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Answer by Anonymous for Hecke equidistribution

2013-06-12T01:20:08Z

A very down-to-earth treatment of this result of Hecke is in Chapter 5 of the nice book Geometric and Analytic Number Theory, by Hlawka, Schoißengeier, and Taschner. By down-to-earth, I mean that they deal directly with this specific case of Hecke's result, and that they prove it using very little -- the method is a modification of the Korevaar--Newman--Zagier approach to the prime number theorem, and so doesn't need any quantitative zero-free region (just a statement that there are no zeros of the appropriate objects on the line $\Re(s)=1$).

Answer by Anonymous for A generalized Möbius function?

2011-12-20T05:15:33Z

The partial sums of $\mu_k(n)$ are estimated in a paper of Addison:

A Note on the Compositeness of Numbers, A. W. Addison, Proceedings of the American Mathematical Society, Vol. 8, No. 1 (Feb., 1957), pp. 151-154,

Article Stable URL: http://www.jstor.org/stable/2032831

Most of the paper treats $\zeta^{\Omega(n)}$, but the (slightly different) function you are interested in is discussed in the closing paragraph.

Answer by Anonymous for A question related to Hilbert's Irreducibility Theorem

2011-12-03T04:12:10Z

The answer is yes, assuming that the two-variable polynomial $f_n(x)T^n + \dots + f_1(x)T + f_0(x)$ is irreducible over $K$.

This follows from the version of Hilbert's irreducibility theorem for number fields proved as Theorem 46 of p.298 of Schinzel's book Polynomials with special regard to reducibility: the relevant passage can be viewed on Google Books

http://books.google.ca/books?id=kca0JqBhnsIC&lpg=PA298&vq=hilbert's%20irreducibility%20theorem&pg=PA298#v=onepage&q&f=false

In fact, if I'm reading it correctly, it looks like one has irreducibility for all rational integers $\alpha$ belonging to an appropriate residue class.

Answer by Anonymous for explicit lower bounds on $|L(1,\chi)|$

2011-11-30T18:32:48Z

This is discussed on page 47 of Narkiewicz's new book (Rational Number Theory in the 20th Century); see

http://books.google.ca/books?id=3SWNZaDM6iMC&lpg=PP1&dq=rational%20number%20theory%20in%20the&pg=PA47#v=onepage&q&f=false

Reference [4268] is to

Metsankyla, T.: Estimations for L-functions and the class numbers of certain imaginary cyclic fields, Ann. Univ. Turku, Ser. AI 140, 1--11 (1970)

[3995] is

Louboutin, Stéphane(F-CAEN) Minoration au point 1 des fonctions L et détermination des corps sextiques abéliens totalement imaginaires principaux. (French) [Lower bound at the point 1 of L-functions and determination of the principal totally imaginary abelian sextic fields] Acta Arith. 62 (1992), no. 2, 109–124.

and [338] is

Barrucand, Pierre; Louboutin, Stéphane(F-CAEN) Minoration au point 1 des fonctions L attachées à des caractères de Dirichlet. (French) [Lower bound at the point 1 of L-functions associated with Dirichlet characters] Colloq. Math. 65 (1993), no. 2, 301–306.

Answer by Anonymous for About integer polynomials which are sums of squares of rational polynomials...

2011-11-30T17:47:06Z

In fact, if $P(x)$ is a polynomial with integer coefficients and if every arithmetic progression contains an integer $n$ for which $P(n)$ is a sum of two rational squares, then $P(x) = u_1(x)^2 + u_2(x)^2$ identically, where $u_1(x)$ and $u_2(x)$ are polynomials with integral coefficients. This follows from a theorem of Davenport, Lewis, and Schinzel; see the Corollary to Theorem 2 in Polynomials of certain special types (Acta Arith. IX, 1964, 107--116).

(In my restatement of their result, I use that being a sum of two rational squares is equivalent to being a sum of two integer squares. This is easy to prove directly from the characterization; alternatively, it follows from a lemma in Serre's book, attributed to Davenport--Cassels, used to prove the three squares theorem. Also, Davenport, Lewis, and Schinzel seem to have an argument similar to Gjergji's implicitly in mind in their proof of the Corollary above. So Gjergji's answer is the "real" one; but maybe this paper will interest others.)

Answer by Anonymous for How are these number-theoretical constants actually distributed?

2011-11-26T02:25:48Z

This paper by Kowalski also looks highly relevant: http://arxiv.org/abs/0805.4682?context=math

Answer by Anonymous for What Dirichlet doesn't tell...

2011-08-05T16:34:41Z

Let me sketch a proof. If you fix a bound $z$, then the events $P(\ell^2)$ for different $\ell \leq z$ are uncorrelated; this is just a consequence of the prime number theorem for progressions and the multiplicativity of Euler's function. This reduces the problem to "understanding the tails"; in other words, we have to show that as $z\to\infty$, the relative upper density of the primes divisible by $\ell^2$ for some $\ell > z$ tends to zero.

Consider the primes $p \leq x$. By the Brun--Titchmarsh inequality, the number of such $p$ for which $p-1$ is divisible by $\ell^2$ for some prime $\ell$ with $z < \ell \leq x^{1/4}$ (say) is $$ \ll \sum_{z< \ell \leq x^{1/4}} \frac{x}{\phi(\ell^2)\log{(x/\ell^2)}} \ll \pi(x) \sum_{z > \ell} \frac{1}{\ell^2} \ll \frac{\pi(z)}{x}. $$ Also, the number of $p$'s with $p-1$ divisible by $\ell^2$ for some $\ell > x^{1/4}$ can be estimated trivially: We just count how many $n \leq x$ are divisible by some $\ell^2$ with $\ell > x^{1/4}$, which is clearly at most $\sum_{\ell > x^{1/4}} \lfloor x/\ell^2\rfloor \ll x^{3/4}$. This is negligible for us. Hence, the relative upper density in the previous paragraph is $\ll 1/z$. So it does indeed tend to zero as $z\to\infty$.

Answer by Anonymous for Multiplicative functions whose Dirichlet series have essential singularities

2011-07-25T04:36:47Z

Hi Robert. For your particular $f$, you'll want to look at this article:

A Note on the Compositeness of Numbers
A. W. Addison
Proceedings of the American Mathematical Society
Vol. 8, No. 1 (Feb., 1957), pp. 151-154
Article Stable URL: http://www.jstor.org/stable/2032831

Addison gives the details for the case when $f(p)=\omega$, where $\omega$ is a primitive $3$rd root of unity, but remarks that the the case of $f(p)=i$ (or in general, any $\zeta_n$) is similar.

Answer by Anonymous for Approximate primitive roots mod p

2011-07-19T05:37:21Z

A result of Erdos and Murty asserts that if $\epsilon(p)$ is any decreasing function tending to zero, then $I(p) \leq p^{1/2-\epsilon(p)}$ for almost all primes $p$ (i.e., all but $o(\pi(x))$ primes $p \leq x$).

Kurlberg and Pomerance (see Lemma 20 in the paper mentioned below) show that for a positive proportion of primes $p$, one has the stronger bound $I(p) \leq p^{0.323}$. This follows from a result of Baker and Harman on shifted primes with large prime factors.

The Erdos--Murty paper is #77 at

http://www.mast.queensu.ca/~murty/index2.html

and the Kurlberg--Pomerance paper is

http://www.math.dartmouth.edu/~carlp/PDF/par13.pdf

See also Theorem 23 of this paper (which is conditional on GRH).

Answer by Anonymous for Greatest common divisor of a^{2^n}-1 and b^{2^n}-1

2011-07-17T01:45:47Z

A comment on one of Joe's questions: Let $B$ be any real number. It is known unconditionally that there are infinitely many $m$ for which $\phi(m)$ is a square and for which the smallest prime factor of $m$ exceeds $B$. One can even take $m$ as a product of two primes here; see, e.g., article 4 from

http://www.integers-ejcnt.org/vol11a.html

or an arXiv preprint of Tristan Freiberg.

If we choose $B$ larger than $|a|$ and $|b|$, then $m \mid \gcd(a^{\phi(m)}-1, b^{\phi(m)}-1)$, and so there is a prime $> B$ in the support of $\gcd(a^{n^2}-1, b^{n^2}-1)$.

Comment by Anonymous

2011-12-11T22:31:08Z

@Theo: I learned about it on math.stackexchange: Two proofs are at math.stackexchange.com/questions/79861/…

Comment by Anonymous

2011-12-11T20:19:31Z

It seems your pessimism is justified. If theta is a rational multiple of $\pi$, then $2\cos(\theta)$ is an algebraic integer. But if $\theta = \arctan(\sqrt{3}/5)$, then $2\cos(\theta) = 5/\sqrt{7}$.

Comment by Anonymous

2011-12-10T05:31:48Z

Timothy, I think you'd enjoy looking at the paper Iterated exponents in number theory, by D.B. Shapiro and S.D. Shapiro: math.osu.edu/~shapiro/IE.pdf

Comment by Anonymous

2011-11-14T20:24:21Z

Two quick comments. It's maybe interesting to note that for prime $q$, it's easy to establish the Monthly bound: Any prime divisor $p$ of $2^q-1$ works. Also, I'm skeptical of the bound attributed to Oesterle. I'd wager that the best bounds, on ERH, are those in ams.org/journals/mcom/1996-65-216/ Note that their upper bound is bigger than $q^2$, and so doesn't settle the OP's problem. The discussion in this paper also suggests the result attributed to Oesterle is based on a misunderstanding or a typo.

Comment by Anonymous

2011-07-25T14:03:36Z

As for your general question, it appears that certain sorts of essential singularities (e.g., the sort that appear in this example) can be handled by the Selberg--Delange method, which is described in Chapter II.5 of Tenenbaum's book.

Comment by Anonymous

2011-07-21T21:13:17Z

I think that's right. The proof of Lemma 20 in that paper shows that you could improve $0.323$ to $\epsilon$ if you knew that there were infinitely many shifted primes $p-1$ with prime factors $> p^{1-\epsilon}$. Of course we think that $p-1$ is infinitely often twice a prime, which is much stronger and which would give the boundedness you originally asked for -- but this still seems hopeless. (However, progress towards this sort of conjecture plays a key role in the proof of the Heath--Brown proof you mentioned, and in the earlier work of Gupta and Murty.)

Comment by Anonymous

2011-07-17T02:04:18Z

Well, I guess I've actually made things too hard here, since $\gcd(a^n-1, b^n-1) \mid \gcd(a^{n^2}-1, b^{n^2}-1)$; hence the support problem is trivial, again by Fermat's little theorem. I'll leave my answer up for now though.