If $x$ and $y$ are solitary numbers satisfying $\gcd(x,y)=1$, under what conditions does it follow that $xy$ is also solitary?

Question

Let $\sigma(z)$ denote the sum of divisors of $z \in \mathbb{N}$. Denote the abundancy index of $z$ by $I(z) = \sigma(z)/z$.

If the equation $I(z)=I(a)$ has the lone solution $z=a$, then $a$ is said to be solitary. Greening's Theorem states that if $\gcd(b,\sigma(b))=1$, then $b$ is solitary. (Note that the converse does not necessarily hold.)

Here is my question:

If $x$ and $y$ are solitary numbers satisfying $\gcd(x,y)=1$, under what conditions does it follow that $xy$ is also solitary?

The latest paper on solitary numbers that I could find on Google seems to be New families of solitary numbers by Paul A. Loomis. However, my present inquiry is not covered in this paper.

Answer 1

This is a partial answer.

Note that all prime numbers are solitary.

Claim 1 : If $x=3$ and $y$ is a prime of the form $6k+1$, then $xy$ is solitary.

Proof : We get $\sigma(xy)=\sigma(3y)=(1+3)(1+6k+1)=8(3k+1)$, so $$\gcd(xy,\sigma(xy))=\gcd(3(6k+1),8(3k+1))=\gcd(6k+1,3k+1)=1$$ It follows that $xy$ is solitary. $\quad\blacksquare$

Claim 2 : If $x=5$ and $y$ is a prime of the form $10k+3$ where $k\ge 1\in\mathbb N$, then $xy$ is solitary.

Proof : We get $\sigma(xy)=\sigma(5y)=(1+5)(1+10k+3)=12(5k+2)$, so $$\gcd(xy,\sigma(xy))=\gcd(5(10k+3),12(5k+2))=\gcd(10k+3,5k+2)=1$$ It follows that $xy$ is solitary. $\quad\blacksquare$

Claim 3 : If $x=5$ and $y$ is a prime of the form $10k+7$, then $xy$ is solitary.

Proof : We get $\sigma(xy)=\sigma(5y)=(1+5)(1+10k+7)=12(5k+4)$, so $$\gcd(xy,\sigma(xy))=\gcd(5(10k+7),12(5k+4))=\gcd(10k+7,5k+4)=1$$ It follows that $xy$ is solitary. $\quad\blacksquare$

---

Added : From Erick Wong's comment, I've got some examples.

Claim : If $x$ is a prime such that $x\equiv 1\pmod 3$ and $y$ is a prime of the form $2x(x+1)k+x-2$, then $xy$ is solitary.

Proof : We get $$\gcd(x,\sigma(y))=\gcd(x,x(2kx+2k+1)-1)=1$$ and $$\gcd(\sigma(x),y)=\gcd(x+1,(x+1)(2kx+1)-3)=1$$ It follows that $xy$ is solitary. $\quad\blacksquare$

P.S. : Since $\gcd(2x(x+1),x-2)=\gcd(x-2,12)=1$ for $x\equiv 1\pmod 3$, there are, by Dirichlet's theorem, infinitely many primes of the form $2x(x+1)k+x-2$.

Some examples for small $x$ :

• If $x=7$ and $y$ is a prime of the form $112k+5$, then $xy$ is solitary.

• If $x=13$ and $y$ is a prime of the form $364k+11$, then $xy$ is solitary.

• If $x=19$ and $y$ is a prime of the form $760k+17$, then $xy$ is solitary.

• If $x=31$ and $y$ is a prime of the form $1984k+29$, then $xy$ is solitary.