Is $\frac{4x^2 - 1}{3x^2}$ an abundancy outlaw? Let $\sigma(x)$ denote the sum of the divisors of $x$.  For example, $\sigma(6) = 1 + 2 + 3 + 6 = 12$.
We call the ratio $I(x) = \sigma(x)/x$ the abundancy index of $x$.  A number $y$ which fails to be in the image of the map $I$ is said to be an abundancy outlaw.
My question is:

Is $$\frac{4x^2 - 1}{3x^2}$$ an abundancy outlaw?

In this paper, it is mentioned that if $r/s$ is an abundancy index with $\gcd(r,s) = 1$, then $r \geq \sigma(s)$.
Note that 
$$\gcd(4x^2 - 1, 3x^2) = \gcd(x^2 - 1, 3x^2) = 1$$
if $x \not\equiv 1 \pmod 3$.
Hence, if $(4x^2 - 1)/3x^2$ is an abundancy index with $x \not\equiv 1 \pmod 3$, then $4x^2 - 1 \geq \sigma(3x^2)$.
Here is where I get stuck.  (I would like to affirmatively answer my question via the contrapositive.)
Lastly, if $(4x^2 - 1)/3x^2$ were an abundancy index $I(y)$, then
$${3x^2}\sigma(y) = y(4x^2 - 1),$$
so that
$${3x^2}\left(\sigma(y) - y\right) = y(x^2 - 1).$$
Since $\gcd(x^2, x^2 - 1) = 1$, I am sure that $x^2 \mid y$ and $(x^2 - 1) \mid \left(3(\sigma(y) - y))\right)$.  So I can write
$$y = kx^2$$
and
$$3(\sigma(y) - y) = m(x^2 - 1).$$
I am not too sure though, how this can help in answering my question.
Update [October 10, 2015 - 1:05 PM]:  Additionally, $I(y) = (4x^2 - 1)/3x^2 < 4/3$, so that $3(\sigma(y) - y) < y$.

A partial answer.
Let $I(y)=\frac{4x^2 - 1}{3x^2}$.
If $\gcd(4x^2 - 1, 3x^2) = 1$ then $3$ is a factor of $y$ contradicting the fact that $I(y)<\frac{4}{3}$.
Therefore $\gcd(4x^2 - 1, 3x^2) = 3$.
