## 6.12.10

### On Sorli's Conjecture on Odd Perfect Numbers

Let $N = {p^k}{m^2}$ be an Odd Perfect Number (OPN) with Euler prime $p$.  Note that:

(1)  $p \equiv k \equiv 1 \pmod 4$
(2)  $\gcd(p, m) = 1$

I also proved (in 2008, p. 112 [Theorem 4.2.5]) the following result:

(3)  $\sigma(p^k)/m^2 <= 2/3$

(The proof begins with the fact that prime-powers are solitary numbers, and then by observing that $\sigma(m^2)/p^k$ is always odd (for odd primes $p$), (3) follows from an "old" result by Dandapat, Hunsucker and Pomerance [1975].  See the list of References below (the very last one) for the hyperlink to the document.)

In particular, this means that:

(4)  $$1 + p^k <= \sigma(p^k) <= (2/3){m^2}$$

But:

(5)  $$\log(\sigma(p^k)) = \log(\sum_{i = 0}^{k}{p^i})$$

But then:

(6)  $$\sigma(p^k) = \sum_{i = 0}^{k}{p^i} = \frac{p^{k + 1} - 1}{p - 1}$$

Therefore:

(7)  $$\log(\sigma(p^k)) = \log(p^{k + 1} - 1) - \log(p - 1)$$

(Note:  We also know that, for the Euler prime $p$, $\sigma(p) = p + 1$ divides $\sigma(p^k)$.  We give a quick proof below.)

Since $p \equiv k \equiv 1 \pmod 4$, we have an even number of addends in the sum:

$$\sigma(p^k) = 1 + p + \ldots + p^{k - 1} + p^{k}$$
$$= (1 + p)(1 + \ldots + p^{k - 1})$$
$$= (1 + p){\sigma(p^{2a})}$$
(since we know that $k \equiv 1 \pmod 4$, then $4 \mid k - 1$), where $a = (k - 1)/2$.

Thus:

(8)  $$\log(\sigma(p^k)) = \log(1 + p) + \log(\sigma(p^{k - 1}))$$

Note that $k - 1 < k$ ( i.e. (8) is a logarithmic relation recursively defining $\sigma(p^k)$ in terms of $\sigma(p^{k - 1})$).

We now state Sorli's conjecture for OPNs:

[Sorli, 2003]  If $N$ is an OPN given in the Eulerian form $N = {p^k}{m^2}$ (i.e. $p$ is the Euler prime and $\gcd(p, m) = 1$), then $k = 1$.

Here we give a characterization for odd numbers $N = {p^k}{m^2}$ to be perfect.

Proof (to appear - perhaps in IJNT):

Assume there is an OPN $N$.  Then $N$ has the Eulerian form $N = {p^k}{m^2}$, with $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$.

Prime powers are deficient because, in general, the abundancy index for prime powers $q >= 2$ satisfies the inequality:

(9)    $1 < I(q^s) < 3/2$   for all exponents s

(Note that (9) follows from the inequality

(10)  $(q + 1)/q <= I(q^s) < (q - 1)/q$

which is true for all primes $q$ and exponents $s$.)

Thus, it is not possible that $m = p$.

Case 1:  Assume $m < p$.  Since $k \equiv 1 \pmod 4$ [in particular, we know that $k \ge 1$], we then have:

$m < p <= p^k$

By the upper bound in (3):  $p^k < (2/3){m^2} < m^2$

But squaring both sides of the inequality $m < p$, we get:

$m^2 < p^2$

We now have the inequality:

$p \le p^k < p^2$

(i.e. the prime-power $p^k$ is "trapped" between two consecutive powers of the prime $p$).

This implies that $1 <= k < 2$.  Since $k \equiv 1 \pmod 4$, we conclude that $k = 1$.

We now consider Case 2.

Case 2:  Assume $p < m$.

From Case 1, we proved the implication

"$m < p^k$ implies $k = 1$".

The contrapositive is "$k > 1$ implies $p^k < m$".  (Note that $m$ is not equal to $p^k$ because of (4).)

Suppose $k > 1$.  By the contrapositive result mentioned, $p^k < m$, and trivially thus: $p < p^k < m$.

Observe that Case 1 gives a criterion for having $k = 1$.

This is because the conditions

$p \le p^k < m$

and

$m < p \le p^k$

are mutually exclusive.

(Of course, the [related] conditions

$k > 1$

and

$k = 1$

are also mutually exclusive.)

Thus, to avoid any contradictions in (any series of) our (logical) implications, since Case 1 holds (for the OPN conjecture) under the assumption of Sorli's conjecture, we must have logical equivalences between:

Case 1:                  $m < p \le p^k$       and      $k = 1$
Case 2:                 $p \le p^k < m$       and      $k \ge 1$

(Note that, in Case 2, we can actually take $k \geq 5$.)

Additionally, in a post over MathOverflow, I was able to show results in the direction of proving the following implication:

MathOverflow (posted over the weekend):  Sorli's Conjecture implies the OPN Conjecture.

That is, if Sorli's Conjecture is proved, then there are no odd perfect numbers.

(I am now hereby editing this post in response to Pace Nielsen's request for "backing up my claim".)

To recap, assuming $k = 1$ will make the sum $I(p^k) + I(m^2)$, which satisfies the inequality

$2.85 = 57/20 < L(p) < I(p^k) + I(m^2) \leq U(p) < 3$

attain the upper bound $U(p) = (3p^2 + 2p + 1)/(p(p + 1))$.

The lower bound $L(p)$ is given by $L(p) = (3p^2 - 4p + 2)/(p(p - 1))$.  Note that:

$I(p) = (p + 1)/p$

and

$I(m^2) = 2/[I(p)] = 2p/(p + 1)$.

Now, let us call a prime $p$ an "admissible" Euler prime if it can be the Euler prime of an OPN.

That is, a prime $p \equiv 1 \pmod 4$ is an "admissible" Euler prime if it comes from the "functional equation" $I(p)I(m^2) = 2$.

On the other hand, given a particular "admissible" Euler prime $p$, by computing $u = U(p)$ and then computing the integers $q$ satisfying $L(q) < u$, you get a $q$ that CANNOT BE EQUAL to $p$. Thus, the "admissible" Euler prime $p$ gives rise to an "inadmissible" Euler prime $q$.  But please do take note that the underlying assumption is that $p = q$, whence you get a contradiction.

Indeed, it suffices to check that the implication

"$L(q) < u = U(p)$ implies $q$ is NOT equal to $p$"

holds for all "admissible" Euler primes $p \ge 5$, just by checking the inequality

$L(q) < U(p) = U(5) = (6/5) + (5/3)$

for $p = 5$.  Why?

This is because $L(p)$ and $U(p)$ are both increasing functions of $p$ in the interval $[5, \infty)$, with $L(p) < U(p)$ for all $p$ (note the strict inequality), with both $L(p)$ and $U(p)$ approaching $3$ as $p$ becomes arbitrarily large.

In particular, note that:

$$L(p) = \frac{3p^2 - 4p + 2}{p(p - 1)} = 3 - \frac{p - 2}{p(p - 1)}$$
and
$$U(p) = \frac{3p^2 + 2p + 1}{p(p + 1)} = 3 - \frac{p - 1}{p(p + 1)}$$

So if $L(q) < U(p)$, then

$$3 - \frac{q - 2}{q(q - 1)} < 3 - \frac{p - 1}{p(p + 1)}$$

which implies that

$$\frac{p - 1}{p(p + 1)} < \frac{q - 2}{q(q - 1)}$$

$$q(p - 1)(q - 1) < p(p + 1)(q - 2)$$

$$pq^2 - pq - q^2 + q < {p^2}q - 2{p^2} + pq - 2p$$

Furthermore, since $q$ and $p$ are integers (in fact, they are prime numbers), then

$$2q - p \geq 4$$

Suppose $2q - p = 4$.  Since $p$ and $q$ are both Euler primes,
$$p \equiv q \equiv 1 \pmod 4$$

Thus, $2q \equiv 2 \pmod 4$ and $p \equiv 1 \pmod 4$ imply that $$2q - p \equiv 1 \pmod 4$$

This contradicts $2q - p = 4$.

Thus, $2q - p \geq 5$.

$2q \geq p + 5$

$2q \geq p + 5 \geq 5 + 5 = 10$

$q \geq 5$

$(p - 1)(q - 1) < (p + 1)(q - 2)$   for all possible pairs $(p, q)$

In particular, WolframAlpha gives the solutions:

(i)    $p > -1, q > (p + 3)/2$

(ii)   $p > -1, q < 1$

(iii)  $p > -1, (p + 3)/2 < q < 1$

We only need to consider Case (i):

(i)    $p > -1, q > (p + 3)/2$

Since we know that the Euler primes $p$ satisfy $p \equiv 1 \pmod 4$, automatically, we have $p \ge 5$, and therefore the first condition is satisfied.

Checking the second condition:

First, we show that it is true that $(p + 3)/2 < p$ if $p \ge 5$.  Assume to the contrary that $(p + 3)/2 \ge p$.  Then we have $p + 3 \ge 2p$.  This leads to the contradiction $p \le 3$.    (Essentially, what we have shown is that, hypothetically, $q < (p + 3)/2 < p$ follows from $p \geq 5$).

Now, we compare $q$ and $p$.  (Recall:  We need to show that $q$ is not equal to $p$.)

Since $p \ge 5$, it follows that $q > (p + 3)/2 \ge (5 + 3)/2 = 8/2 = 4$.  Thus, $q \ge 5$.

It thus remains to tackle the case $p = q = 5$.

It is (actually) a more fundamental problem to establish that $q = (p + 3)/2$ leads to a contradiction when $p$ and $q$ are both Euler primes.  This is because:

If $p$ and $q$ are both Euler primes, then $p \equiv q \equiv 1 \pmod 4$.

Yet $q = (p + 3)/2$ implies that

$$q = (p + 3)/2 \equiv (1 + 3)/2 \equiv 4/2 = 2 \pmod 4$$

Another way of saying this is that if $p$ and $q$ are both Euler primes, then the equality $q = (p + 3)/2$ cannot happen since it will contradict $p \equiv q \equiv 1 \pmod 4$.  That is why it was necessary to have the strict inequality $q > (p + 3)/2$.

But then again it begs the question:

Why must $p$ and $q$ necessarily be distinct Euler primes, if OPNs are to exist?

In other words, it does appear (heuristically) that Euler primes are in one-to-one correspondence with OPNs, in the same way that Mersenne primes are in one-to-one correspondence with even perfect numbers (EPNs).

For the case $q < p$

WolframAlpha gives the solutions:

(i)    $p > 3, (1/2)(p + 3) < q < p$

(ii)   $p > 3, q < 1$

(iii)  $-1 < p \le 1, q < p$

(iv)   $1 < p \le 3, q < 1$

For the case $p < q$

WolframAlpha gives the solutions:

(i)    $p > 3, q > p$

(ii)   $-1 < p \le 1, q > (1/2)(p + 3)$

(iii)  $-1 < p \le 1, p < q < 1$

(iv)   $1 < p \le 3, q > (1/2)(p + 3)$

(v)    $p < -1, (1/2)(p + 3) < q < 1$

For the case $p = q$

WolframAlpha gives the solutions:

(i)    $p > 3, q = p$

(ii)   $-1 < p < 1, q = p$

Integer solution: $p = 0, q = 0$

Assuming WolframAlpha's numerical methods of computation are correct for $p \ne q$, we need to tackle this remaining case:  $p = q = 5$.

Lastly, in order to prove that the OPN conjecture is true (in full generality), we need to consider the case:

(Case 2)  $N$ is an OPN in the Eulerian form $N = {p^k}{m^2}$ with $k > 1$ AND $p^k < m$.

In other words, we now have the following characterization theorem for OPNs:

Theorem [December 2010]:  If $N$ is an OPN with the Eulerian form $N = {p^k}{m^2}$, then $p^k < m$ and $k >= 5$.

The Theorem gives sufficient and necessary conditions for odd numbers of the form
$N = {p^k}{m^2}$ to be perfect.

On some final notes: The even perfect numbers also have a form similar to the OPNs:

$M$ is an even perfect number if and only if

$$M = {2^{r - 1}}(2^r - 1) = (1/2){M_r}(M_r + 1)$$

where $M_r$ is a Mersenne prime with exponent $r$.

(The "exponent" $k'$ on the Mersenne prime [ i.e. if you write it as ${M_r}^{k'} = (2^r - 1)^{k'}$ ] is of course $k' = 1$.)

Note that $(M_r + 1)/2 < M_r$.

Additionally, $M_r \equiv 3 \pmod 4$ for all Mersenne primes with exponent $r$.  (Compare that with $p^k \equiv 1 \pmod 4$ for all Euler primes with exponent $k$.)

Lastly, it is known that the set of perfect numbers is a (proper) subset of the set of triangular numbers.

In the coming days/weeks, I will try to tackle the remaining case (Case 2) for OPNs.

References:

(A)   Ronald Sorli [2003], "Algorithms in the Study of Multiperfect and Odd Perfect Numbers"
(B)   Arnie Dris [2008], "Solving the Odd Perfect Number Problem: Some Old and New Approaches"
(C)   MathOverflow post [2010],  "On Sorli’s Conjecture Re: OPNs (Circa 2003)"
(F)   Melvin Bernard Nathanson [2000], "Elementary Methods in Number Theory"
(G)  G. G. Dandapat, J. L. Hunsucker, and Carl Pomerance [1975], "Some New Results on Odd Perfect Numbers"

Arnie Dris said...

Additionally, this "enhanced" Euclid-Euler model for the "bifurcation" (at the abundancy index value of 2) between even and odd perfect numbers predicts that

SQRT(M_r + 1) < M_r

where M_r is the Mersenne prime with exponent r. (We use "the" because the Mersenne primes are in one-to-one correspondence with the even perfect numbers [i.e. there is exactly one Mersenne prime corresponding to every even perfect number].)

Conjecturally, we can therefore predict (using this same "enhanced" Euclid-Euler model for perfect numbers) that the Euler primes are in one-to-one correspondence with the odd perfect numbers [i.e. there is exactly one Euler prime corresponding to every odd perfect number]. Heuristically, this appears to be true on the basis of the following facts:

(1) Prime powers are solitary numbers. (i.e. I(x) = I(p^k) only has one solution and that is x = p^k where I(n) is the abundancy index of the number n, and I(n) = Sigma(n)/n).

(2) If N = (p^k)(m^2) is an OPN, then by a theorem of Erdos [1959] (Remarks on number theory, II, Some problems on the Sigma function, Acta Arithmetica, vol 5, pp. 171-177), we have:

The number of solutions of the equation

I({m_1}^2) = I({m_2}^2)

satisfying {m_1}^2 < {m_2}^2 <= x equals

Cx + o(x) where C >= (8/147). (Note that C here is the same as the (natural) density of friendly integers.) This means that, for arbitrarily large {m_1}^2 and {m_2}^2, there will be at least (10^(150))(8/147) solutions to I({m_1}^2) = I({m_2}^2), a number which is obviously greater than 1. (Here, I have used the lower bound for OPNs by Brent, Cohen and te Riele [1991]. However, a draft paper off the following link

http://www.lri.fr/~ochem/opn/

contains some results on OPNs (by Pascal Ochem and Michael Rao) which, among other things, claim that N > (10)^(1500). This will produce the lower bound m^2 > (10)^(750) which is even bigger than the lower bound for N by Brent, et. al.

The preceding observations essentially say that, for a number m^2 coming from the square component of N = (p^k)(m^2), such a number would have to be friendly (in the sense of the equation I(x) = I(m^2) having at least two solutions).

(3) The results from (1) and (2) can be summarized as follows:

The map J : Z+ --> Q+ x Q+ given by

J(N) = (X, Y)

where N = (p^k)(m^2) is an OPN, and

X = I(p^k) = Sigma(p^k)/(p^k)
Y = I(m^2) = Sigma(m^2)/(m^2)

with

X in the interval (1, 5/4) = (1, 1.25)
Y in the interval (8/5, 2) = (1.6, 2)
X + Y in the interval (57/20, 3) = (2.85, 3)

is neither surjective nor injective.

(4) Heuristically, (3) means that distinct Euler primes give rise to distinct OPNs (i.e. an OPN is determined by its Euler prime). Thus, I conjecture that:

The Euler primes (or equivalently, the Euler factors p^k) are in one-to-one correspondence with the odd perfect numbers N = (p^k)(m^2).

I have already verified that the Mersenne primes do satisfy the inequality

SQRT(M_r + 1) < M_r

and will make a post related to that in a while.

Arnie Dris said...

I am still clarifying my thoughts on the "heuristics" of why the Euler primes are in fact in one-to-one correspondence with the odd perfect numbers.

That is, I am claiming that no two odd perfect numbers can "share" the same Euler prime, in the sense that:

If two OPNs can have the same Euler prime, then

I({N_1}) = I({N_2}) = 2 and {N_1} < {N_2}

with

{N_1} = (p^k)[(M_1)^2] and
{N_2} = (p^k)[(M_2)^2],

then I({M_1}^2) = I({M_2}^2).

CONJECTURE: Squares are solitary numbers.

for more information regarding an attempt in this direction. The link contains a post I made to the Math Forum @ Drexel while writing up my masters thesis in the year 2006.)

Arnie Dris said...

Update (December 8, 2010 - 8:27 PM PH time): It is indeed true that the Euler primes are in one-to-one correspondence with the OPNs. I will be making a follow-up post to this one. The link to the newest post on this "conjectured" one-to-one correspondence will be posted here.

Arnie Dris said...

Here's the link: http://arnienumbers.blogspot.com/2010/12/are-euler-primes-in-one-to-one.html. Additionally, I had it all along: The verification that q < p = 5 amounts to solving L(q) < u = U(5) and showing that the only possible integer solutions to the resulting inequality are q = 0, or q = 1, which are both neither prime, nor equal to 5.

Arnie Dris said...

With respect to my very first comment in this post, the "enhanced"/"extended" Euclid-Euler model for perfect numbers only predicts that

SQRT((1/2)(M_r + 1)) < M_r

but I think I will be able to prove the slightly stronger inequality:

SQRT(M_r + 1) < M_r

as I had originally commented. Everything else I said in that post is correct, subject to this modification.