Timeline for An integrality question about expressing an integer as a product of numbers below $n$
Current License: CC BY-SA 3.0
9 events
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| Dec 14, 2013 at 23:28 | comment | added | Greg Martin | Yeah, that's harder. I searched quickly for counterexamples, with $n=9$ (actually I ignored $j=5$ and $j=7$) and $X=\sum x_j=2,3,4,5,6$. There are a few integers less than $9^6$ that cannot be written as the product of $X=6$ integers not exceeding $9$; but none of them seem to arise from a nonnegative linear combination of the lattice points corresponding to $\{1,2,3,4,6,8,9\}$. I suspect if there were such a missing integer (with $X$ larger) of the form $2^m3^n$ with $m<2n$, then we might be in business for a counterexample. The closest I found (for $X=6$) was $2^{11}3^5$. | |
| Dec 14, 2013 at 22:34 | comment | added | Lucia | @GregMartin: Ok let me try one last time! Suppose the $x_j$ are rational. Then the situation for numbers really corresponds to the one for polytopes. Is there a counterexample here? | |
| Dec 14, 2013 at 22:31 | comment | added | Lucia | @GregMartin: Great that example works for the numbers case. The point of course is that while relations in the polytope situation give relations among numbers, there are more relations that one gets in the integer situation. So what about the problem as phrased for polytopes. | |
| Dec 14, 2013 at 22:27 | comment | added | Greg Martin | @Lucia: took me a little longer, but I broke this one too >:) | |
| Dec 14, 2013 at 22:27 | history | edited | Greg Martin | CC BY-SA 3.0 |
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| Dec 14, 2013 at 21:01 | comment | added | Lucia | @TheMaskedAvenger: That will only show that there is a representation with $\sum a_j$ being not more than $\sum x_j$ plus $n$. It's not at all clear that you can get $\sum a_j$ to be at most $\sum x_j$ (so that you can pile the difference on the $1$ exponent). | |
| Dec 14, 2013 at 20:59 | comment | added | The Masked Avenger | If I read Lucia's intent correctly, the matter is simply bookkeepping of exponents of primes dividing N. In this simpler case, I think the x_i can be "piled on" the largest prime power, and any difference placed on 1's exponent. | |
| Dec 14, 2013 at 20:35 | comment | added | Lucia | Thank you that is an excellent answer to my original question, but I had overlooked a subtlety as pointed out by TheMaskedAvenger. I really had in mind that $N$ is genuinely composed only of primes up to $n$ (so that one is in the polytope situation). Your answer came in as I was editing the question after TheMaskedAvenger's comment. Sorry about that. | |
| Dec 14, 2013 at 20:20 | history | answered | Greg Martin | CC BY-SA 3.0 |