Why is proving that $10$ is solitary considered very difficult?

Question

The title says it all.

We denote the sum of the divisors of $x$ by $\sigma(x)$. The ratio $I(x)=\sigma(x)/x$ is called the abundancy index of $x$. If $I(m)=I(n)$, then $\{m,n\}$ is called a friendly pair. A number which is not a member of a friendly pair is called solitary.

So:

Why is proving that $10$ is solitary considered very difficult?

Is there no (known) partial converse to Greening's Theorem (i.e., $\gcd(N, \sigma(N)) = 1 \implies N$ is solitary)? What is the main obstruction?

OEIS sequence A014567 contains the following statement from Dean Hickerson:

It is easy to show that if $N$ and $\sigma(N)$ are relatively prime then $N$ is solitary. But the converse is not true; for example, $18$, $45$, $48$ and $52$ are solitary. Probably also $10$, $14$, $15$, $20$, $22$ and many others are solitary, but I do not think that will ever be proved.

Update (November 11, 2015): I am aware of the paper Does Ten Have a Friend?, which gives some necessary conditions for a number to be a friend of $10$. This question is essentially an offshoot of the following earlier MSE posts:

1. Can distinct odd perfect numbers $N = {p^k}{m^2}$ share the same Euler factor $p^k$?
2. If $\sigma(N) = aN + b$, where $\gcd(a, b) = 1$, does it follow that at least one of $N$'s factors is solitary?
3. Is $p(p + 1)$ always a friendly number for $p$ a prime number?