Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a tough one today.

Show that if $n$ is an odd perfect number, then not all of $3$, $5$, and $7$ are divisors of $n$.

Any and all help is appreciated. Thanks very much.

share|improve this question
I don't think this is known. –  Chris Eagle Nov 14 '12 at 2:39
Is this an exercise from somewhere? It's relatively well known that if an odd perfect number $n$ is not divisible by $3,\ 5$ or $7$ then $n$ has a large amount of prime factors, but I'm not aware that your statement holds. –  EuYu Nov 14 '12 at 2:39
Wikipedia says it's true and gives a reference to a 1949 paper by Kühnel. Wolfram MathWorld says it's a result by Catalan from 1888. –  lhf Nov 14 '12 at 2:43
@lhf Interesting. The reference provided is dead (even the doi link is broken) but luckily I found this reference which contains a surprisingly compact proof. –  EuYu Nov 14 '12 at 2:46
Oh, are we supposed to read the question as meaning that no odd perfect number is divisible by all of $3, 5, 7$? That's a rather perverse reading, but it does at least make the claim true. –  Chris Eagle Nov 14 '12 at 4:12

1 Answer 1

up vote 6 down vote accepted

I did look at the link, but of course the first thing I thought of was Euler's form of an odd perfect number from which this follows immediately. Euler proved that the prime decomposition $n=q^\alpha p_1^{2e_1}\cdots p_m^{2e_m}$ where $q\equiv \alpha\equiv 1\operatorname{mod} 4$, i.e. a perfect square times one extra prime to a power.

Taking this into account, if $n$ is divisible by $3, 5,$ and $7$, then $3$ and $7$ must occur to at least the power $2$.

The abundancy index $I(n)=\frac{\sigma(n)}{n}$ gets larger if you increase powers or number of primes (i.e. for any $k>1$, $I(kn)>I(n)$, see Laatsch's article).

Thus $2=I(n)>I(5\cdot 3^2\cdot 7^2)=\frac{1+5}{5}\cdot\frac{1+3+3^2}{3}\cdot\frac{1+7+7^2}{7}>2$, a contradiction.

share|improve this answer
The denominators should be $3^2$ and $7^2$, so that the product is $98.8/49$, very close to $2$, not drastically larger. –  Zack Wolske Aug 11 '14 at 16:21
+1, good catch @ZackWolske! =) –  Jose Arnaldo Dris Aug 28 '14 at 10:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.