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 May 7th, 2016, 05:50 AM #1 Newbie   Joined: May 2014 From: India Posts: 7 Thanks: 0 Numbers x such that the sum of the divisors is a perfect square. Hello I am reading "The Theory of Numbers, by Robert D. Carmichael" and stuck in an exercise problem, Find numbers x such that the sum of the divisors of x is a perfect square. I know sum of divisors of a x=$\displaystyle x=p_1^{{\alpha}_1}.p_2^{{\alpha}_2}...p_n^{{\alpha }_i}$ is Sum of divisors =$\displaystyle =\prod{\frac{p_i^{{\alpha}_i+1}-1}{p_i-1}}$ But couldn't proceed further on how resolve the product in to X2 It will be helpful if someone supply some hints
 May 7th, 2016, 07:42 PM #2 Senior Member   Joined: Feb 2012 Posts: 144 Thanks: 16 I think you are counting 1 several times.
 May 8th, 2016, 07:11 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond 1, 3, 15, 105, 945 ... depending on how you factor them and using only proper divisors.
 May 9th, 2016, 03:40 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,655 Thanks: 2087 Also 12, as 1 + 2 + 3 + 4 + 6 = 16 = 4². However, the problem states "divisors", not "proper divisors". As the problem doesn't ask for all numbers, it would suffice to provide 1, 3 and 22 (see here).
 May 9th, 2016, 12:43 PM #5 Newbie   Joined: May 2014 From: India Posts: 7 Thanks: 0 Thanks everyone; I have seen the OEIS series, skipjack. I was wondering whether there is any general form for such numbers, as OEIS doesn't seem to mention it. Last edited by skipjack; December 13th, 2016 at 10:53 PM.

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