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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

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

which is a contradiction.

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"
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