A number is said to be Perfect if it is equal to the sum of its proper positive factors, i.e. the sum of its factors excluding the number itself. Summing

*all*of the factors of a number \(n\)

*— i.e. including the number \(n\) itself — is known as the sum-of-divisors function:*

\[\sigma \left( n \right) = \sum\limits_{d\left| n \right.} d \]

A number \(n\) is thus said to be Perfect when \(\sigma \left( n \right)\) = 2n. For example, the factors of 28 — 1, 2, 4, 7, 14 and 28 — sum to 28 + 28 = 56, thus \(\sigma \left( {28} \right)\) = 2 \(\times \) 28 and so 28 is a Perfect number. In contrast, and by way of illustration, the factors of 42 — 1, 2, 3, 6, 7, 14, 21 and 42 — sum to 96, or 2.2857... \(\times \) 42, thus \(\sigma \left( {42} \right) \ne\) 2 \(\times \) 42 and so 42 is not a Perfect number. In fact, 42 is said to be an Abundant number because the sum of its proper positive divisors is greater than the number itself, i.e. \(\sigma \left( n \right) > 2n\). When the sum of a number's proper positive divisors is less than the number itself, i.e. \(\sigma \left( n \right) < 2n\), the number is said to be Deficient. For example, the factors of 50 — 1, 2, 5, 10, 25 and 50 — sum to 93, or 1.86 \(\times \) 50, and thus \(\sigma \left( {50} \right)\) < 2 \(\times \) 42. As such:

\[\begin{array}{l}\sigma \left( n \right) < 2n \Rightarrow n\;{\rm{is}}\;{\rm{deficient}}\\\sigma \left( n \right) = 2n \Rightarrow n\;{\rm{is}}\;{\rm{perfect}}\\\sigma \left( n \right) > 2n \Rightarrow n\;{\rm{is}}\;{\rm{abundant}}\end{array}\]

Perfect numbers have intrigued us since Euclid, who observed via Proposition 36 in Book IX of his

*Elements*that a number of the form (2\(^{p - 1}\))(2\(^p\) − 1) is a perfect number whenever 2\(^p\) − 1 is prime (i.e. what subsequently became known as a Mersenne prime) [1]. The fact that every

*even*perfect number is of this type, i.e. that every

*even perfect number can be written in the form (2\(^{p - 1}\))(2\(^p\) − 1) whenever 2\(^p\) − 1 is prime, was proposed by René Descartes in his famed correspondence with Marin Mersenne (specifically in his letter to Mersenne of 15 November 1638) [2], and later proven in 1849 by Leonhard Euler [3]. Descartes was, indeed, 'among the first to consider the existence of odd perfect numbers' (Greathouse and Weisstein, 2012).*

As of writing, of the fifty perfect numbers that have been found, all are even, and it is not known if there are infinitely many or, indeed, whether any odd perfect numbers exist — although Pascal Ochem and Michaël Rao showed (2012) that no number up to 10\(^{1500}\) is an odd perfect number.

One number

*has*been found, however, that

*would*have been an odd perfect number if only one of its factors was prime rather than a 'spoof prime' (i.e. a composite number wrongly assumed to be prime). This number — 198,585,576,189 — was found in 1638 by Descartes, documented in his 15 November letter to Mersenne, and is known as a Descartes Number, or, as an 'odd spoof perfect number' (note that the spoof prime in question is 22021, i.e. 19\(^2\) \(\times \) 61).

\[198,585,576,189 = {3^2} \times {7^2} \times {11^2} \times {13^2} \times 22021\]

Descartes showed that if 22021 was prime, the proper factors of 198,585,576,189 [4] would sum to 198,585,576,189 — making it the only odd perfect number ever found. As it is, because 22021 is not prime, or because 22021 is a spoof prime, the proper factors of 198,585,576,189 do not sum to 198,585,576,189 (they actually sum to 227,441,894,589). This makes 198,585,576,189 an odd

*spoof*perfect number, which in itself is the only such number ever found!

To elaborate: Descartes showed that an equivalence to the sum-of-divisors function using the prime factorisation of a number is the following [5]:

\[\sigma \left( n \right) = \prod\limits_{{p^a}\left\|{\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} {\rm{ }}\]

where \(p\) is a distinct prime factor. This means that a number's factors can be summed using only the number's prime factors. Taking 28 again as an example, we have seen that it is a perfect number because

\[\sigma \left( 28 \right) = \sum\limits_{d\left| 28 \right.} d = 1 + 2 + 4 + 7 + 14 + 28 = 56 = 2n\]

We can also observe this result using the prime factors of 28, i.e. 2\(^2 \times \) 7:

\[\prod\limits_{{p^a}\left\|{\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} = \frac{{{2^{2 + 1}} - 1}}{{2 - 1}} \times \frac{{{7^{1 + 1}} - 1}}{{7 - 1}} = \frac{7}{1} \times \frac{{48}}{6} = 56 = 2n\]

And so, similarly, the sum of the divisors of Descartes' number:

\[\begin{array}{c}\begin{align}\sigma \left( 198,585,576,189 \right) &= \prod\limits_{{p^a}\left\| {\;n} \right.} {\frac{{{p^{a + 1}} - 1}}{{p - 1}}} \\ \\&= \frac{{{3^{2 + 1}} - 1}}{{3 - 1}} \times \frac{{{7^{2 + 1}} - 1}}{{7 - 1}} \times \frac{{{{11}^{2 + 1}} - 1}}{{11 - 1}} \times \frac{{{{13}^{2 + 1}} - 1}}{{13 - 1}} \times \frac{{{{22021}^{1 + 1}} - 1}}{{22021 - 1}}\\ \\&= 13 \times 57 \times 133 \times 183 \times 22022\\ &= 13 \times \left( {3 \times 19} \right) \times \left( {7 \times 19} \right) \times \left( {3 \times 61} \right) \times \left( {2 \times 7 \times 11 \times 11 \times 13} \right)\\ &=2 \times {3^2} \times {7^2} \times {11^2} \times {13^2} \times \left( {{{19}^2} \times 61} \right)\\ &= 2 \times {3^2} \times {7^2} \times {11^2} \times {13^2} \times 22021\\ &= 2 \times 198,585,576,189\end{align}\end{array}\]

Hence, if 22021 were prime rather than 19\(^2\) \(\times \) 61, Descartes' number 198,585,576,189 would be an odd perfect number because \(\sigma \left( n \right)\) = 2n.

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