Is it possible to simplify this nested GCD? Is it possible to simplify this nested GCD?

$$\gcd\bigg(\gcd(m^2,\sigma(m^2)),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg)$$

Here, $\gcd(m^2,\sigma(m^2))>1$ and $\sigma(m^2)$ is the sum of divisors of $m^2$.
I tried using WolframAlpha, but it appears to evaluate the GCD erroneously to
$$\gcd\bigg(\gcd(m^2,\sigma(m^2)),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg) = 1.$$
This is because I know from a published result that the following must hold for the problem that I am considering:
$$\gcd\bigg(\gcd(m^2,\sigma(m^2)),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg) > 1.$$
Updated (March 2 2019)
I tried to evaluate the simpler expression
$$\gcd(m^2,\sigma(m^2))$$
using WolframAlpha, and obtained
$$\gcd(m^2,\sigma(m^2)) = 1,$$
which I know to be false.  Hence, it appears that my problem cannot be solved using WolframAlpha alone.
I have therefore removed the wolfram-alpha and computer-algebra-systems tags.
 A: Here is my attempt:
By GCD associativity,
$$\gcd\bigg(\gcd(m^2,\sigma(m^2)),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg)=\gcd\bigg(\sigma(m^2),\gcd\left(m^2,\frac{m^2}{\gcd(m^2,\sigma(m^2))}\right)\bigg).$$
Next, since
$$\frac{m^2}{\gcd(m^2,\sigma(m^2))} \mid m^2,$$
then
$$\gcd\left(m^2,\frac{m^2}{\gcd(m^2,\sigma(m^2))}\right)=\frac{m^2}{\gcd(m^2,\sigma(m^2))},$$
so that we obtain
$$\gcd\bigg(\sigma(m^2),\gcd\left(m^2,\frac{m^2}{\gcd(m^2,\sigma(m^2))}\right)\bigg)=\gcd\bigg(\sigma(m^2),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg).$$
Using the formula $\gcd(na,nb)=n\gcd(a,b)$, we get
$$\gcd\bigg(\sigma(m^2),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg)=\frac{\gcd\bigg(\sigma(m^2)\gcd(m^2,\sigma(m^2)),m^2\bigg)}{\gcd(m^2,\sigma(m^2))}=\frac{\gcd\bigg(\gcd(m^2 \sigma(m^2),(\sigma(m^2))^2),m^2\bigg)}{\gcd(m^2,\sigma(m^2))}.$$
Again, by GCD associativity, we obtain
$$\frac{\gcd\bigg(\gcd\left(m^2 \sigma(m^2),(\sigma(m^2))^2\right),m^2\bigg)}{\gcd(m^2,\sigma(m^2))}=\frac{\gcd\bigg(\gcd(m^2 \sigma(m^2),m^2),(\sigma(m^2))^2\bigg)}{\gcd(m^2,\sigma(m^2))}.$$
Now, since
$$m^2 \mid m^2 \sigma(m^2)$$
we get
$$\gcd(m^2 \sigma(m^2),m^2) = m^2$$
so that we finally have
$$\frac{\gcd\bigg(\gcd(m^2 \sigma(m^2),m^2),(\sigma(m^2))^2\bigg)}{\gcd(m^2,\sigma(m^2))} = \frac{\gcd\bigg(m^2, (\sigma(m^2))^2\bigg)}{\gcd(m^2,\sigma(m^2))} = \frac{\bigg(\gcd(m,\sigma(m^2))\bigg)^2}{\gcd(m^2,\sigma(m^2))}.$$
