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 $3(\sigma(y) - y) < y$.
