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May 7th, 2016, 06:50 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 =$\displaystyle =\prod{\frac{p_i^{{\alpha}_i+1}1}{p_i1}}$ 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, 08:42 PM  #2 
Senior Member Joined: Feb 2012 Posts: 144 Thanks: 16 
I think you are counting 1 several times.

May 8th, 2016, 08:11 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,912 Thanks: 1110 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, 01:43 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 11:53 PM. 

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divisors, numbers, perfect, square, sum, umbers 
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