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Sum of divisors

For each positive integer n, let \(\sigma(n)\) denote the sum of all its positive divisors (including 1 and itself).

Determine all positive integers \(n > 1\), such that there doesn't exist any positive integer m that satisfies: \(n < m^2 \leq \sigma(n) \).

Note by Tomás Carvalho
1 year, 11 months ago

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For each positive integer \(n\), let \(\sigma(n)\) be the sum of its divisors (including 1 and \(n\)). Determine all positive integers \(n > 1\) such that there doesn't exist any positive integer \(m\) that satisfies \(n < m^2 \le \sigma(n)\).

We claim the only such numbers are the prime numbers and the square of prime numbers, excluding 3.

First, if \(p\) is a prime number, then \(\sigma(p) = p+1\) and \(\sigma(p^2) = p^2+p+1\). Then any such \(m\) satisfying \(p < m^2 \le \sigma(p)\) must satisfies \(m^2 = p+1\), or \(p = m^2 - 1\). Except for \(p = 3\), this is a composite number (it has factors \(m-1, m+1\)). Any such \(m\) satisfying \(p^2 < m^2 \le \sigma(p)\) must satisfy \(m \ge p+1\) because \(m^2 > p^2\), but this leads to \(m^2 = p^2+2p+1 > p^2+p+1 = \sigma(p)\), contradiction.

Now, \(n = 3\) is a special case. Otherwise, if \(n\) is not a prime number or the square of one, then \(n\) has at least two additional distinct factors \(a,b\) which multiply to \(n\). By AM-GM inequality, \(\frac{a+b}{2} \ge \sqrt{ab}\) or \(a+b \ge 2\sqrt{n}\); since \(a \neq b\), equality is not achieved, so \(a+b > 2\sqrt{n}\). Thus \(\sigma(n) \ge 1+a+b+n > n+2\sqrt{n}+1 = (\sqrt{n}+1)^2\). But there must be a square number \(m^2\) between \((\sqrt{n}^2, (\sqrt{n}+1)^2]\); this satisfies \(n = \sqrt{n}^2 < m^2 \le (\sqrt{n}+1)^2 \le \sigma(n)\), contradiction.

Ivan Koswara - 1 year, 11 months ago

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@Ivan Koswara wow! nice solution ... how did yo think of tht ... Never seen such a problem .... Could yo provide links to such problems ..... Thanks :)

Abhinav Raichur - 1 year, 11 months ago

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It's an Olympiad Problem, I don't know the link to this one in particular but if you search Math Olympiads online I'm sure you'll find many challenging problems. (If you're looking for even harder ones search IMO problems)

Tomás Carvalho - 1 year, 11 months ago

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@IvaThank you very much, it's a very nice solution. I had been able to prove it doesn't work for primes except for 3 and for squares of primes, but i just couldn't prove that it works for every other number. Thanks! ;)

Tomás Carvalho - 1 year, 11 months ago

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@Ivan Koswara Oh and if you like these problems, here's a harder one (I have literally made no progress on it): Let n be a positive integer. We mean to paint every integer from 1 to n in blue or red. We say that the coloration is "balanced" if the following property is verified: if a and b are integers between 1 and n, painted in different colors, then, if a+b is also an integer between 1 and n, a+b is painted in blue, and if ab is also an integer between 1 and n, ab is painted in red. Let N be the number of "balanced" colorations of the integers from 1 to n. Let P be the number of prime numbers p with the property that the largest integer smaller or equal to n/p is a power of p with a postive integer exponent. Prove that: N=1+n+P.

Tomás Carvalho - 1 year, 11 months ago

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@Aareyan Manzoor Can you help me solve these two other problems I've posted? Thank you.

Tomás Carvalho - 1 year, 11 months ago

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(23)<(24)= (an>af) can you help me to solve that problems? thank you

Young Ma - 6 months ago

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wow i like math can you teach me?

Young Ma - 6 months ago

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