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# Thread: Sum of divisors of prime

1. 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 $\displaystyle x = p_1^{{\alpha}_1}.p_2^{{\alpha}_1}...p_n^{{\alpha}_1}$ 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 $\displaystyle X^2$

It will be helpful if someone supply some hints

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