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Is there an odd integer $x < 105$ for which it is known that $x \nmid N$, if $N$ is an odd perfect number?

I have asked the same question in MSE, but did not get any answers. I was wondering if anybody here has anything to share regarding this.

Thank you!

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  • $\begingroup$ Can't you include 105 in your range? $\endgroup$ – Włodzimierz Holsztyński Aug 12 '14 at 1:16
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    $\begingroup$ That is exactly the point, @WlodzimierzHolsztynski. We already know that $105 \nmid N$ (see e.g., this MSE question), where $N$ is an odd perfect number. What I am interested to know about would be: Is there an odd integer $x$ smaller than $105$ for which it is known that $x \nmid N$? $\endgroup$ – Jose Arnaldo Bebita-Dris Aug 12 '14 at 9:28
  • $\begingroup$ Note that $105 = 3\cdot5\cdot7$. I guess another way to ask my question would be: Is there any special reason why, if $x$ is an odd integer satisfying $x \nmid N$ where $N$ is an odd perfect number, it seems that $\omega(x) \geq 3$, where $\omega(x)$ is the number of distinct prime divisors of $x$? (See e.g., MSE1, MSE2, and MSE3.) $\endgroup$ – Jose Arnaldo Bebita-Dris Aug 12 '14 at 9:34
  • $\begingroup$ J.A.D., thank you. $\endgroup$ – Włodzimierz Holsztyński Aug 13 '14 at 3:29
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No, such a result would be a major breakthrough regarding our knowledge on odd perfect numbers.

A few years ago there was some confusion, since due to careless reading and citing of the article "Every odd perfect number has a prime factor which exceeds $10^6$" by Cohen and Hagis the impression arose that they had proved just such a theorem (which they never claimed).

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  • $\begingroup$ Thank you for your answer @Jan-ChristophSchlage-Pu. I guess what Cohen and Hagis really proved is that the largest prime factor of an odd perfect number exceeds ${10}^6$? $\endgroup$ – Jose Arnaldo Bebita-Dris Aug 12 '14 at 13:06
  • $\begingroup$ Additionally, can you comment more on why you say that "such a result would be a major breakthrough regarding our knowledge on odd perfect numbers"? $\endgroup$ – Jose Arnaldo Bebita-Dris Aug 12 '14 at 13:08
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    $\begingroup$ Cohen and Hagis proved that the largest prime factor of an odd perfect number is $>10^6$. Of course, the proof is indirect, so it begins with "Let $n$ be an odd perfect number with all prime divisors $\leq 10^6$...". Then a sequence of properties for $n$ are deduced, one of which reads "Lemma: $n$ is not divisible by 3, 5, 7, ..." . This Lemma was later taken out of context and cited as "Let $n$ be an odd perfect number. Then $n$ is not divisible by 3,5,7,...". $\endgroup$ – Jan-Christoph Schlage-Puchta Aug 13 '14 at 13:29
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    $\begingroup$ There is a lot of literature proving the non-existence of odd perfect numbers satisfying certain additional restrictions. Practically all of them require some starting point of the form "One of these primes divides $n$" or "One of these primes does not divide $n$". Sometimes this starting point is easy to get (e.g. if you bound the number of distinct prime factors), but quite often getting the starting point is the most difficult part. $\endgroup$ – Jan-Christoph Schlage-Puchta Aug 13 '14 at 13:49
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    $\begingroup$ For example, McDaniel (On the divisibility of an odd perfect number by the sixth power of a prime. Math. Comp. 25 (1971), 383–385) showed that an odd perfect number is either divisible by the sixth power of a prime, or it is not divisible by any prime $<100$. The proof is quite easy, once one has shown that if $n$ is not divisible by the sixth power of a prime, then $n$ is not divisible by some small prime, but getting there is pretty difficult. $\endgroup$ – Jan-Christoph Schlage-Puchta Aug 13 '14 at 13:53
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This is too long to be posted as a comment, as I just wanted to share some of my recent thoughts on why it is difficult to obtain such an integer $x \nmid N$, where $N = {q^k}{n^2}$ is a (hypothetical) odd perfect number.

Since a multiple of an abundant number is also abundant, and since the smallest odd abundant number is $945$, we know that $${3^3}\cdot{5}\cdot{7} = 945 \nmid N.$$ In fact, we know that $$3\cdot5\cdot7 = 105 \nmid N,$$ although $105$ is deficient.

Note that $$\{\{3 \mid N\} \land \{5 \mid N\} \land \{7 \mid N\}\} \implies \{105 \mid N\}.$$ By the contrapositive, $$\{105 \nmid N\} \implies \{\{3 \nmid N\} \lor \{5 \nmid N\} \lor \{7 \nmid N\}\}.$$

However, all existing computational projects which implement factor chains (see this MSE post for a very down-to-earth sample) to check whether $3 \mid N$ or otherwise have not terminated (i.e., AKA "resulted to a contradiction"). Likewise, it is currently unknown whether the smallest possible Euler prime $q = 5$ does divide an odd perfect number. (It is known though, by work of Iannucci, that $q = 5$ implies $k = \nu_{q}(N) = 1$. Iannucci proved that $q = 5$ implies $k = \nu_{q}(N) = 1$ under some additional assumptions.)

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