Let $\omega(n)$ denote the number of distinct prime factors of a positive integer $n$, and let $N$ be an odd perfect number. It is not difficult to show that $\omega(N)\ge3$. In fact, Nocco already proved this in 1863. Showing that $\omega(N)\neq3$ requires a little more effort, but the proof is still fairly short. I will present some version of the proof below; the original one is due to Peirce in 1830 (which, interestingly, seems to have been published before Nocco's proof).

The two key ideas here are the geometric series and the factor chain method. Let $N=\displaystyle\prod_{i=1}^k p_i^{e_i}$ be the prime factorization of $N$. Then it follows that

$$\sigma(N)=\prod_{i=1}^k \sum_{j=0}^{e_i} p_i^j=\prod_{i=1}^k \frac{p_i^{e_i+1}-1}{p_i-1}.$$

This formula is useful, because each factor is necessarily greater than one when $e_i$ is positive, which is the case for all prime factors $p_i$. Hence $\sigma(N)\ge\sigma\left(\displaystyle\prod_{i\in{I}} p_i^{e_i}\right)$, where $I\subseteq\{1,2,\ldots,k\}$.

The second idea is perhaps even more fundamental, because we have yet to use the property of perfectness. Since $\sigma(N)=2N$, every divisor of $\sigma(N)$ must also divide $2N$; the other way around is not as interesting. In particular, every factor $\frac{p_i^{e_i+1}-1}{p_i-1}$ must divide $2N$. This result is really powerful; for example, observing that $2^1||2N$ yields Euler's well-known form of an odd perfect number after some algebra.

With these tools in our toolbox, the proof is almost immediate. Let $p^\alpha$, $q^\beta$, and $r^\gamma$ be three odd prime powers with $p<q<r$ and assume that $N=p^\alpha q^\beta r^\gamma$ is an odd perfect number. By the same token as above, we see that

$$\begin{align*}\sigma(N)=\sigma(p^\alpha q^\beta r^\gamma) &= \frac{p^{\alpha+1}-1}{p-1} \cdot \frac{q^{\beta+1}-1}{q-1} \cdot \frac{r^{\gamma+1}-1}{r-1} < \frac{p^\alpha p}{p-1} \cdot \frac{q^\beta q}{q-1} \cdot \frac{r^\gamma r}{r-1}\\\implies& \frac{\sigma(N)}{N} = \frac{\sigma(p^\alpha q^\beta r^\gamma)}{p^\alpha q^\beta r^\gamma} < \frac{p}{p-1} \cdot \frac{q}{q-1} \cdot \frac{r}{r-1}.\end{align*}$$

Suppose that $p=5$. Then $\frac{\sigma(N)}{N}<\frac{5}{4}\cdot\frac{7}{6}\cdot\frac{11}{10}=\frac{77}{48}<2$, which is impossible. Hence $p=3$.

Suppose that $q=7$. Then $\frac{\sigma(N)}{N}<\frac{3}{2}\cdot\frac{7}{6}\cdot\frac{11}{10}=\frac{77}{40}<2$, which is impossible. Hence $q=5$.

Suppose that $r=17$. Then $\frac{\sigma(N)}{N}<\frac{3}{2}\cdot\frac{5}{4}\cdot\frac{17}{16}=\frac{255}{128}<2$, which is impossible. Hence $r \le 13$.

In consequence, we have three possible cases to consider: $r=7$, $r=11$, and $r=13$.

In the first case, we have $N=3^\alpha 5^\beta 7^\gamma$. By the factor chain method, $3^1||N$ implies $4|2N$ and $7^1||N$ implies $8|2N$, neither of which is possible. Hence $\alpha\ge2$ and $\gamma\ge2$. It follows that

$$\frac{\sigma(N)}{N} \ge \frac{\sigma(3^2\cdot5\cdot7^2)}{3^2\cdot5\cdot7^2} = \left(1 + \frac{1}{3} + \frac{1}{3^2}\right)\left(1 + \frac{1}{5}\right)\left(1 + \frac{1}{7} + \frac{1}{7^2}\right)=\frac{494}{245}>2,$$

which is impossible, but we already knew that $105=3\cdot5\cdot7$ cannot divide an odd perfect number.

In the second case, we have $N=3^\alpha 5^\beta 11^\gamma$. Suppose that $\beta=1$. It follows that

$$\frac{\sigma(N)}{N} = \frac{\sigma(3^\alpha\cdot5\cdot11^\gamma)}{3^\alpha\cdot5\cdot11^\gamma} < \frac{3}{2} \cdot \frac{6}{5} \cdot \frac{11}{10} = \frac{99}{50} < 2,$$

which is impossible (could we know that beforehand?). Hence $\beta\ge2$. By the factor chain method, $\alpha\ge4$ and $\gamma\ge4$. It follows that

$$\frac{\sigma(N)}{N} \ge \frac{\sigma(3^4\cdot5^2\cdot11^4)}{3^4\cdot5^2\cdot11^4} = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4}\right)\left(1 + \frac{1}{5} + \frac{1}{5^2}\right)\left(1 + \frac{1}{11} + \frac{1}{11^2} + \frac{1}{11^3} + \frac{1}{11^4}\right)=\frac{99851}{49005}>2,$$

which is impossible, but we already knew that $825=3\cdot5^2\cdot11$ cannot divide an odd perfect number.

In the third case, we have $N=3^\alpha 5^\beta 13^\gamma$. $\beta=1$ implies deficiency, so $\beta\ge2$. Now, $\alpha=2$ also implies deficiency, so $\alpha\ge4$. By the factor chain method, we must have $\gamma\ge2$, which this time implies abundance. Hence the assumption leads to a contradiction; furthermore, $8775=3^3\cdot5^2\cdot13$ cannot divide an odd perfect number. We are finally done.

Unfortunately, proving that $\omega(N)\neq4$ is not this easy by hand. We can similarly show that $p=3$, but then we already have two possible choices for $q$. Of course, one could just write a program and loop through all the possibilities, but Sylvester did it successfully in 1888, unlikely by the aid of computers. Hence I wonder if this algorithm can be significantly improved, or if there are other more efficient techniques that achieve the same goal.

  • $\begingroup$ Sylvester's 1888 result is here: archive.org/stream/collectedmathem04sylvrich#page/n653/mode/2up But it proves only what you have proved above. $\endgroup$
    – Charles
    Jun 10, 2013 at 18:59
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    $\begingroup$ Wait, that was premature -- he improved his result to 5 within the same year: archive.org/stream/collectedmathem04sylvrich#page/n659/mode/2up $\endgroup$
    – Charles
    Jun 10, 2013 at 19:05
  • $\begingroup$ @Charles Sylvester proved many things in 1888. The proof for $\omega(N)\neq4$ should be this one: Sur l'impossibilité de l'existence d'un nombre parfait impair qui ne contienne pas au moins 5 diviseurs premiers distincts. $\endgroup$
    – Librecoin
    Jun 10, 2013 at 19:05
  • $\begingroup$ I just beat you to it. $\endgroup$
    – Charles
    Jun 10, 2013 at 19:06
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    $\begingroup$ I will attempt to solve this, but if I cannot and no one else does I will post this as an answer. $\endgroup$
    – Charles
    Jun 10, 2013 at 19:12

1 Answer 1


Here is Sylvester's original proof: http://archive.org/stream/collectedmathem04sylvrich#page/n659/mode/2up


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