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Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. If $\sigma(N)=2N$ (equivalently, when $I(N)=2$) then $N$ is called a perfect number. Both $\sigma$ and $I$ are multiplicative functions.

It is easy to prove the following:

If $I(n) = 5/3$, then $5 \nmid n$.

Proof

Suppose that $I(n)=5/3$. By this answer, we know that $n$ must be an odd square.

Hence $3^2 \mid n$ (since $n$ is a square and $3\sigma(n)=5n$, where we notice that $\gcd(3,5)=1$, so that $3 \mid n$). If it is also the case that $5 \mid n$, then $5^2 \mid n$ (again, since $n$ is a square), so that we obtain $$1.\overline{666} = \frac{5}{3}=I(n) \geq I(3^2)I(5^2) = \frac{13}{9}\cdot\frac{31}{25} = \frac{403}{225} = 1.79\overline{1},$$ which is a contradiction.

Hence, $5 \nmid n$, and $5n$ would be an odd perfect number if there exists an $n$ such that $I(n)=5/3$: $$I(5n) = I(5)I(n) = \frac{6}{5}\cdot\frac{5}{3} = 2.$$

Here is my question in this post:

If $\sigma(n)/n = 5/3$, does it also follow that $3 \nmid \sigma(n)$?

MY ATTEMPT

Assume that $3 \nmid \sigma(n)$. We can apply the $\sigma$ function to both sides of $3\sigma(n)=5n$, thereby yielding $$\sigma(3\sigma(n))=\sigma(5n).$$ Since $3$ and $5$ are primes, $5 \nmid n$, and $3 \nmid \sigma(n)$ by assumption, then $$\gcd(3,\sigma(n))=\gcd(5,n)=1.$$ Using the fact that $\sigma$ is multiplicative, we get $$\sigma(3)\sigma(\sigma(n))=\sigma(3\sigma(n))=\sigma(5n)=\sigma(5)\sigma(n)$$ so that we obtain $$I(\sigma(n))=\frac{\sigma(\sigma(n))}{\sigma(n)}=\frac{\sigma(5)}{\sigma(3)}=\frac{6}{4}=\frac{3}{2}.$$ This last equation implies that $\sigma(n)=2$, which violates the fact that $n$ is an odd square.

We therefore conclude that $3 \mid \sigma(n)$.

Jose Arnaldo Bebita Dris
  • 8,471

2 Answers2

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MY ATTEMPT

Assume that $3 \nmid \sigma(n)$. We can apply the $\sigma$ function to both sides of $3\sigma(n)=5n$, thereby yielding $$\sigma(3\sigma(n))=\sigma(5n).$$ Since $3$ and $5$ are primes, $5 \nmid n$, and $3 \nmid \sigma(n)$ by assumption, then $$\gcd(3,\sigma(n))=\gcd(5,n)=1.$$ Using the fact that $\sigma$ is multiplicative, we get $$\sigma(3)\sigma(\sigma(n))=\sigma(3\sigma(n))=\sigma(5n)=\sigma(5)\sigma(n)$$ so that we obtain $$I(\sigma(n))=\frac{\sigma(\sigma(n))}{\sigma(n)}=\frac{\sigma(5)}{\sigma(3)}=\frac{6}{4}=\frac{3}{2}.$$ This last equation implies that $\sigma(n)=2$, which violates the fact that $n$ is an odd square.

We therefore conclude that $3 \mid \sigma(n)$.

Jose Arnaldo Bebita Dris
  • 8,471
  • Aha! From this closely related MSE question, we have that if $I(n) = (p+2)/p$ where $p$ is an odd prime, then $$\gcd(n,\sigma(n)) = \frac{n}{p} = \frac{\sigma(n)}{p+2} = \frac{s(n)}{2} = \frac{D(n)}{p - 2},$$ where $s(x)=\sigma(x)-x$ is the aliquot sum of the positive integer $x$, and where $D(x)=2x-\sigma(x)$ is the deficiency of $x$. – Jose Arnaldo Bebita Dris Feb 07 '23 at 11:21
  • In particular, since both $3 \mid n$ and $3 \mid \sigma(n)$ are true, then $3 \mid \gcd(n,\sigma(n))$ holds. Since $p=3$ in the present problem, then this means that $$6 \mid s(n), 3 \mid D(n).$$ In fact, since $p=3$, then $\gcd(n,\sigma(n))=D(n)$ ought to hold. – Jose Arnaldo Bebita Dris Feb 07 '23 at 11:24
  • We infer that $s(n)$ must be abundant. – Jose Arnaldo Bebita Dris Feb 07 '23 at 11:26
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Another (shorter) approach:

Since $3\sigma(n)=5n$ and $n$ is an odd square, then $3^2 \mid n$. If $3 \nmid \sigma(n)$, then $3^2 \nmid \text{LHS}$, contradicting $3^2 \mid \text{RHS}$.

Jose Arnaldo Bebita Dris
  • 8,471