GH from MO
Reputation
515/400 score
 3h comment On the number of divisors in a given range @Arul: There is no such bound, i.e. you can get $(\log N)^{1000}$ divisors if you wish. In fact the number of divisors in the given range can be much larger than polylogarithmic, but I have not attempted to determine the maximum order of this quantity. 7h comment On the number of divisors in a given range Yes. If $n\geq t$ and $n\in\mathbb{Z}$, then $n\geq\lceil t\rceil$. Similarly, if $n\leq t$ and $n\in\mathbb{Z}$, then $n\leq\lfloor t\rfloor$. Hence my last display is clear with $\leq$ around $Mp_1p_2p_3p_4p_5$. However, we talk about a product of different primes, so strict inequality also follows (namely for $p_2$, $p_3$, $p_4$ the intermediate inequalities are strict). Note also that $\leq$ is sufficient in the last display. 7h revised On the number of divisors in a given range deleted 6 characters in body 8h comment Goldbach for certain classes of $n$ @martin: The proof is in his notes (I guess). He has many unpublished results, actually. Smaller values like $c=1/100$ appeared in the literature (by other authors), I am sure you can find them with Google or MathSciNet or Zentralblatt. 8h revised Goldbach for certain classes of $n$ edited tags 8h answered Goldbach for certain classes of $n$ 8h revised On the number of divisors in a given range added 1 character in body; edited title 8h answered On the number of divisors in a given range 11h revised rational numbers and triangular numbers edited tags 2d comment Analytic continuation for $L$-functions of elliptic curves It is not true that $L(s,\chi)$ is a product of Dirichlet $L$-functions when $K$ is abelian over $\mathbb{Q}$. For example, let $\chi$ be a quadratic Hecke character of $K$, and let $L/K$ be the quadratic extension attached to $\chi$. Then $\zeta_L(s)=\zeta_K(s)L(s,\chi)$, and the condition is equivalent to $L/\mathbb{Q}$ being abelian, which is stronger than $K/\mathbb{Q}$ being abelian. Another example is provided by any classical CM elliptic curve $E/\mathbb{Q}$. Indeed, in this case $L(s,E)=L(s,\chi)$ with $K/\mathbb{Q}$ quadratic, but $L(s,E)$ does not factor into Dirichlet $L$-functions. 2d revised Rational points on the “quintic circle” $x^5 + y^5 = 7$ edited tags Nov 21 awarded Good Answer Nov 21 revised No nonconstant coprime polynomials $a(t)$, $b(t)$, $c(t) \in \mathbb{C}[t]$ where $a(t)^3 + b(t)^3 = c(t)^3$ edited tags Nov 21 comment No nonconstant coprime polynomials $a(t)$, $b(t)$, $c(t) \in \mathbb{C}[t]$ where $a(t)^3 + b(t)^3 = c(t)^3$ Perhaps you can add that a nonconstant holomorphic map from the Riemann sphere $\mathbb{C}\mathbb{P}^1$ to an elliptic curve $E=\mathbb{C}/\Lambda$ would lift to a nonconstant holomorphic map $\mathbb{C}\mathbb{P}^1\to\mathbb{C}$, which is impossible by Liouville's theorem (using that $\mathbb{C}\mathbb{P}^1$ is compact). Nov 18 comment Proving Legendre's Sum of 3 Squares Theorem via Geometry of Numbers @IgorRivin: Yes, probably this is why I thought I was wrong here. At any rate, a modern exposition of Legendre's proof would be welcome (I recall Andre Weil wrote about this in his history book, but I am too lazy and busy to check). Thanks for your comment! Nov 18 revised Primes $P_{2n-1}$ that are $2$ mod $3$ edited tags Nov 18 answered Proving Legendre's Sum of 3 Squares Theorem via Geometry of Numbers Nov 18 revised Where did the term “additive energy” originate? edited tags Nov 18 comment No Siegel-Landau zeros for $\mathrm{GL}(n)$ Continued: Finally, it should also be mentioned that Hoffstein-Ramakrishnan proved the absence of Siegel zeros for all non-selfdual cusp forms on $\mathrm{GL}_n$ over a number field. Nov 18 comment No Siegel-Landau zeros for $\mathrm{GL}(n)$ Your question is very good, but there is some confusion in your post. First, a Maass form is also a cusp form. Second, Hoffstein-Lockhart proved a Siegel type ineffective lower bound for $L(1,\mathrm{ad}^2\pi)$, but they did not prove the absence of Siegel zeros. Third, Hoffstein-Ramakrishnan proved the absence of Siegel zeros for all non-dihedral $\mathrm{GL}_2$ cusp forms, even over number fields. Note that for dihedral forms the problem is equivalent to the case of $\mathrm{GL}_1$, which is unresolved.