundant numbers.

    Proof.

    Consider the integers n = 945k, where k is any positive integers not divisible by 2,3,5, or 7.

    Since 945 = (3^3)(5)(7), so gcd(945,k)=1, implies that  \sigma (n) = \sigma (945) \sigma (k) = \sigma (3^3) \sigma (5) \sigma (7) \sigma (k) = \frac {3^4-1}{3-1} \frac {5^2-1}{5-1} \frac {7^2-1}{7-1} \sigma (k) = 1920 \sigma (k)

    Therefore I have  \sigma (945k) = 1920 \sigma (k)

    Now, I claim that  \sigma (945k) =  1920 \sigma (k) > (2)945k = 1890k \ \ \ \ \ \forall k \in \mathbb {N} , and I'm using induction to prove it.

    Certainly, 1920 \sigma (1) = 1920 > 1890 , so the claim is true when k=1.

    Suppose the claim is true for k=t, we then have  1920 \sigma (t) > 1890t

    Now, I want to show that (1920) \sigma (t+1) = 1890 (t+1) , how would I do that?

    Thanks.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Prove that there exist infinitely many odd abundant numbers.

    Proof.

    Consider the integers n = 945k, where k is any positive integers not divisible by 2,3,5, or 7.

    Since 945 = (3^3)(5)(7), so gcd(945,k)=1, implies that  \sigma (n) = \sigma (945) \sigma (k) = \sigma (3^3) \sigma (5) \sigma (7) \sigma (k) = \frac {3^4-1}{3-1} \frac {5^2-1}{5-1} \frac {7^2-1}{7-1} \sigma (k) = 1920 \sigma (k)
    All you need to show is that \sigma (n) > 2n. Now \sigma(n) = 1920 \sigma(k) and 2n = 1890k. Since \sigma(k) > k it immediatelly completes the proof.
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