How Can You Find Numbers Whose Sum of Divisors is a Perfect Square?

[mathworker]

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 into \(\displaystyle X^2\)

It will be helpful if someone supply some hints :)

 

[jvicens]

First, let's rewrite the formula for the sum of divisors in a different way:

Sum of divisors = (1 + p_1 + p_1^2 + ... + p_1^{{\alpha}_1}) * (1 + p_2 + p_2^2 + ... + p_2^{{\alpha}_2}) * ... * (1 + p_n + p_n^2 + ... + p_n^{{\alpha}_n})

Now, let's focus on just one of the factors in the product, say (1 + p_i + p_i^2 + ... + p_i^{{\alpha}_i}). We can rewrite this as:

(1 + p_i + p_i^2 + ... + p_i^{{\alpha}_i}) = (1 + p_i + p_i^2 + ... + p_i^{{\alpha}_i-1}) * (1 + p_i^{{\alpha}_i})

Note that the first factor on the right side is just the sum of the divisors of p_i^{{\alpha}_i-1}. So, if we can find a number x such that the sum of its divisors is a perfect square, then we can also find a number x*p_i^{{\alpha}_i} such that the sum of its divisors is also a perfect square.

Now, let's look at the second factor on the right side, (1 + p_i^{{\alpha}_i}). This is just a geometric series with a common ratio of p_i. So, we can rewrite it as:

(1 + p_i^{{\alpha}_i}) = (p_i^{{\alpha}_i+1} - 1) / (p_i - 1)

If we set this equal to a perfect square, say k^2, we get the equation:

(p_i^{{\alpha}_i+1} - 1) / (p_i - 1) = k^2

Solving for p_i, we get:

p_i = (k^2 + 1) / (k - 1)

Now, we can plug in different values of k and find corresponding values of p_i and x. For example, if we let k = 2, we get p_i = 3 and x = 9. This means that the sum of divisors of