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If $n=\prod\limits_{i=1}^k p_i^{\alpha_i}$, is a perfect number, with $p_i$ distinct primes, $\alpha_i\geq1$, then for each $i$ there is a $j \in \{1,\ldots,k\}$ such that $p_i \mid (p_j^{\alpha_j+1}-1)$. $j$ is quite clearly not equal to $i$.

Proof Note that $\sigma(n)\phi(n)=n\prod\limits_{i=1}^k \frac{p_i^{\alpha_i+1}-1}{p_i}$.

Now $n$ is perfect so $\sigma(n)=2n$, i.e.

$2\phi(n)=\prod\limits_{i=1}^k \frac{p_i^{\alpha_i+1}-1}{p_i}$.

Lhs is an integer so must the rhs be. $\square$

It's rather trivial for even perfect numbers.

e.g. $n=2^{p-1}(2^p-1)$, $p$ prime and $2^p-1$ a mersenne prime. Then $2^p-1\mid 2^{p-1+1}-1$ and $2 \mid (2^{2p}-2^{p+1}+1-1)$.

I noticed this while studying yesterday, just curious where this might have been in use already? Or even if it's any use at all...

Note: The condition is necessary but not sufficient since $n=40$ satisfies it and is not perfect. i.e. $40=2^3\cdot5$ and $5\mid2^4-1$ and $2\mid 5^2-1$. Apart from being even and $5$ not a mersenne prime, $2\cdot\phi(40)=32\neq36=\frac{2^4-1}{2}\cdot\frac{5^2-1}{5}$.

I mean in principal you could deduce some possibly other not very helpful things like, given eulers form for odd perfect numbers $n=q^\alpha \prod\limits_{i=1}^k p_i^{2\epsilon_i}$, $q\equiv\alpha\equiv1 \pmod{4}$ that $q \mid p_i^{2\epsilon_1+1}-1$ for some $i$, and hence $ord_q(p) \mid 2\epsilon_i+1$. $ord_q(p)$ stands for the multiplicative order of $p$ modulo $q$. And also if $p_i^{2\epsilon_1+1}-1=kq$ then $k \equiv 0,2 \pmod{4}$ for $p_i\equiv 1,3 \pmod{4}$.

That's all I could think of off the top of my head...

Update: Also if $q=5$ and $3 \nmid n$ then $\alpha\geq 5$

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  • $\begingroup$ Somewhat related - On the index of an odd perfect number. $\endgroup$ – Jose Arnaldo Bebita-Dris Sep 1 '14 at 12:29
  • $\begingroup$ Regarding your last update: That is easy to prove. Suppose to the contrary that $q=5$, $3 \nmid n$ and $\alpha = 1$. Then $q+1=\sigma(q)=\sigma(q^k) \mid \sigma(N) = 2N$, so that $(q+1)/2 \mid N$. Since $q=5$ and $\gcd(q,q+1)=1$, then $3=(q+1)/2 \mid n^2$, which implies that $3 \mid n$. This is a contradiction. Hence it must be the case that $\alpha \geq 5$, when $q = 5$ and $3 \nmid n$. $\endgroup$ – Jose Arnaldo Bebita-Dris Mar 2 '18 at 7:36
  • $\begingroup$ @JoseArnaldoBebitaDris Thanks for the comment, it's been a while since I've thought about number theory, I was doing a module at the time :) $\endgroup$ – snulty Mar 2 '18 at 20:29
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Yes, it's pretty well-known. I think it's usually not even called out explicitly. Going off memory you might find it in Hare's paper on odd perfect numbers, or else in one of Ianucci's papers on the same topic.

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  • $\begingroup$ I would assume as much. Just a curiosity I didn't know about. I'll look them up so. Thanks for the references! (+1) for taking the time $\endgroup$ – snulty May 21 '14 at 21:50

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