Is it possible to simplify this nested GCD?

Question

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.

Answer 1

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