# Tag Info

14

Klueners and Malle have a database of number fields of degree $\leq 19$ that (tries) to include every Galois group and every possible signature. Examining this database shows that for composite $n$ with $4 \leq n \leq 18$, $X_{n} = \{ k : 0 \leq k \leq n \text{ and } k \equiv n \pmod{2} \}$. EDIT: If $n = 2k$ is even, then $\mathbb{Z}/2\mathbb{Z} \wr \... 13 We give a uniform approach to$p \leq 61$by applying analytic discriminant bounds to the Hilbert class field. To be sure this is not entirely "conceptual", but then some computation is needed even to deal with$p < 36$using Minkowski. If$p = 4k+1$is prime then$K = {\bf Q}(\sqrt{p})$has odd class number$h$, so either$h=1$or$h \geq 3$. If$h \...

10

I assume you mean $K_1=\mathbb Q(\sqrt{d_1})$. 1) If by $K$ you mean $\mathbb Q(\sqrt{d_1d_2})$, then there is no simple relation between $h$ and $h_1$ and $h_2$. 2) If by $K$ you mean the quartic biquadratic field $\mathbb Q(\sqrt{d_1},\sqrt{d_2})$, a theorem of Herglotz says that $h=h_1h_2h_3/2^j$, where $h_3$ is the class number of $\mathbb Q(\sqrt{... 6 Jensen On the number of real roots of a solvable polynomial includes a proof of: Loewy’s theorem. Let$K$be a real number ﬁeld and$f(X)$an irreducible polynomial in$K[X]$of odd degree$n$. If$p$is the smallest prime divisor of$n$and the Galois group of$f(X)$over$K$is solvable, then$r(f) = 1$or$n$or satisﬁes the inequalities$p ≤ r(f)≤ n−p ...

5

It seems worth noting that the equation in the title does have infinitely many solutions in positive integers, as for all $n$ it is satisfied by $$a={n(n^2-3)\over2},\ b=n^2-1,\ c=n^2-3.$$ The number of solutions of this form with $a\le k$ will be on the order of $\root3\of{2k}$, but Dmitry has found solutions not of this form.

5

See Section VIII.7 in Fröhlich-Taylor, Algebraic Number Theory, Cambridge University Press.

5

As Fedor Petrov says, this looks incorrect. The presence of $$2/\pi = \prod_{n}\left(1-\frac{1}{(2n)^2}\right)$$ (known as the Wallis product) makes me think Sylvester is mistakenly comparing it to this somehow, with the missing implicit step being 'every prime is at least some even number', so $M\geq W$, since if $2n\leq p$ then we can replace the $(1-1/(... 4 I also do not understand. The infinite product of$(1-1/p^2)$equals$6/\pi^2<2/\pi$. 4 This is the Polylogarithm, valid for arbitrary complex order$s$and for all complex arguments$a$with$|a| < 1$; it can be extended to$|a| \ge 1$by the process of analytic continuation. It is related to the Lerch zeta function. The functional equation is originally due Jonquiere in 1889. The fullest investigation of the functional equation I can ... 4 Here's the Landau paper at the Internet Archive: https://archive.org/details/archivdermathem37unkngoog/page/n324/mode/1up Here's the Watson paper at the EuDML: https://eudml.org/doc/168581, which links to the GDZ for a scan. 3 I think the 'identical equation' Sylvester has in mind is $$\sum_q \mu(q) \frac{x^q}{1-x^{2q}} = x+x^5+x^{13}+x^{17}+x^{25}+x^{29}+\cdots$$ where the left-hand sum is over all natural numbers$q$divisible only by primes of the form$4s+3$and the right hand side is the sum of all powers$x^r$where$r$is divisible only by primes of the form$4s+1$. (... 3 See Fesenko and Vostokov - Local fields and their extensions. (i) is Proposition 3.3(2). (ii) is Proposition 3.2(1). (iii)(1) is Proposition 3.5(1) (and, yes,$b$may be chosen as a uniformiser). For (iii)(2), I think you meant$e \mid \lvert k\rvert - 1$, not$p \mid \lvert k\rvert - 1$(which is impossible). Then ($k$, hence)$K$contains an$e$th ... 3 Your question may be get closed, but before that happens, I will leave an answer that might help you. Taken very literally the answer to your question is likely no. It is known that there can be no algorithm to decide whether a Diophantine equation has an integer solution (Hilbert's 10 problem), which makes it unlikely that there would be a sheaf theoretic ... 3 Any$A$-module homomorphism$B/b\to A/a$gives, by composing with$B\to B/b$, a homomorphism$B\to A/a$. Using the surjection$A\to A/a$and the fact that$B$is$A$-projective, the map lifts to give a homomorphism$B\to A$. So, the only real assumption needed is that$B$is$A$-projective. 2 Take$x=1$and$y^2=z$. Then conditions are that$z/4-1/16$,$10+9z$and$5+16z$are squares. By multiplying these conditions we get the elliptic curve$u^2 = (z/4-1/16)(10+9z)(5+16z)$. It has torsion group$Z/2Z \times Z/4Z$and rank$1$. A point$P$of infinite order corresponds to$z=-1/3$. One of the torsion points$T$corresponds to$z=1/4$and this ... 2 Here is a possible explanation from Chapter 1 section 4 of "The Geometry of Schemes" by Eisenbud and Harris. The apparently abstract idea of the functor of points has its root in the study of solutions of equations. Let$X = \text{Spec}( R)$be an affine scheme, where$R = \mathbb{Z}[x_1, x_2,...]/(f_1, f_2,...)$. If T is any other ring (one should ... 2 Here's another infinite family. Let$x,y$be positive integers such that$x^2-2y^2=\pm1$– there are infinitely many such pairs. Let$a=x^2$,$b=2y^2$,$c=xy$, then a little algebra will show that$(a,b,c)$satisfy the equation in the title. E.g.,$x=3$,$y=2$leads to$(9,8,6)$, and$x=7$,$y=5$yields$(49,50,35)$, two triples already found by Dmitry, ... 2 Some complementary heuristics, too long for a comment. Again start from the fact that the class number$h$of$K = {\bf Q}(\sqrt{p})$is odd if$p$is a prime of the form$4k+1$. This time we compare with Dirichlet's class number formula, which here gives $$L(1,\chi_p) = \frac{2\log \epsilon}{\sqrt p} h,$$ where the character$\chi_p$is the Legendre ... 2 It is not easy to give all the details so I'll give a sketch in the case of unramified prime For$p$an unramified prime number,$Q\subset O_K$a prime ideal above$p$, those of$O_k$are of the form$P_g =g(Q)\cap O_k$for$g\in G$with norm$N(P_g)=p^{f_g}$for some integers$f_g\sigma$a Frobenius such that$\forall a\in O_K,\sigma(a)\equiv a^p\bmod ...

2

There's probably a more elementary reference, but, according to Bushnell, Henniart, and Kutzko - Local Rankin–Selberg convolutions for $\operatorname{GL}_n$, (6.1.2), if $m$ is the level of $\pi$, then the conductor of $\pi$ depends on a choice of additive character $\psi$, which will be trivial on $\mathfrak p^{c(\psi)}$ but not on \$\mathfrak p^{c(\psi) - 1}...

1

Serre deals with problems of this type in the paper: Serre -Divisibilité de certaines fonctions arithmétiques The fact you want should follow from the results in Sections 1 and 2. Alternatively, this should also follow from the more general Proposition 2.2 in: https://arxiv.org/abs/1810.06024

Only top voted, non community-wiki answers of a minimum length are eligible