11.12.10

Pre-Final Stage: Towards An Elementary Proof of the OPN Conjecture

Recap:

Essentially, I have been trying to prove the following implication in my last five (5) blog posts here:

If (distinct) Euler primes $P$ and $Q$ satisfy $5 = P < Q$, then Sorli's conjecture implies that there are no OPNs.

Given the following assertion:

CONJECTURE:  The Euler primes are in one-to-one correspondence with the OPNs.

consider how you will go about in proving it.

Assume to the contrary that there is no such bijection between the Euler primes and the OPNs.

Then you either have:

(1)  At least two different Euler primes (say $P$ and $Q$) will map to the same OPN.  (This is, of course, ruled out by Euler's theorem for OPNs.)
(2)  At least two OPNs will map to the same Euler prime.  (This WAS the hard part.)

That is, assuming there are at least two OPNs, we want to show that they cannot share the same Euler prime.

In terms of the abundancy index, given two OPNs $N_1$ and $N_2$ with

$${N_1} = {P^{a}}{{M_1}^2}$$

and

$${N_2} = {Q^{b}}{{M_2}^2}$$

since $I({N_1}) = I({N_2}) = 2$, we get

$$I({P^{a}}{{M_1}^2}) = I({Q^{b}}{{M_2}^2})$$

So now, assume to the contrary that $P = Q$.

Then you get:

$$I({{M_1}^2}) = I({{M_2}^2})$$

whereupon I hinted at Conjecture II which concerns the solitude or otherwise of squares.  Additionally, in a comment to a previous post, I did mention that I proved (in the year 2008, as part of the results in my MS thesis) that the map $f : {\mathbb{N}}^{+} \rightarrow {\mathbb{Q}}^{+} \times {\mathbb{Q}}^{+}$ given by

$N = {q^k}{n^2}$ is an OPN
$f(N) = (X, Y)$
where
$X = I(q^k)$ is the abundancy index of $q^k$, and
$Y = I(n^2)$ is the abundancy index of $n^2$.

is neither injective nor surjective (and so, in particular, it cannot be bijective).

Since $I(N) = I(q^k)I(n^2) = 2$, $f$ is nothing but a map from ${\mathbb{N}}^{+}$ to rational points on the hyperbola $XY = 2$.

Furthermore, $A = \{X\}$ is the set of abundancy indices of the solitary numbers $q^k$, where $q$ is the Euler prime of $N$ (i.e., $q^k$ is also called the Euler factor of $N$).

Lastly, $B = \{Y\}$ is the set of abundancy indices of the friendly numbers $n^2$, where $n$ is related to the Euler factor via the equation $n = \sqrt{N/{q^k}}$.

There is an algebraic way to prove that the Euler primes are distinct.  Here is the argument:

(1)  Assume to the contrary that the Euler primes are not distinct (i.e. $P = Q$).
(2)  Use an inequality (i.e. a relation) between $P$ and $Q$ to show that $P$ is NOT EQUAL to $Q$.
(3)  Hence, conclude that $P$ is not equal to $Q$, for all (possible) Euler primes $P$ and $Q$.

And thereby get:  The Euler primes are distinct.

At step (2), we needed to establish three (very important) sub-steps:

(2.a)  Lemma 1:  $U(P) \leq L(Q)$ for all possible Euler-prime pairs $(P,Q)$ with $Q < P$.
(2.b)  Lemma 2:  If $L(Q) < U(P)$ then $P \leq Q$.
(2.c)  Lemma 3:  If $L(Q) < U(5)$ then $P = Q = 5$.

and, of course, the additional (and final) step that:

$P$ and $Q$ cannot be $5$.

Consider how we could go about with proving that $P = Q = 5$ cannot be the case, if we wanted to show that OPNs do not exist.

(That is, if we can eliminate the first case for the lowest possible Euler prime p = 5, then by "induction" via the method that I have been ranting about so far, we will be able to eliminate all the higher possible Euler primes $p > 5$.)

Thus, in this direction, suppose to the contrary that the Euler prime takes the value $p = 5$.

By Lemma 3:  If $L(x) < U(5)$, we expect to get $x = 5$ as the only "solution" (in the sense of being an "admissible" Euler prime).

Using WolframAlpha to check this inequality:

$$[(3x^2 - 4x + 2)]/[x(x - 1)] = L(x) < U(5) = (6/5) + (5/3) = 43/15$$
$$[(3x^2 - 4x + 2)]/[x(x - 1)] < 43/15$$

gives you the 2D plot:

and the 1D plot:

The integer solutions are:

$$x = 3, x = 4, x = 5$$

where of course the only solution corresponding to an "admissible" Euler prime is $x = 5$.

Recall the following lattice of relations [automorphisms (?)] for the functions
$L(x)$ and $U(x)$:

Note that $L(x) < U(x)$ for all $x$ in $[4, \infty)$.

$$U(2) = U(3) = L(3) = 17/6$$

In particular, we have the following chain of inequalities and equalities from the lattice diagram just given:

$$\ldots U(x - 1) = L(x) < U(x) = L(x + 1) < U(x + 1) = L(x + 2) \ldots$$
$$\ldots < U(x + 2) = L(x + 3) < U(x + 3) \ldots$$

Since this chain holds for all $x$ in $[4, \infty)$, letting $x = P$ gives you:

$$\ldots U(P - 1) = L(P) < U(P) = L(P + 1) < U(P + 1) = L(P + 2) \ldots$$
$$\ldots < U(P + 2) = L(P + 3) < U(P + 3) \ldots$$

In particular:

$L(x) = U(x)$    if and only if $x = 3$

$$L(x + 1) = U(x)$$ for all $x$ NOT in the set $\{-1, 0, 1\}$

If $L(Q) < U(5)$ then $Q = 5$.

$$17/6 = U(2) = L(3) = U(3) = L(4) < U(4) = L(5) < U(5) = L(6) \ldots$$
$$\ldots < U(6) = L(7) < U(7) = L(8) < U(8) = L(9) < U(9) \ldots$$

If $L(Q) < U(5)$, then we go down the chain:

$$17/6 = U(2) = L(3) = U(3) = L(4) < U(4) = L(5) < U(5)$$

Observe that
$$17/6 < 2.84 < 2.85 = 57/20 < I(q^k) + I(n^2).$$

So now, let us try to compute for the values of $p$ and $q$ such that:

$$L(p) = p$$

and

$$U(q) = q$$

{  i.e. we intend to compute for the fixed points of $L(x)$ and $U(x)$  }.

Again, using WolframAlpha:

The fixed point of $L(x)$ is given in closed form as:

and in approximate form as:

The fixed point of $U(x)$ is given in closed form as:

and in approximate form as:

Notice that $U(q) = q < p = L(p)$.