Perfect but for one factor

A number is perfect when its divisors, itself included, add up to twice the number: $\sigma(N) = 2N$. Numbers that are their own parts followed the even ones from Euclid’s construction to Euler’s proof that the construction gives them all, and then recorded what is known about the odd case: none has been found, none has been ruled out, and an odd perfect number would have to be larger than $10^{1500}$ with at least ten distinct prime factors.

A list of conditions is hard to picture. There is a better way to see what an odd perfect number would be like, and it is a near miss that René Descartes found in 1638 and sent to Marin Mersenne. It is the number

$$N = 3^2 \cdot 7^2 \cdot 11^2 \cdot 13^2 \cdot 22021 = 198{,}585{,}576{,}189,$$

and it satisfies $\sigma(N) = 2N$ if 22021 is treated as a prime. It is not a prime. But the calculation that goes wrong goes wrong in exactly one place, and everything else about the number is what an odd perfect number would need.

A number that is perfect by miscounting #

The divisor sum is multiplicative: for a number factored into prime powers, $\sigma$ is the product of the divisor sums of the prime powers — which the shape of a number’s divisors draws as a rectangle whose sides are the prime powers’ own divisor lists — and for a prime power $\sigma(p^e) = 1 + p + \cdots + p^e$. So checking whether $N$ is perfect is a matter of computing five small sums and multiplying.

The first four give $13$, $57 = 3 \cdot 19$, $133 = 7 \cdot 19$ and $183 = 3 \cdot 61$. If 22021 were prime its divisor sum would be $22022 = 2 \cdot 7 \cdot 11^2 \cdot 13$. The product of all five is $2 \cdot 3^2 \cdot 7^2 \cdot 11^2 \cdot 13^2 \cdot 19^2 \cdot 61$, and $19^2 \cdot 61$ is exactly 22021. So the product is $2N$.

The structure of the coincidence is the whole lesson. Each prime power’s divisor sum produces new primes — 19 from $7^2$ and from $11^2$, 61 from $13^2$ — and a perfect number needs every prime that appears in $\sigma(N)$ to appear in $N$ with exactly the right exponent. The primes the sums produce must be absorbed, and whatever they produce in turn must be absorbed, until the chain closes. Descartes closed it by absorbing $19^2 \cdot 61$ into a single “prime” whose own divisor sum, $22022$, happens to supply the 2, the 7, the $11^2$ and the 13 that the rest needs. With 22021 factored honestly, $\sigma(N)$ is about $2.145N$, and the chain does not close.

Why a miscount can look like perfection #

The multiplicativity is itself a consequence of unique factorisation. The divisors of $N$ are exactly the products of one divisor from each prime power, one way each, because every divisor factors into primes in exactly one way; summing them therefore gives the product of the prime powers’ sums. Declare a composite number prime, and the divisors of the result are miscounted — the divisors 19, 61, $19^2$, $19 \cdot 61$ of 22021 are left out, and its divisor sum is recorded as $1 + 22021$ instead of $23{,}622$. The bookkeeping of $\sigma$ is otherwise followed perfectly.

So a spoof is a statement about a world in which factorisation is slightly wrong, and Descartes’ number shows that in such a world odd perfect numbers exist. That is the precise sense in which it is informative: every argument about odd perfect numbers that does not use the primality of the factors applies to it, and it satisfies the conclusion those arguments are trying to rule out.

Why the even ones close so easily #

The even perfect numbers are the contrast. Euclid’s numbers $2^{k-1}(2^k - 1)$ are perfect whenever $2^k - 1$ is prime, and numbers that are their own parts showed Euler’s converse: every even perfect number has this form. The chain of divisor sums closes in one step. $\sigma(2^{k-1}) = 2^k - 1$ is a Mersenne prime, which is already a factor; $\sigma(2^k - 1) = 2^k$ supplies the power of two, with one factor of 2 left over for the doubling. Two prime powers, each supplying exactly what the other needs.

An odd number has no power of two to play either role. Its divisor sums produce primes that are not yet factors, those primes must be admitted with the right exponents, and their divisor sums produce more. The multiperfect numbers close such chains for ratios of 3, 4 and higher, always with a power of 2 at the start, and the chains run to dozens of primes. An odd perfect number would be a chain of the same kind with no power of 2 anywhere, and Descartes’ number is a short chain that closes only because one link is a composite in disguise.

The form an odd perfect number must have #

Some of what makes Descartes’ number work is forced, and Euler proved it. For odd $p$, $\sigma(p^e) = 1 + p + \cdots + p^e$ is a sum of $e + 1$ odd numbers, so it is odd exactly when $e$ is even. An odd perfect number has $\sigma(N) = 2N$, which contains exactly one factor of 2. So exactly one of its prime powers has an even divisor sum, and that sum must contain 2 exactly once — it must leave remainder 2 on division by 4.

The table computes $\sigma(p^e)$ modulo 4 and finds the remainder 2 exactly when $p$ and $e$ both leave remainder 1 on division by 4. An odd perfect number has the form $N = p^e m^2$ with $p \equiv e \equiv 1 \pmod 4$ and $p$ not dividing $m$: one special prime to an exponent of the form $4k + 1$, and every other prime to an even power. Descartes’ number has exactly this shape. Its special prime is the fake one, $22021 \equiv 1 \pmod 4$, to the first power, and its other primes are all squared.

That is not a coincidence of Descartes’ choice. Any odd number satisfying $\sigma(N) = 2N$ under any assignment of “primes” whose divisor sums multiply correctly must have this shape, since Euler’s argument uses only the arithmetic of $1 + p + \cdots + p^e$ modulo 4. A spoof and a real odd perfect number obey the same constraint.

Odd numbers find abundancy hard #

The second constraint is the size of the target. A ratio nobody else has called $\sigma(n)/n$ the abundancy of $n$, and a perfect number is exactly a number of abundancy 2. Why a quarter of numbers overshoot found that about one number in four has abundancy above 2 — but almost all of those are even, because the factor 2 alone contributes $\sigma(2)/2 = 3/2$ and higher powers of 2 nearly double the abundancy by themselves. Odd numbers have no such help. Up to 20,000 only 43 odd numbers reach 2 at all, and the first, 945, needs the primes 3, 5 and 7.

The abundancy of $n$ is at most the product of $p/(p-1)$ over its distinct primes, the limit approached as the exponents grow. Using the smallest odd primes gives the most abundance per prime: $\tfrac32 \cdot \tfrac54 \cdot \tfrac76 = 2.1875$ already exceeds 2 with three primes. Without the prime 3, the product needs seven primes to reach 2; without 3 and 5, fifteen. An odd perfect number must reach 2 exactly, and with the exponents constrained by Euler’s form most of its primes are squared or higher — and much more detailed versions of this counting, combined with enormous computer searches, give the proved lower bound: Pace Nielsen showed in 2015 that an odd perfect number has at least ten distinct prime factors.

Descartes’ number uses five: 3, 7, 11, 13 and the pretend prime. That it reaches 2 with five is possible only because the fifth “prime” is really three prime factors, $19 \cdot 19 \cdot 61$, doing the work of more.

Every odd number that one more factor would complete #

Descartes’ construction can be run backwards as a search. Take any odd $m$, and ask for a prime $q$ not dividing it such that $m \cdot q$ is perfect. Since $\sigma(mq) = \sigma(m)(q + 1)$, the condition is $\sigma(m)(q + 1) = 2mq$, which can be solved for $q$:

$$q = \frac{\sigma(m)}{2m - \sigma(m)}.$$

For most $m$ this is not a whole number. The search runs over every odd $m$ below ten million, computes $\sigma(m)$ for all of them at once by a sieve, and keeps the ones where $q$ is whole. There are 80. For 79 of them $q$ is even, so $m \cdot q$ is even — some of these are genuine even perfect numbers or relatives of them, like $m = 3$ with $q = 2$, which gives 6. An even $q$ other than 2 is never prime, so apart from 2 itself each of these is a spoof of an even number, and the even perfect numbers are already completely described by Euclid and Euler. Exactly one has an odd $q$: $m = 9{,}018{,}009 = 3^2 \cdot 7^2 \cdot 11^2 \cdot 13^2$ and $q = 22021$. It is Descartes’ number, found again, and $q$ is not prime, so it is not an odd perfect number.

That the search finds nothing else below ten million is consistent with what is known more generally. An odd number that is perfect when one composite factor is treated as prime is now called a Descartes number, and William Banks, Ahmet Güloğlu, Wesley Nevans and Filip Saidak showed in 2008 that any other one not divisible by 3, if it is free of cubed primes, must have more than a million distinct prime factors. Descartes’ number is the only one known.

How close the odd numbers come #

The near misses among honest odd numbers show the same architecture from the other side. The closest below ten million is $442{,}365 = 3 \cdot 5 \cdot 7 \cdot 11 \cdot 383$, which overshoots 2 by fourteen parts in a million. The pattern repeats down the list: the smallest odd primes supply most of the abundancy, and one or two larger primes fine-tune the product towards 2. Getting the fine-tuning exact requires the larger prime’s own divisor sum to supply exactly the primes the rest is missing, as 22022 did for Descartes, and among real primes that never happens in the range searched.

The sign of the miss matters too. Overshooting is easy to arrange and undershooting nearly as easy; what is hard is landing between them, because as the product of divisor sums is adjusted by swapping one prime for another, it jumps by a factor like $(p+1)/p$ divided by $(p'+1)/p'$, and exact equality needs the numerator and denominator of every factor to cancel against the others. For an odd number the cancellations have to be arranged among odd primes only, and the table is a record of how close they come in the range searched: fourteen parts in a million, never nought.

None of these numbers has Euler’s form, and none is close in the sense that matters. An abundancy of $2 + 10^{-5}$ is a real number close to 2; perfection is an equation between whole numbers, $\sigma(N) = 2N$, and a miss of fourteen parts in a million is a miss of six units in $\sigma(442{,}365)$. The spoof is the better near miss because it misses in structure rather than in size.

Finitely many for each number of primes #

The chain picture also explains the one general finiteness theorem in the subject. Leonard Dickson proved in 1913 that for each $k$ there are only finitely many odd perfect numbers with exactly $k$ distinct prime factors — in fact finitely many odd numbers of any fixed abundancy with $k$ primes. The reason is that the abundancy $\prod p/(p-1)$, approached as exponents grow, pins down the smallest prime; given it, the next is confined to a finite range; and so on down the chain. Carl Pomerance made the bound explicit in 1977, and Pace Nielsen brought it down in 2003 to $N < 2^{4^k}$.

A bound of that shape is enormous for $k = 10$, and it cannot settle the question. But it turns the problem for each $k$ into a finite search over chains of the kind Descartes’ number is built from, and the lower bounds on the number of primes — five, from James Joseph Sylvester in the 1880s, up to Nielsen’s ten in 2015 — are the record of those searches being completed for each smaller $k$. Every new lower bound is a proof that no chain with that many primes closes honestly, though for five primes one closes dishonestly.

The iteration of the sum of the parts taken again sends a number to the sum of its proper divisors and follows it, and a perfect number is a fixed point of that map. The chains here are a different iteration of the same function — not following one number’s orbit but asking which factorisations are closed under taking divisor sums — and they are what make perfection a finite question for each number of primes.

What the search cannot show #

The search to ten million is small against the known bound: an odd perfect number would exceed $10^{1500}$, and no search reaching it can ever be run directly. The searches that established that bound do not enumerate numbers; they enumerate the possible chains of prime powers and divisor sums, the same absorbing chains that make Descartes’ number work, and prove that every chain either closes too early or grows past the bound.

The Descartes numbers themselves are searched the same way, by chains rather than by enumeration, and Descartes’ example was found by hand. The figure’s search finds it because it happens to lie below ten million; it is evidence that the construction is rare near the bottom, not a statement about the whole of the integers.

And the abundancy figures are about sizes, not exact equations. They show why an odd perfect number needs many primes, which is a necessary condition, and they cannot show that the many primes can be arranged to close the chain exactly — which is the whole question.

Still open: an odd one, or a proof there is none #

The question is two thousand years old and has not changed: is there an odd perfect number? The constraints keep accumulating — more than $10^{1500}$, at least ten distinct primes, at least 101 prime factors with multiplicity, a largest prime factor above $10^8$ — and each one makes the hypothetical number stranger without ruling it out. Most people who work on it believe there is none, on heuristic grounds: the chains of divisor sums that would have to close become astronomically unlikely as the numbers grow.

The spoofs sharpen the question in a useful direction. A proof that no odd perfect number exists would have to use the fact that primes are prime — Descartes’ number shows that everything else about perfection can be satisfied by an odd number. In 2022 a group led by Pace Nielsen at Brigham Young University studied spoofs systematically, allowing composite and even negative “primes”, and found families with the same shape; any impossibility proof has to break down on those families and succeed on real factorisations. No proof of that kind is in sight.

A factorisation wrong in one place #

Descartes’ number is a perfect number in every respect but one. It has the form Euler proved an odd perfect number must have, its divisor sums chain together and close, and the product comes out at exactly twice the number. The one thing wrong is that a number called prime is not.

That makes it the most informative object in the subject. It shows that the parity constraint, the size constraint and the chaining are all satisfiable by an odd number, so none of them alone can be the reason odd perfect numbers do not exist, if they do not. Whatever the reason is, it lives in the one property Descartes allowed himself to ignore: that when a divisor sum produces a prime, the prime has a divisor sum of its own.