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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.

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    $\begingroup$ I don't think this is known. $\endgroup$ – Chris Eagle Nov 14 '12 at 2:39
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    $\begingroup$ 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. $\endgroup$ – EuYu Nov 14 '12 at 2:39
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    $\begingroup$ 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. $\endgroup$ – lhf Nov 14 '12 at 2:43
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    $\begingroup$ @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. $\endgroup$ – EuYu Nov 14 '12 at 2:46
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    $\begingroup$ 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. $\endgroup$ – Chris Eagle Nov 14 '12 at 4:12
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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 $2=I(n)>I(5\cdot 3^2\cdot 7^2)=\frac{1+5}{5}\cdot\frac{1+3+3^2}{3^2}\cdot\frac{1+7+7^2}{7^2}>2$, a contradiction.

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