Timeline for Partitioning the set of divisors into two sets with equal product
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Aug 26, 2013 at 5:26 | comment | added | Theo Johnson-Freyd | I think this is a great puzzle, but it does not appear to be about research mathematics, and would be better suited for M.SE. | |
| Aug 26, 2013 at 0:56 | answer | added | Linus Hamilton | timeline score: 5 | |
| Aug 26, 2013 at 0:02 | comment | added | Gerry Myerson | Numbers whose divisors multiply to a square are tabulated at oeis.org/A048943 --- it says, $n$ is a 4th power, or $n$ has at least two prime factors with odd multiplicities, or $n$ has at least one prime factor with multiplicity 3 modulo 4. | |
| Aug 25, 2013 at 19:42 | comment | added | Anthony Quas | So I guess that what I should have said is each prime should be raised to a power that is congruent to 2 or 3 modulo 4. | |
| Aug 25, 2013 at 19:38 | review | Close votes | |||
| Aug 26, 2013 at 10:01 | |||||
| Aug 25, 2013 at 19:35 | comment | added | Greg Martin | The product of all the divisors of $n$ equals $n^{\tau(n)/2}$, where $\tau(n)$ is the number of divisors of $n$. (This can be easily proved with a pairing argument.) If the divisors of $n$ can be partitioned in the desired way, it follows that the product of all divisors of $n$ is a perfect square. Therefore a necessary condition is that $n^{\tau(n)}$ be a perfect 4th power; in other words, either $\tau(n)$ is a multiple of 4, or $n$ is a 4th power. I suspect that this necessary condition is also sufficient. | |
| Aug 25, 2013 at 19:31 | comment | added | Greg Martin | not so: for 16, we have {1,2,16} and {4,8}. | |
| Aug 25, 2013 at 19:23 | comment | added | Anthony Quas | each prime has to be raised to an odd power | |
| Aug 25, 2013 at 18:39 | history | asked | user38138 | CC BY-SA 3.0 |