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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.

Jose Arnaldo Bebita Dris
  • 8,471
  • Am I right in thinking that the nested GCD in question can be simplified using GCD associativity as follows? $$\gcd\bigg(\gcd(m^2,\sigma(m^2)),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg) = \gcd\bigg(\sigma(m^2),\gcd\left(m^2,\frac{m^2}{\gcd(m^2,\sigma(m^2))}\right)\bigg) = \gcd\bigg(\sigma(m^2),\frac{m^2}{\gcd(m^2,\sigma(m^2))}\bigg)?$$ – Jose Arnaldo Bebita Dris Mar 02 '19 at 12:02
  • The $\gcd$ isn't greater than one, and I believe WolframAlpha is not wrong, I tested for $m=1,; 2,;3,;\cdots,;7$ and both WolframAlpha results were right, of course this isn't a proof, but at least gives a counter example to your claim that the $\gcd$ is greater than $1$ – Bruno Andrades Mar 02 '19 at 14:37
  • @BrunoAndrades: Thank you for your comment.

    I used Sage Cell Server and found the following examples for $\gcd(m^2,\sigma(m^2))>1$ for $m < 100$:

    $$14, 21, 39, 42, 57, 63, 70, 77, 78, 84, 93, 98, 99$$

     – Jose Arnaldo Bebita Dris Mar 02 '19 at 22:40
  • Well that was unexpected, I guess neither WolframAlpha or the OP is right – Bruno Andrades Mar 02 '19 at 22:42
  • @BrunoAndrades: I used the following GP code in Sage Cell Server to produce the above examples: for (x=1, 100,if(gcd(x^2,sigma(x^2))>1,print(x))). – Jose Arnaldo Bebita Dris Mar 02 '19 at 22:43
  • I also used the following Mathematica code to double-check the exact same results by Sage Cell Server: Select[Range[10^2], GCD[#^2, DivisorSigma[1, #^2]] > 1 &] – Jose Arnaldo Bebita Dris Mar 02 '19 at 22:44
  • @BrunoAndrades: Well, actually, what makes the computations more difficult is the fact that it is known that $m > {10}^{375}$, for the particular problem that I am considering. – Jose Arnaldo Bebita Dris Mar 02 '19 at 22:48

1 Answers1

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

Jose Arnaldo Bebita Dris
  • 8,471