2
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For example:
$6$ has this property since proper divisors of $6$ are: $2$ and $3.

From this thread: What does the product of all proper divisors equal to?

My attempt was:
If $n = p_1^{a_1} \times p_2^{a_2} \times ... \times p_k^{a_k}$
Then $n = n^{\frac{\tau(n)}{2} - 1}$. Where $\tau(n) = (a_1 + 1) \times (a_2 + 1) \times ... \times (a_k + 1)$

So is it good enough to stop here, or we can express $n$ in a better formula?

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  • $\begingroup$ You missed a factor 2 in the exponent from Arturo Magidin's answer to the earlier question. This leads naturally to Sivaram Ambikasaran's answer to this one. $\endgroup$ – Ross Millikan Feb 28 '11 at 2:14
  • $\begingroup$ @Ross Millikan: Opp! My bad. Thanks for pointing out. $\endgroup$ – Chan Feb 28 '11 at 2:16
5
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$n = n^{\tau(n)/2 - 1}$ implies $\tau(n)=4$ and so $n$ is a product of two primes or the cube of a prime.

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  • $\begingroup$ Nice thought! Thanks for this clever hint. $\endgroup$ – Chan Feb 28 '11 at 2:18
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Any number of the form $n = p_1 \times p_2$ where $p_1,p_2$ are primes.

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  • $\begingroup$ Ambikasaran: Thank you. $\endgroup$ – Chan Feb 28 '11 at 2:19
  • 1
    $\begingroup$ You've missed $n=p^3$. $\endgroup$ – lhf Feb 28 '11 at 14:44

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