Odd perfect numbers and egyptian fraction conjecture

Question

I have reached the conclusion that the non-existence of odd perfect numbers is related to the following

Conjecture Let it be $R=\{d_1,d_2,...,d_n\}$ the set of distinct proper divisors less than $\sqrt{O}$ of some odd positive integer $O$, such that $d_1<d_2<...<d_n$, and such that for some divisor $d_k\in R$ holds that $d_k\leq d_n<3d_k$. Then, it exist no solution to the egyptian fraction $$\frac{2}{d_k}+\frac{1}{d_1}+\frac{1}{d_2}+...=1$$

The process to reach the conjecture is the following (for you to check it is correct):

Let $P$ be some odd perfect number.

Let $R=\left\{ d_{1},d_{2},...,d_{n}\right\}$ be the set of proper divisors of $P$ less than $\sqrt{P}$ excluding $1$, and $T=\left\{ \frac{P}{d_{1}},\frac{P}{d_{2}},...\frac{P}{d_{n}}\right\}$ be the set of proper divisors of $P$ greater than $\sqrt{P}$ excluding $P$.

As $P$ is a perfect number,

$$1+d_{1}+d_{2}+...+d_{n}+\frac{P}{d_{n}}+...+\frac{P}{d_{2}}+\frac{P}{d_{1}}=P (1)$$

Operating,

$$1+d_{1}+d_{2}+...+d_{n}=P-\frac{P}{d_{1}}-\frac{P}{d_{2}}-...-\frac{P}{d_{n}}$$

$$1+d_{1}+d_{2}+...+d_{n}=P\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}\right)$$

$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$

As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer. Therefore, $P$ is a perfect number only if both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$.

It is trivial to show that one of the two expression is less than $\sqrt{P}$, and therefore belongs to $R$, and the other is greater than $\sqrt{P}$ and belongs to $T$.

It is clear that $1+d_{1}+d_{2}+...+d_{n}$ is greater than the greatest element of $R$, so subsequently we can state that

$$\left(\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}\right)=d_{k}\in R (2)$$

$$1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}\in T (3)$$

The proof showing that $d_k\leq d_n<3d_k$ is a bit long, so I skip it for the shake of briefness. And operating with $(2)$, the conjecture follows. I have found (thanks to some help provided here in MathStackExchange) examples of odd positive integers for which $(3)$ alone holds, but not any example of an odd positive integer for which $(2)$ holds with the restrictions exposed.

Any help on how to attack this conjecture would be welcomed. Otherwise, if someone is able to show any counterexample to the conjecture (that is, provide some odd positive integer for which $(1)$ holds), it would be more than welcomed to save efforts proving it. I have not seen this conjecture elsewhere, so it is possible that it is mistaken.

Thanks in advance!

Edit

I post the proof showing that $d_k\leq d_n<3d_k$, as it is fundamental for the conjecture and for the solution I have provided for verification.

From (3), and operating,

$$1+\sum_{j=1}^{n}d_{j}=\frac{P}{d_{k}}$$

$$P=d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

As by definition we have that $d_{n}<\sqrt{P}$, just substituting we can set that

$$d_{n}<\sqrt{d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)}$$

$$d_{n}^{2}<d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$

$$\frac{d_{n}}{d_{k}}<\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}} (4)$$

Other hand, taking (1) and dividing by $\sqrt{P}$, we get that

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{P}{d_{n}\sqrt{P}}+...+\frac{P}{d_{2}\sqrt{P}}+\frac{P}{d_{1}\sqrt{P}}=\sqrt{P}$$

$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{\sqrt{P}}{d_{n}}+...+\frac{\sqrt{P}}{d_{2}}+\frac{\sqrt{P}}{d_{1}}=\sqrt{P}$$

As $\forall d_{j}\,\frac{\sqrt{P}}{d_{j}}>1$, and $1+\sum_{j=1}^{n}d_{j}>\sqrt{P}$, then we get inmediately that $n<\sqrt{P}-1$; otherwise, the sum of all the terms would be greater than $\sqrt{P}$ and $P$ would not be a perfect number. Thus, we can affirm that

$$n\leq\sqrt{P}-2 (5)$$

As the minimum gap between consecutive elements of $R$ is $2$, the maximum sum of elements of $R$ with the minimum gap between them is

$$d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

Therefore, we can establish that

$$1+\sum_{j=1}^{n}d_{j}\leq d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$

$$1+\sum_{j=1}^{n}d_{j}\leq nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)$$

Subsequently, we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\frac{nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq n-\frac{\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$

Substituting $n$ by the inequality obtained in (5), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-2-1\right)^{2}+\left(\sqrt{P}-2-1\right)\right)}{d_{n}}$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)\right)}{d_{n}} (6)$$

Operating with the numerator of the third term of the right handside of (6), we get that

$$\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)=$$

$$P+9-6\sqrt{P}+\sqrt{P}-3=$$

$$P-5\sqrt{P}+6$$

As by definition $d_{n}<\sqrt{P}$, we can affirm that

$$\frac{P-5\sqrt{P}+6}{d_{n}}>\sqrt{P}-5+\frac{6}{\sqrt{P}}$$

Therefore, substituting at (6), we get that

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<\sqrt{P}-2-\left(\sqrt{P}-5+\frac{6}{\sqrt{P}}\right)$$

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3-\frac{6}{\sqrt{P}}$$

Subsequently,

$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3$$

$$1+\sum_{j=1}^{n}d_{j}<3d_{n}$$

Finally, substituting in (4), we get the desired result

$$\frac{d_{n}}{d_{k}}<3$$

Answer 1

(1) You are trying to prove that if $P$ is an odd perfect number, then both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$. As Arnie Bebita-Dris pointed out, you have an error in the following part :

$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer.

Take $(a,b)=(5,\frac 57)$ for which $P=\frac ab$ and $a\ (\lt P)$ are integers, and $\frac 1b$ is not an integer.

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(2) You are trying to prove that if $P$ is an odd perfect number, then $d_n\lt 3d_k$. Your proof starts with $1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}$ which you have not proved. This means that you have not proved that if $P$ is an odd perfect number, then $d_n\lt 3d_k$.

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(3) In your proof for the conjecture, I think that you have correctly proved that $p_k$ is a composite number.

However, you have an error in the following part :

$$d_{j}=\frac{d_{k}\left(1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)\right)}{d_{k}-2}$$As $d_{k}$ and $d_{k}-2$ are odd integers, it follows that $\gcd\left(d_{k},d_{k}-2\right)=1$. As each $d_{j}$ is some odd positive integer, then necessarily for each $d_{j}$ we have one of the following two options: either $d_{k}=3$, or $d_{k}-2$ divides $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$.

You are implicitly assuming that $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$ is an integer, which you have not proved.

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(4) I'm going to prove the following claims :

Claim 1 : If the conjecture is true, then $n$ is even larger than $4$.

Claim 2 : $(d_1,d_2,d_3)$ is either

• $(p,p^2,p^3)$ where $p$ is a prime number

• $(p,p^2,q)$ where $p,q$ are prime numbers

• $(p,q,p^2)$ where $p,q$ are prime numbers

• $(p,q,r)$ where $p,q,r$ are prime numbers

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Claim 1 : If the conjecture is true, then $n$ is even larger than $4$.

Proof :

Seeing $$\sum_{\substack{j=1\\j\neq k}}^{n}\left(\prod_{\substack{s\in S\\s\neq j}}d_{s}\right)+2\prod_{\substack{s\in S\\s\neq k}}d_{s}=\prod_{s\in S}d_{s}$$ in $\text{mod $2$}$ where $S=\left\{ 1,2,...,n\right\}$, we have $(n-1)+0\equiv 1\pmod 2$, so $n$ has to be even.

If $n=2$, then $$\frac{2}{x}+\frac 1y=1\implies (x-2)(y-1)=2\implies (x-2,y-1)=(1,2),(2,1)$$ implying $(x,y)=(3,3),(4,2)$ which don't satisfy our conditions.

If $n=4$, then, according to this answer, there is no such $O$. $\quad\blacksquare$

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Claim 2 : $(d_1,d_2,d_3)$ is either

• $(p,p^2,p^3)$ where $p$ is a prime number

• $(p,p^2,q)$ where $p,q$ are prime numbers

• $(p,q,p^2)$ where $p,q$ are prime numbers

• $(p,q,r)$ where $p,q,r$ are prime numbers

Proof :

Suppose that $d_1$ is a composite number. Then, there is a prime number $p$ satisfying $p\lt d_1$ and $p\mid O$. This contradicts that $d_1$ is the smallest divisor of $O$ larger than $1$. So, $d_1$ is a prime number.

Next, suppose that $d_2$ has two distinct prime factors $p\lt q$. Then, $p,q$ are divisors of $O$ satisfying $1\lt p\lt q\lt d_2$. This contradicts that $d_2$ is the second smallest divisors of $O$ larger than $1$. It follows that $d_2$ is of the form $q^{k}$ where $q$ is a prime number. If $q=d_1$, then $k=2$. If $q\not=d_1$, then $k=1$.

If $d_1=p,d_2=p^2$, then $d_3$ is either a prime number or $p^3$.

If $d_1=p,d_2=q$, then $d_3$ is of the form $s^k$ where $s$ is a prime number. If $s=p$, then $k=2$, If $s\not= p$, then $k=1$. $\quad\blacksquare$

Answer 2

I have been able to find a potential proof of the conjecture, I post it for you to check and tell me if you think it is right:

Let us define set $S=\left\{ 1,2,...,n\right\}$ . Operating with the egyptian fraction, we get that

$$\frac{\sum_{\substack{j=1\\j\neq k}}^{n}\left(\prod_{\substack{s\in S\\s\neq j}}d_{s}\right)+2\prod_{\substack{s\in S\\s\neq k}}d_{s}}{\prod_{s\in S}d_{s}}=\frac{\prod_{s\in S}d_{s}}{\prod_{s\in S}d_{s}} (1)$$

$$\sum_{\substack{j=1\\j\neq k}}^{n}\left(\prod_{\substack{s\in S\\s\neq j}}d_{s}\right)+2\prod_{\substack{s\in S\\s\neq k}}d_{s}=\prod_{s\in S}d_{s}$$

It is easy to see that this implies the following:

For each $d_{j\in S}$, we have that

$$d_{j}\mid\prod_{\substack{s\in S\\s\neq j}}d_{s}$$

This property, considered jointly with the condition $d_k\leq d_n<3d_k$ , has the following direct implication:

$d_{k}$ is some composite number

Proof. If $d_{k}$ was some prime number, as $d_{k}\mid\prod_{\substack{s\in S\\s\neq k}}d_{s}$ and all the proper divisors of $O$ are distinct, then some $d_{s\neq k}$ must be some odd composite number multiple of $d_{k}$. As the minimum possible multiple of $d_{k}$ distinct of $d_{k}$ is $3d_{k}$, then we have that some $d_{s\neq k}\geq3d_{k}$. However, as we have that $d_{n}<3d_{k}$, there can not exist any $d_{s\neq k}\geq3d_{k}$. Subsequently, $d_{k}$ must be composite.

Thus, $d_{k}=d_{i}d_{j}$, where $d_{i}$ and $d_{j}$ can be either prime numbers or composite numbers, but in any case such that $3d_{i}\leq d_{k}$ and $3d_{j}\leq d_{k}$.

Now, we are in a position to prove that

if $d_{k}\leq d_{n}<3d_{k}$, it can not exist any solution to the egyptian fraction $\frac{2}{d_{k}}+\frac{1}{d_{1}}+\frac{1}{d_{2}}+...=1$.

Proof. For each $d_{j\in S}$, dividing each of the terms of (1) by all $d_{s\neq k,j}$, we get that

$$d_{j}d_{k}=2d_{j}+d_{k}+d_{j}d_{k}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right) (2)$$

Operating with (2), we get that

$$d_{j}\left(d_{k}-2\right)=d_{k}\left(1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)\right)$$

$$d_{j}=\frac{d_{k}\left(1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)\right)}{d_{k}-2}$$

As $d_{k}$ and $d_{k}-2$ are odd integers, it follows that $\gcd\left(d_{k},d_{k}-2\right)=1$. As each $d_{j}$ is some odd positive integer, then necessarily for each $d_{j}$ we have one of the following two options: either $d_{k}=3$, or $d_{k}-2$ divides $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$. For each $d_{j}$ we can discard the first option, as $3$ is a prime number and $d_{k}$ is some composite number. Looking at the second option left, it can be noticed that $\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}<1$, because precisely the original egyptian fraction states that $\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)+\frac{1}{d_{j}}+\frac{2}{d_{k}}=1$. But this implies that, if $d_{k}-2$ divides $1+d_{j}\left(\sum_{\substack{i=1\\i\neq k,j}}^{n}\frac{1}{d_{i}}\right)$, then necessarily

$$d_{k}-2<1+d_{j}$$

$$d_{k}-3<d_{j}$$

As $d_{k}-3$ is an even number, then

$$d_{k}-2\leq d_{j} (3)$$

As this must be true for each $d_{j}$, we reach a contradiction between the bound set in (3) and the fact that $d_k$ is composite; according to the bound, $d_{k}$ cannot be some composite number of two other divisors $d_{i}$ and $d_{j}$ and must be some prime number. Subsequently, it follows that the egyptian fraction with the condition $d_k\leq d_n<3d_k$ can not exist.

Answer 3

There are at least two (2) gaps in your reasoning, that I have been able to identify so far.

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First, the fact that $$\bigg(1 + d_1 + d_2 + \ldots + d_n\bigg)\Bigg(\dfrac{1}{1 - \dfrac{1}{d_1} - \dfrac{1}{d_2} - \ldots - \dfrac{1}{d_n}}\Bigg) = P$$ does NOT imply that $$1 + d_1 + d_2 + \ldots + d_n = a \mid P$$ and $$\dfrac{1}{1 - \dfrac{1}{d_1} - \dfrac{1}{d_2} - \ldots - \dfrac{1}{d_n}} = b \mid P.$$

What you do have is the biconditional $$a \mid P \iff b \mid P$$ but NOT the conjunction $$(a \mid P) \land (b \mid P).$$

Indeed, you have neither proved $a \mid P$ nor $b \mid P$ in the OP.

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Second, the inequality $$\frac{P - 5\sqrt{P} + 6}{d_n} \geq \sqrt{P} - 5 + \frac{6}{\sqrt{P}}$$ does NOT follow from $$d_n < \sqrt{P}$$ basically because if we set $c = d_n, e = \sqrt{P}$ and $d = -5 < 0$, then $$c < e \implies 1 < \frac{e}{c} \implies d > \frac{de}{c}.$$

So, your proof for the claimed upper bound $d_n < 3d_k$ is FLAWED.