Divisor Square Sum

For a positive integer $n$, let $\sigma_2(n)$ be the sum of the squares of its divisors. For example,
$$\sigma_2(10) = 1 + 4 + 25 + 100 = 130.$$
Find the sum of all $n$, $0 \lt n \lt 64\,000\,000$ such that $\sigma_2(n)$ is a perfect square.

We use the relation $\sigma_2(n) = \sum_{d|n}d^2$ where $d$ are the divisors of $n$. We are looking for $n$ such that $\sigma_2(n)$ is a perfect square. Let’s first explore how the sum of squares of divisors would look like for different kinds of $n$.

For a prime number $p$, the divisors are just $1$ and $p$, so $\sigma_2(p) = 1 + p^2$ , which is not a perfect square unless $p = 1$ or $2$.

For any number $n=p^k$ (where $p$ is a prime number and $k$ an integer), the divisors of $n$ are $1, p, p^2, … , p^k$ and so $\sigma_2(n) = 1 + p^2 + p^4 + … + p^{2k}$. This is a geometric series with sum $= \frac{p^{2k+2}-1}{p^2-1}$. We see that for an odd prime $p$, this is hardly ever a perfect square. However, if $p=2$, we see that the sum is a perfect square when $k=1,3,5,7$. Thus, $n= 2, 8, 32, 128$ are answers.

If $n=p^k*q^l$ (where $p$ and $q$ are primes and $k,l$ are integers), the divisors would be combinations of powers of $p$ and $q$, and the sum of squares of divisors wouldn’t be so “simple”, and it will not be a perfect square.

We can cover rest of the numbers by using the formulas above and adding them up in some way. However, as we can see, the sum of squares of divisors of a number $n = p^k$ where $p$ is a prime number is already hard to be a perfect square, let alone for larger $n$ which have more factors. So it seems that we already covered all possible $n$ such that $\sigma_2(n)$ is a perfect square, i.e., $n = 2, 8, 32, 128$.

Summing up these four numbers, we get $2+8+32+128 = 170$.

Therefore, the sum of all $n$, $0 \lt n \lt 64\,000\,000$ such that $\sigma_2(n)$ is a perfect square is $170$.
This problem can be solved using more advanced number theory, involving Dirichlet convolutions and multiplicative functions, but this solution should be enough for most purposes.

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