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10

This is false. Let $p$ be an odd prime, let $\ell$ be another prime, and let $m$ be a small prime divisor of $p^{\ell}-1$, that doesn't divide $p-1$. Let $n= 1 + \frac{ p^{\ell}-1}{m}$. Then $n-1$ is a multiple of $p-1$, is not a multiple of $p^{\ell}-1$, and is not a multiple of $p^{k}-1$ for any other $k$ because $p^{\ell}-1$ is not a multiple of $p^{k}-... 10 Geyer in Unendliche algebraische Zahlkörper, über denen jede Gleichung auflösbar von beschränkter Stufe ist, Satz 1.13 and the paragraph after that, gives a full characterization of which abelian profinite groups occur as absolute Galois groups: They are either$\mathbb{Z}/2\mathbb{Z}$or$\prod_p\mathbb{Z}_p^{c(p)}$for some cardinal numbers$c(p)$. Side ... 8 For simplicity, let's consider the case of holomorphic modular forms over$\mathbb Q$of squarefree level and trivial nebentypus. Then one knows from Iwaniec, Henryk; Luo, Wenzhi; Sarnak, Peter. Low lying zeros of families of$L$-functions. Publications Mathématiques de l'IHÉS, Tome 91 (2000) pp. 55-131. an asymptotic formula for the dimensions of the ... 6 It is conjectured that the number of twin primes between$x$and$2x$has order of magnitude$x/\log^2 x$. If this is true, then$F(s)$diverges for any$s<1$. See also this related MO post. 5 Yes. Note that for large$k$we have$1^m+\dots+(k-1)^m\geqslant k^m$. Choose minimal$k$with such property. Then$k>2$and$2k^m>2(k-1)^m>1^m+\dots+(k-1)^m$by minimality. 4 Well, the abscissa of convergence is$1$, though I see no way to deduce anything about it from the Polignac conjecture.$\sigma_P \le 1$is obvious from$|\zeta_P(s)| \le \zeta(\Re s)$, while for$\sigma_P \ge 1$it is enough to prove that$\zeta_P(1) = \infty$: $$\zeta_P(1) \ge \sum_{k=1}^\infty \frac{1}{2^{k+1}}\sum_{n = 2^k}^{2^{k+1}}\frac{1}{g_n} \gg \... 4 I think I found an answer to the question above: Let k(a,b) be a (positive definite \ge 0, symmetric) kernel on \mathbb{N}\times \mathbb{N} such that if k^*(a,b) is a function on G \times G then we have:$$k^*(a,b) = k(a',b')$$where a'=\frac{a}{\gcd^*(a,b)}, b'=\frac{b}{\gcd^*(a,b)}, then k^* is a kernel on G\times G. Proof: Since k ... 4 Here is a way to argue without showing directly that the polynomial must have degree 2. It was explained to me by Borys Kadets (all further mistakes are, of course, my contribution). Lemma. If a set of primes S of density \frac{1}{2} admits such polynomial then some subset S'\subset S with \#(S\setminus S')<\infty admits a monic quadratic ... 4 The negative solution of Hilbert's tenth problem means that there is no algorithm that would recognize the solvable polynomial equations. Hence the "clean partition" you are looking for does not exist. Of course, one can partition the polynomials any way one likes. The partition solvable/non-solvable/undecidable is a legitimate one, except that one "does ... 4 The power series F(x) is closely related to the series of the "exponential reversion of Fibonacci numbers"$$R(x)=\sum_{n\ge1}r_n\frac{x^n}{n!}$$(the r_n are A258943, quoted in a comment). In fact it appears that, again in the notation of Henri Cohen,$$a_{n+1}=nr_n,$$equivalently$$F'(x)=xR'(x).$$So if the Fibonacci numbers are encapsulated by$$x=\... 3 Obviously, abscissa of convergence is at most 1, because the series is dominated by the series for$\zeta(s)$. In fact, it is equal to 1. Indeed, assume that your series converges for some$s=1-\varepsilon$,$\varepsilon>0$. Then the sum $$\sum_{X<n<2X} (ng_n)^{-s}$$ is bounded by constant independent of$X$. Let us prove this is not the case. ... 3 This is a partial answer to Q2, in particular a full answer to the question asked by François Brunault in the comments: By the way, if K is ferile, will A(K)⊗Q be necessarily be infinite-dimensional? Here, A is a non-trivial abelian variety over K, and fertile = large = ample = for every geometrically integral smooth K-variety the set of K-rational ... 3 The problem of this question is qualitatively and quantitatively different in some ways from that considered by Andrew Bremner and myself. If we take the cubic, with$N$as a fixed constant, it is possible to show that the related elliptic curve is \begin{equation*} E_N:G^2=H^3+((35N+18)H+4(1260N+1441))^2 \end{equation*} with the formulae linking$(H,G)$to ... 3 Denote by$\mathbb{A}$the adeles over a number field$F$. An cuspidal automorphic representation$\pi$of$\text{GL}_2(\mathbb{A})$now factors by the tensor product theorem as$\pi\cong \prod_v\pi_v$, where$\pi_v$are smooth irreducible unitary representations of$\text{GL}_2(F_v)$. The Ramanujan-Petersson conjecture now states that the representations$\...

2

Probably not. I don't see an easy proof in general. Similar questions have been asked in Arithmetic Dynamics (look for dynamical analogues of Mordell-Lang). Clearly, the answer is no for $n=2$. For $n=3$, if you had such a situation then you would get infinitely many points on a fixed number field on the curves of the form $P(x_1^{m^k},x_2^{m^k},x_3^{m^k})=0$...

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If $x=1+p^my$ ($1\le y\le p^{t-m}$) then $$\psi \left(\frac{ax+bx^{-1}}{p^t} \right)=\psi \left(\frac{a(1+p^my)+b(1+p^my)^{-1}}{p^t} \right)=\psi \left(\frac{a+b}{p^t} \right)\psi \left(\frac{f(y)}{p^{t-m}} \right),$$ where $f(y)=ay+b(-y+y^2p^m-y^3p^{2m}+\cdots)$. The sum $$\sum_{y=1}^{p^{n}} \psi \left(\frac{f(y)}{p^{n}} \right)$$ can be calculated ...

1

From the way you phrase your question, I suspect that you misunderstand something. Being supercuspidal is a local condition. If $\pi$ is an automorphic representation of $\mathrm{GL}_2$ over $\mathbb{Q}$, then $\pi_p$ might be supercuspidal for some ramified primes $p$, and non-supercuspidal for some other ramified primes $p$. The Ramanujan conjecture is a ...

1

If it is required that $D$ be interior to $\triangle ABC$, then there are smaller examples than the one shown by Noam Elkies. Each of the following two has the bonus that it is an orthic system and can be embedded on the integer lattice with one edge parallel to each axis. Another scalene example is slightly smaller (of area 1014). If we relax the above ...

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