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Math Help - Sum of positive divisor

  1. #1
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    Sum of positive divisor

    Need help on the following:

    Show a positive integer n is composite iff \sigma(n) > n + \sqrt{n}.
    \sigma(n) denote the sum of positive divisor of n.

    Thank you for your help.
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  2. #2
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    Quote Originally Posted by kleenex View Post
    Need help on the following:

    Show a positive integer n is composite iff \sigma(n) > n + \sqrt{n}.
    \sigma(n) denote the sum of positive divisor of n.

    Thank you for your help.
    Say n>1.

    n is prime if and only if \sigma (n) = n+1. Thus, n is compositive if and only if \sigma (n) > n+1 \mbox{ or }\sigma (n) < n+1. But we cannot have \simga(n) < n+1 because \sigma (n) > n+1 (since n and 1 are always divisors). Thus, n is compositive if and only if \sigma (n) > n+1. Thus, if \sigma (n) > n+\sqrt{n} >n+1 and so it is compositive.
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  3. #3
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    Thank you,

    I was thinking:
    Let d|n where 1<d<n so 1<n/d<n. If d \leq \sqrt n then n/d \geq \sqrt n but then I don't know what next. But your way is easier than my.

    On the other hand you got syntax error.
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  4. #4
    Super Member PaulRS's Avatar
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    OK, to complete your proof you have to show that if a number n doesn't have a prime divisor less or equal than \sqrt[]{n} it's prime

    Indeed, suppose it doesn't have have a prime divisor less or equal than \sqrt[]{n}, then n has at least two prime divisors p_1,p_2 (if it's composite) such that p_1\cdot{p_2}\leq{n}.

    But p_1,p_2>{\sqrt[]{n}} so the product p_1\cdot{p_2}>n and this is absurd

    So every composite number n has at least one prime divisor p such that p\leq{\sqrt[]{n}}

    But you can only achieve the equality if n is a perfect square, and in that case \sigma(n)\geq{n+\sqrt[]{n}+1}>n+\sqrt[]{n} so the inequality holds

    Otherwise it's not a perfect square and it has a prime divisor p such that p<\sqrt[]{n}
    So \frac{n}{p}>\sqrt[]{n} is a divisor of n and therefore \sigma(n)>n+\sqrt[]{n}
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