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Hi, I apologize if there is already an (obvious) answer to my question, but please bear with me for the moment as I find it hard to see a good way to answer this question:

In the same way that the Mersenne primes uniquely determine the even perfect numbers (i.e. in the "bijective" sense), do the Euler primes likewise uniquely determine the odd perfect numbers?

That is, while an Euler prime obviously maps to a single odd perfect number, it is not obvious to me that an odd perfect number could only map to a single Euler prime.

This is because:

(1) In the most general case, it is not known whether squares are friendly or solitary.

(2) It is not even known if the smallest possible Euler prime 5 does divide an odd perfect number.

(3) It is conceivable that two distinct numbers $P$ and $Q$ which satisfy a particular number-theoretic equation do not necessarily have to satisfy $\gcd(P, Q) = 1$.

I hope you guys can help me out on this one. Establishing this very last gap will provide a proof for the implication:

Assuming Sorli's Conjecture for OPNs is true, then there are no OPNs.

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    $\begingroup$ "an Euler prime obviously maps to a single odd perfect number" completely not obvious for me! I know many Euler primes, but no odd perfect numbers. $\endgroup$
    – Fedor Petrov
    Commented Dec 10, 2010 at 18:33
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    $\begingroup$ @Arnie, so I think the 3rd paragraph in your question is backward. Each OPN (if there are any) corresponds to exactly one Euler prime, while it is not obvious (to either of us) that a given prime can be the Euler prime of at most one OPN. $\endgroup$
    – Gerry Myerson
    Commented Dec 10, 2010 at 21:54
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    $\begingroup$ I don't get it - p is the only prime factor with an odd exponent. $\endgroup$
    – Franz Lemmermeyer
    Commented Dec 11, 2010 at 11:32
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    $\begingroup$ @Arnie: I think Franz's point -- which I agree with -- is the following. Your statement about the "correspondence" between odd perfect numbers and Euler primes seems backwards: according to Euler, every odd perfect number $N$ has a unique Euler prime -- i.e., a unique prime number $p$ with $\operatorname{ord}_p(N) \equiv 1 \pmod 4$. You are asking -- aren't you? -- whether the same prime number can be an Euler prime for more than one odd perfect number. $\endgroup$
    – Pete L. Clark
    Commented Dec 12, 2010 at 2:13
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    $\begingroup$ @Arnie: I'm afraid I don't understand why "because I want to prove that there are no odd perfect numbers" is a proof strategy for your uniqueness statement. I didn't ask whether you thought it was true -- if there are no odd perfect numbers, it is vacuously true -- but rather why you thought it is any easier to prove than the no odd perfect numbers conjecture itself. $\endgroup$
    – Pete L. Clark
    Commented Dec 13, 2010 at 22:12

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I tried to get in touch with Dean Hickerson and here's what he got to say regarding the problem of determining the status of squares with respect to solitude or friendliness:

Notice that all of the squares from $1$ to $121$ are solitary (since they satisfy $gcd(n, \sigma(n)) = 1$). (See OEIS - A014567.) Does my observation hold true in general?

No. The smallest square for which $\gcd(n, \sigma(n))$ is not equal to $1$ is $196$: $\gcd(196, \sigma(196)) = 7$. (But it's easy to show that $196$ is solitary.)

In $1995$ I found a square that isn't solitary; I don't know if there are any smaller ones: $26334^2 = 693479556 = 2^2 3^4 7^2 11^2 19^2$. There are at least $5$ other numbers with the same abundancy index:

$$ 8640 = {2^6}\cdot{3^3}\cdot{5}$$

$$ 52416 = {2^6}\cdot{3^2}\cdot{7}\cdot{13}$$

$$ 71814642425856 = {2^{13}}\cdot{3^4}\cdot{11^3}\cdot{31}\cdot{43}\cdot{61}$$

$$ 2168446760665473024 = {2^{13}}\cdot{3^{10}}\cdot{11}\cdot{23}\cdot{43}\cdot{107}\cdot{3851}$$

$$ 5321505362711814144 = {2^{13}}\cdot{3^6}\cdot{11}\cdot{23}\cdot{43}\cdot{137}\cdot{547}\cdot{1093}$$

I had expected my conjecture to fail for even squares. Notwithstanding, have you also found any odd squares which are not solitary?

No, I haven't. It's easy to find odd squares for which $\gcd(n, \sigma(n))$ is not equal to $1$; e.g. if $n = {21}^2 = 441$ then $\gcd(n, \sigma(n)) = 3$. But $441$ is solitary.

Hence, it appears that the status of odd squares is still an open problem.

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  • $\begingroup$ An odd perfect number (OPN) $N$ given in the so-called Eulerian form $N={q^k}{n^2}$ (with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$) can only have exactly one Euler factor $q^k$. On the other hand, assuming the existence of at least two OPNs, I am considering the problem of whether these two (distinct) OPNs can share the same Euler factor $q^k$. Given the abundancy index constraints, we have the equations $I(q^k) = \frac{1}{I({n_1}^2)} = \frac{1}{I({n_2}^2)}$. Thus, the related problem on the solitary status (or otherwise) of odd squares. This appears to be intractable, Pete said. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Jun 11, 2012 at 19:20
  • $\begingroup$ It is trivial to prove that the equations $$\gcd(n^2,\sigma(n^2))=\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}=\frac{\sigma(N/q^k)}{q^k}$$ hold. In particular, using a result from Dris (2012), we have $$\frac{\sigma(N/q^k)}{q^k} \geq 3$$ so that we obtain $$\gcd(n^2, \sigma(n^2)) \neq 1.$$ Thus, Greening's Theorem (a sufficient but not necessary condiiton for a number to be solitary) fails to establish the solitude of $n^2$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Sep 11, 2016 at 18:50

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