# A property for perfect numbers

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 $\alpha\geq 5$

• Somewhat related - On the index of an odd perfect number. – Jose Arnaldo Bebita-Dris Sep 1 '14 at 12:29
• 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 $3 \nmid n$. – Jose Arnaldo Bebita-Dris Mar 2 '18 at 7:36
• @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 :) – snulty Mar 2 '18 at 20:29