Are all perfect squares also solitary numbers?

Question

A perfect square is a number of the form N = x^2, where x is an integer. Solitary numbers are numbers which are not friendly. Friendly numbers, on the other hand, are numbers which constitute a friendly pair (m, n), where sigma(m)/m = sigma(n)/n and sigma is the sum-of-all-positive-divisors function. You can refer to the MathWorld pages for Friendly and Solitary Numbers for more information. From the MathWorld initial list of solitary numbers, it appears that 1, 4, 9, 16, ... are all solitary. Can you prove that all perfect squares are solitary?

Answer 1

Is this an open question?
Here's the info from Wikipedia:
All prime numbers are solitary. It is not known if prime powers are solitary but if they are not, they must be greater than 10^1500 and the power greater than 1400.
So it looks like all prime squares are indeed solitary.
Can this be extended to composite squares?
Update: A sufficient(but not necessary) condition
that n be solitary is (σ(n), n) = 1,
where σ(n) is the sum of the divisors of n.
I have checked all squares less than 40 with PARI
and found that this condition holds for all
n² except n = 14, 21 and 39.
So the first "doubtful" square is 14² = 196.
Its index is 57/28.
I hope to write a program sometime soon
to check this for other squares.

Answer 2

This is a very cool math problem. First, I noticed that the easiest friendly pairs can be if the m and n are related; such as if n = 2m or something. However, this is not applicable to perfect numbers since one of their divisors (the square root of it) appears only once when divisor sigma. So, they can never equal the divisor sigma of m / m. This is the first time I've heard of these solitary numbers and that's all I could think of. I'm sure you probably thought so already. Sorry I couldn't give a better answer.