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comment Cardinality of the prime divisor set of a k-power sum
This follows from the finiteness of solutions to the $S$-unit equation. See for example the main theorem in math.uwaterloo.ca/pure-mathematics/sites/ca.pure-mathematics/… . Maybe there's an easier proof ...
1d
comment Graduate program applications that require questionnaires and other non-letter material
By the way, even the streamlined process we now have for MathJobs might be under threat -- this year, three universities made me upload letters for job candidates at their website rather than getting it themselves from Math Jobs. (Rochester, Vermont and North Carolina). I hope this infection doesn't spread, or that people will take steps in advance to resist it before too late.
Dec
12
comment How prove this polynomial inequality from a book
See Siegel's book "Lectures on the geometry of numbers" (pages 27 and 28).
Dec
5
comment Which natural numbers are a square minus a sum of two squares?
If $n>3$ is $3\pmod 4$ then it is easy to check that there is a representation of the desired form (indeed with $c=1$). So these numbers (and squares times these numbers) must be excluded from the idoneal list. For other integers, the problem seems pretty much equivalent to the ``one class per genus"/idoneal numbers problem.
Dec
5
comment Which natural numbers are a square minus a sum of two squares?
Do you allow $b$ or $c$ to be zero?
Dec
2
comment Is more alternating always better?
@User43408: If you are indeed Ken Perko, welcome to MO! Your suggested edit doesn't seem to fit completely with this question. You could ask that as a separate question if you see fit.
Nov
27
comment Do we know that 'most' finite groups are Galois groups of number fields?
@DerekHolt: I see my error. It was that $2^{k+1} < 3\times 2^k$ and the groups of order $2^{k+1}$ swamp the groups of order $3\times 2^k$.
Nov
27
comment Do we know that 'most' finite groups are Galois groups of number fields?
Since one can take a direct product of groups of order $2^k$ with say $C_3$, it doesn't seem correct to say that most groups are $2$-groups. However, I would guess that most groups are solvable: the orders of nonsolvable groups seems to be understood, see oeis.org/A056866 .
Nov
27
comment Isomorphism problem for two radical extensions
@FelipeVoloch: Thanks for the clarification, but surely those initially voting to close can vote to reopen if the question is edited and made correct. I'm also puzzled by how the answer came in after closure -- my guess would be that the answer was already in some draft form before the question was closed.
Nov
27
comment Isomorphism problem for two radical extensions
If the question is good enough to attract a response from User74230, it's good enough for this site. Voting to reopen.
Nov
26
comment lower-bound for $Pr[X\geq EX]$
That looks very nice! Well done!
Nov
26
comment lower-bound for $Pr[X\geq EX]$
Right, in terms of $\alpha$ only. (I think in some ranges (either $\alpha\to 0$ or $\alpha \to 1$) one in fact gets a uniform bound even independent of $\alpha$.) Note: I don't have any real applications for this, just curiosity!
Nov
26
comment lower-bound for $Pr[X\geq EX]$
The problem is to get lower bounds independent of $n$. From Pokrovskiy's work (which is quite involved, but gives more) I think I can do this. But maybe you can see a simpler way!
Nov
25
comment lower-bound for $Pr[X\geq EX]$
Here's Feige's paper: wisdom.weizmann.ac.il/~feige/Others/newmarkov.pdf
Nov
25
comment lower-bound for $Pr[X\geq EX]$
@fedja: Of course the example above, doesn't really have to do with numbers or primes. Suppose the $X_i$ are all $0$ with probability $1-\alpha$ and some value $a_i$ with probability $\alpha$. Then the expected value of $\sum X_i$ is $\alpha\sum a_i$. Estimate the probability that $\sum X_i$ is at least this expectation. A good bound for this would follow from the Manickam, Miklos, Singhi conjecture, and for certain ranges of $\alpha$ this would now follow from thework of Pokrovskiy arxiv.org/pdf/1308.2176.pdf . But perhaps I'm seeing more to this question than OP intended!
Nov
25
comment lower-bound for $Pr[X\geq EX]$
@fedja: Good point. Here's an example of what I have in mind. Factor a square-free number $N$ as a product of primes $p_1\cdots p_n$. Now take $a_i=\log p_i/\log N$, and $X_i=0$ with probability $1-\alpha$ and $a_i$ with probability $\alpha$. Then the expected value of $\sum X_j$ is $\alpha$, and the problem is to bound a weighted sum of divisors of $N$ that are at least $N^{\alpha}$. This problem was studied by Alladi, Erdos and Vaaler and Soundararajan, and when $\alpha$ is rational there is a lower bound depending only on $\alpha$ and not on $n$.
Nov
25
comment lower-bound for $Pr[X\geq EX]$
So I had in mind the situation that the variables are all independent. It would be good for OP to clarify if that's to be assumed or not. The situation when the variables are independent is related to problems in combinatorial number theory (studied by Alladi, Erdos and Vaaler), and to the Manickam, Miklosh and Singhi conjecture in combinatorics (on which there has been interesting progress lately).
Nov
24
comment lower-bound for $Pr[X\geq EX]$
This is actually a non-trivial and very interesting question. It should not be closed. I'll try to add an answer with references to related work if I get a chance.
Nov
23
comment Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
Seva: It doesn't seem to me that $p$ is assumed to be $1\pmod 8$. When $q$ divides $a$ (which is the example you raised) then the biquadratic symbol is determined by $(\frac{2}{q})$. I'm not an exper, and it may be safest to consult Lemmermeyer's book.
Nov
23
comment Biquadratic reciprocity for $p\equiv 1\pmod 4$ and $q\equiv 3\pmod 4$
The sign doesn't matter. If you multiply $(\frac{\sigma(b+\sigma)}{q})$ and $(\frac{-\sigma(b-\sigma)}{q})$ together, you get $(\frac{-\sigma^2(b-\sigma^2)}{q}) = (\frac{a^2}{q})=1$.