Do odd numbers $n$ satisfying $\gcd(n, \sigma(n)) > \sqrt{n}$ have a special form?

Question

Let $\sigma(n)$ denote the sum of divisors of the positive integer $n$.

Using Sage Cell Server, I was able to get the following odd numbers $n < 5000$ satisfying $\gcd(n, \sigma(n)) > \sqrt{n}$:

$$117 = {{3}^2}\cdot{13}$$ $$135 = {{3}^3}\cdot{5}$$ $$585 = {{3}^2}\cdot{5}\cdot{13}$$ $$775 = {{5}^2}\cdot{31}$$ $$819 = {{3}^2}\cdot{7}\cdot{13}$$ $$891 = {{3}^4}\cdot{11}$$ $$1287 = {{3}^2}\cdot{11}\cdot{13}$$ $$1305 = {{3}^2}\cdot{5}\cdot{29}$$ $$1485 = {{3}^3}\cdot{5}\cdot{11}$$ $$1989 = {{3}^2}\cdot{13}\cdot{17}$$ $$2295 = {{3}^3}\cdot{5}\cdot{17}$$ $$2793 = {3}\cdot{{7}^2}\cdot{19}$$ $$3515 = {5}\cdot{19}\cdot{37}$$ $$4095 = {{3}^2}\cdot{5}\cdot{7}\cdot{13}$$ $$4455 = {{3}^4}\cdot{5}\cdot{11}$$ $$4655 = {5}\cdot{{7}^2}\cdot{19}$$

Is this surprising? Or is there an underlying explanation for this phenomenon?

Updated Question (Added October 30, 2018)

Under what suitable conditions does it follow that $3 \mid n$, if $n$ is an odd number satisfying $\gcd(n,\sigma(n)) > \sqrt{n}$?

Note that it follows from the inequality $\gcd(n,\sigma(n)) > \sqrt{n}$ that $n$ must be composite.

Answer 1

Claim : There are infinitely many odd numbers $n$ such that $$3\not\mid n\ \ \qquad\text{and}\ \ \qquad \gcd(n,\sigma(n))\gt\sqrt n$$

Proof :

For $$n=p^{2a}\sigma(p^{2a})$$ where $p$ is a prime larger than $3$ and $a$ is a positive integer such that $(p,a)\not\equiv (1,1)\pmod 3$, we have $$2\not\mid n,\qquad\qquad 3\not\mid n$$ and $$\gcd(n,\sigma(n))=\gcd(p^{2a}\sigma(p^{2a}),\sigma(p^{2a})\sigma(\sigma(p^{2a})))\ge \sigma(p^{2a})\gt \sqrt{p^{2a}\sigma(p^{2a})}=\sqrt n\ .\quad\blacksquare$$

---

Some small examples :

• $n=5^{2}\sigma(5^{2})=5^2\cdot 31$ which you've already shown.

• $n=5^{4}\sigma(5^{4})=5^4\cdot 11\cdot 71$

• $n=5^{6}\sigma(5^{6})=5^6\cdot 19531$

• $n=7^4\sigma(7^4)=7^4\cdot 2801$

• $n=7^6\sigma(7^6)=7^6\cdot 29\cdot 4733$

• $n=7^{10}\sigma(7^{10})=7^{10}\cdot 1123\cdot 293459$

• $n=11^2\sigma(11^2)=7\cdot 11^2\cdot 19$

• $n=11^4\sigma(11^4)=5\cdot 11^4\cdot 3221$

• $n=11^6\sigma(11^6)=11^6\cdot 43\cdot 45319$

---

Added :

For $p\equiv 1\pmod 3$, we have $$\sigma(p^{2a})=1+p+\cdots +p^{2a}\equiv 1+1+\cdots +1\equiv 2a+1\not\equiv 0\pmod 3$$ for $a\not\equiv 1\pmod 3$.

For $p\equiv -1\pmod 3$, we have $$\sigma(p^{2a})=1+p+\cdots +p^{2a}=1-1+1-1+1-\cdots -1+1\equiv 1\pmod 3$$

Answer 2

All else being equal, a larger count of divisors will result in a larger sum of divisors, and a larger sum of divisors will result in a larger gcd. The count of divisors will the the product of one more than the powers of each distinct prime. I.e. if $n=\Pi p^{e_i}$, then the count of divisors of $n$ will be $\Pi (e_i+1)$. So to give yourself the best chance of having gcd >$\sqrt n$, you should have the exponents of primes as high as possible while keeping $n$ as small as possible. This then suggest using small primes, as that will result in larger powers with smaller resulting $n$. And the smallest odd prime number is 3. If you drop the odd requirement, your list would probably be dominated by powers of 2. And notice that more than half of your numbers are divisible by 5, the next smallest odd prime.

Answer 3

(Not a complete answer.)

Note that, for $p$ prime and $k \in \mathbb{N}$, we have $$\gcd(p^k,\sigma(p^k))=1 \not\gt \sqrt{p^k},$$ so that $\gcd(n,\sigma(n)) > \sqrt{n}$ implies $n$ is composite.

Perhaps the correct way to phrase the question is as follows:

If $n$ is an odd composite number satisfying $\gcd(n,\sigma(n)) > \sqrt{n}$, does it follow that $3 \mid n$? Under what suitable conditions does it follow that $3 \mid n$, if $n$ is an odd composite number satisfying $\gcd(n,\sigma(n)) > \sqrt{n}$? Under what suitable conditions does it follow that $3 \mid n$, if $n$ is an odd number satisfying $\gcd(n,\sigma(n)) > \sqrt{n}$?

Answer 4

Too long for a comment.

Claim : There is an odd integer $n$ such that $$\gcd(n,\sigma(n))\gt\sqrt n,\qquad 3^2\mid n,\qquad 3^3\not\mid n\qquad\text{and}\qquad 13\not\mid n$$

Proof :

For $n=3^2\times 5^4\times 11\times 71$, we get $$\sigma(n)=13\times 781\times 12\times 72=2^5\times 3^3\times 11\times 13\times 71$$from which we have $$\begin{align}\gcd(n,\sigma(n))& =\gcd(3^2\times 5^4\times 11\times 71,2^5\times 3^3\times 11\times 13\times 71) \\\\&=\gcd(3^2\times 11\times 71,3^3\times 11\times 71) \\\\&=3^2\times 11\times 71\end{align}$$

It follows that $$\begin{align}(\gcd(n,\sigma(n)))^2-n&=3^4\times 11^2\times 71^2-3^2\times 5^4\times 11\times 71 \\\\&=3^2\times 11\times 71\times (3^2\times 11\times 71-5^4) \\\\&=3^2\times 11\times 71\times 6404 \\\\&\gt 0\qquad\square\end{align}$$