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May 7th, 2016, 05:50 AM   #1
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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
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May 7th, 2016, 07:42 PM   #2
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I think you are counting 1 several times.
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May 8th, 2016, 07:11 PM   #3
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1, 3, 15, 105, 945 ... depending on how you factor them and using only proper divisors.
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May 9th, 2016, 03:40 AM   #4
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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).
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May 9th, 2016, 12:43 PM   #5
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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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