17

This question grabbed my attention a couple of years ago and I've just put a paper on the arXiv with new upper bounds for $k=5$. I began by computing lots of data, then teased out the structure of particularly well-performing configurations. The headline is that $f(5,n) = O(n^{-4/3})$, improving to $O(n^{-7/3})$ infinitely often.
The first picture that ...

16

For simplicity, I will assume that $f$ is a cusp form (hence $k\geq 12$ and $k$ is even).
The answer to question (a) is negative. It was proved independently by Rankin (1939) and Selberg (1940) that the Dirichlet series
$$\sum_n\frac{|a_n(f)|^2}{n^s}$$
has a simple pole at $s=k$. Hence infinitely many coefficients $a_n(f)$ are nonzero.
Regarding question (b),...

16

I think that there is indeed some possibility to lower the bound, and this is something I've looked at seriously a few times. I spent a semester (in 2019) with the Computational Number Theory Group here at BYU trying to do it, without success. Let me outline some of the difficulties we found.
Issue 1. The main technique used in establishing the current ...

15

(I'm posting this as an answer since I prefer to mantain my anonymity, thus I cannot comment.)
I'm also from a third-world country. At the age of 24, when I was finishing my master's thesis, I was diagnosed with bipolar disorder, and I was drinking way too much. It took me years to recover from alcoholism and to handle my depression. There were three things ...

14

See Chapter 15 ("Oscillation of error terms") in Montgomery-Vaughan: Multiplicative number theory I. See especially Theorems 15.2-15.3 and 15.11.
Added by Steven Clark and GH from MO. For convenience, we copied below the relevant theorems and some additional text from the book. As usual,
$$M(x):=\sum\limits_{n\le x}\mu(n)$$
is the Mertens function.
...

12

This specific question is probably not addressed in the literature; let's try to figure it out!
Let $K$ be an algebraic extension of $\mathbb Q_p$ such that the tilt of $\widehat{K}$ is isomorphic to $\mathbb F_p((t^{1/p^\infty}))$. We can observe the following:
Tilting preserves residue fields, so necessarily $K$ has residue field $\mathbb F_p$, i.e. is ...

12

A variant of the Siegel-Walfisz theorem states that there is a constant $c>0$ with the following property. For any $A>0$ and $q\leq(\log x)^A$, we have
$$\displaystyle \sum_{\substack{n\leq x\\n \equiv a \bmod q}}\mu(n)\ll_A x\exp\left(-c\sqrt{\log x}\right).$$
See Exercise 13 for Section 11.3 of Montgomery-Vaughan: Multiplicative number theory I.

12

There is a completely elementary way to see that the answer to a) is negative - if $f$ had only finitely many Fourier coefficients, it would extend to an entire function on $\mathbb C$. But for any $\pmatrix{a&b\\c&d}\in\Gamma$ we get
$$f\left(\frac{b}{d}\right)=f\left(\frac{a\cdot 0+b}{c\cdot 0+d}\right)=(c\cdot 0+d)^kf(0)=d^kf(0)$$
which means that ...

12

Write the coordinates $a_{ij}$ of one embedding into $GL_n$ as polynomial functions, defined over $\mathbb Q$, in the coordinates $b_{ij}$ of a different embedding into $GL_n$. We can do this because, by definition, embeddings give an isomorphism of algebraic varieties. (I guess we should allow the inverse of the determinant as one of our coordinates. I will ...

12

1. Normality of $\lambda^*$ in base $2$ is equivalent to the Chowla's conjecture. It is widely believed to be true, but very far from being proved.
2. Ergodicity of the Liouville function for Cesàro averages on $([N])_{N\in\mathbb{N}}$ (cf. Definition 2.5 in arXiv:1611.09338) is also equivalent to the Chowla's conjecture. Thanks to Will Sawin for clarifying ...

11

To produce an obvious counterexample to (1), fix a finite field $F$ of characteristic $p$, $n\ge 3$ and $L=\bigcup_m F[x_1^{p^{-m}},\dots,x_n^{p^{-m}}]$. This is a perfect field. The symmetric group $S_n$ acts by permuting the variables. Let $K$ be the fixed point field. Then $L$ is a finite Galois extension of $K$ with Galois group $S_n$.
A counterexample ...

10

Yes, this is always the case.
If $\alpha=\sqrt{n}$, then $|\mathrm{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})|=2$ (since there are no intermediate fields) so $\mathrm{Gal}(\mathbb{Q}(\alpha)/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}$ since that is the only group with two elements, thus it is abelian.
By the Kronecker-Weber theorem, that means that $\mathbb{Q}(\alpha)$ is ...

9

For (3), the answer is no, because there is no irreducible element in $\overline{\mathbb{Z}}$: for example, every element is a square.
Similarly for (1), the answer is no because there are infinite ascending chains of ideals: for example, $(x) \subset (x^{1/2}) \subset (x^{1/4}) \subset \cdots$, where $x$ is not a unit (e.g. $x=2$) and $\{x^{1/2^n}\}$ is a ...

8

We show the following.
Theorem. For any $n \geq 6$, there is a permutation $\sigma \in S_n$ such that $\prod_{k=1}^n k^{\sigma(k)}$ is a square (respectively, a cube).
Proof. Let us first handle the case of squares. It is equivalent to find a subset $A$ of $\{1,\ldots,n\}$ with cardinality $r = \lceil \frac{n}{2} \rceil$ such that the product of the elements ...

8

You seem to mean two slightly different things by the $p=\ell$ case. Your first theorem is about a field of characteristic $p$, and the second is about $\mathbb C_p$, which is a field of characteristic zero. Let's talk first about the significance of this.
For a field of actual characteristic $p$, the proof of "Artin vanishing" for mod $p$ ...

8

Let $s\in\mathbb{C}$ be any point with $\Re s>0$. The Dirichlet series of $\eta(s)$ converges, hence $L(s)$ converges if and only if $L^*(s)$ converges.
For $\sigma_0>1/2$, it is also known that $L(s)$ converges in the half-plane $\Re s>\sigma_0$ if and only if $\zeta(s)$ has no zero in that half-plane.
Combining the previous two statements, it ...

8

I claim that for even $n\in \{0,2,4,\ldots, 4\cdot 5^n-2\}$ each remainder of $F_n$ modulo $5^n$ is realized at most twice (thus exactly twice), and the same for odd $n\in \{1,3,5,\ldots, 4\cdot 5^n-1\}$. Denoting $u=(1+\sqrt{5})/2$, $v=(1-\sqrt{5})/2$ we have Binet formula $F_n=(u^n-v^n)/(u-v)$. If $k,m$ are even, then $$F_k-F_m=\frac{u^k-u^{-k}-u^m+u^{-m}}{...

7

There is no good upper bound for $p_k$ in the following sense: for every $k$, there exists a constant $c_k>0$ such that $p_k>p_0-c_k$ holds for infinitely many primes $p_0$. This was proved by Maynard (2013). You will find a concrete value for $c_k$ in Maynard's work, which was decreased by Polymath8b (2014). The best conjectured value would follow ...

7

Elements of $M$ have inverses which are integer-valued polynomial maps, so the version with integer-valued polynomial maps forms a group.
In fact, we only need to assume surjectivity to guarantee the existence of a polynomial inverse, so the monoid of surjective integer-valued polynomials from the plane to the plane is also a group.
I don't know if this ...

7

Such attempted generalizations of ABC to four or more variables
often fail to specializations of the identity
$$
(x^2+xy-y^2)^3 + (x^2-xy-y^2)^3 = 2 (x^6 - y^6).
\label{1}\tag{*}
$$
One can use elliptic curves to make both $x^2 + xy - y^2$ and $x^2 - xy - y^2$
"powerful" (of the form $A^2 B^3$), which makes each of the four terms
$(x^2+xy-y^2)^3$, ...

answered Apr 11 at 5:07

Noam D. Elkies

68.1k1212 gold badges244244 silver badges326326 bronze badges

7

The conjecture is true. That is, if the integral cubic polynomial
$$P(Z)=a_3 Z^3+a_2 Z^2+a_1 Z+1$$
is irreducible in $\mathbb{Z}[Z]$ (hence also in $\mathbb{Q}[Z]$ by Gauss's lemma), then there are only finitely many positive integer solutions of the equation
$$(xz+1)(yz+1)=P(z).$$
1. First we consider the case when $x\mid a_3$ or $y\mid a_3$. By symmetry, ...

6

I'm not sure I understand the point of the question, but the answer is yes: For example, $K=\overline{\mathbb{F}_p}(t)$ and $L=\overline{\mathbb{Q}}(t)$ are both known to have absolute Galois group free profinite on countably many generators, but $R=\overline{\mathbb{Q}}[t]$ is a noetherian proper subring of $L$ with quotient field $L$.

6

If $a$ is a totally positive real cyclotomic number, then it is a sum of two squares of real cyclotomic numbers.
It suffices to check that the equation $x^2+ y^2 - a z^2=0$ has solutions in real cyclotomic numbers. It has solutions in a particular real cyclotomic number field $F$ if it has solutions everywhere locally. This equation has solutions locally if ...

6

I think we can describe $P$ a bit more, using Dirichlet’s unit theorem.
Since $\overline{\mathbf{Z}} = \varinjlim_{[K:\mathbf{Q}] < \infty} \mathcal{O}_K$, the same is true for the units. Now Dirichlet tells us that taking logs of the archimedean valuations $v \mid \infty$ of $K$ gives us an embedding:
$$\mathcal{O}_K^\times/\mu_\infty(K) \hookrightarrow \...

5

Although Magma's MordellWeilShaInformation can't find the fifth generator for the $\mathbb{Z}/8\mathbb{Z}$ curve, it can find the last generator for an isogenous $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$ curve by searching up to height $10^5$ on the $4$-coverings.
SetClassGroupBounds("GRH");
E24 := EllipticCurve([1, 0, 0, -...

5

1 - No, the problem is NP-hard even if degree of all polynomials is 2.
2 - See, for example, paper Solving Systems of Polynomial Equations over
GF(2) by a Parity-Counting Self-Reduction by Björklund et al.

5

The ring $\overline{\mathbb{Z}}$ is called the ring of algebraic integers. You can find information about prime ideals, e.g., in https://math.stackexchange.com/questions/156231/non-zero-prime-ideals-in-the-ring-of-all-algebraic-integers.

5

It is possible to find equations for $\pi : X_1(N) \to X_0(N)$ using work of Yifan Yang. In the article Defining equations of modular curves, he explains an algorithm to obtain equations for the modular curves $X_1(N)$ and $X_0(N)$. Let me explain how to use his results to get an equation for $\pi$.
Yang shows that there exist two modular units $F,G$ on $X_1(...

4

If you could solve this problem in polynomial time, then NP would be contained in BPP, which is viewed as being approximately as unlikely as P = NP. Too see this, pick your favorite encoding of SAT into diophantine equations on $\{0,1\}^n$ (for instance, you can take $f$ to be a sum of squares of expressions corresponding to individual clauses), and apply ...

4

Not that I'm a great fan of Mathematica, but Wolfram Alpha yields the same thing as Will Jagy got with PARI:
Input: 48807585839879^3 + 399757627176339^3
Result: 63999999999999992173445324722427124758794658
I think that the issue is that the default precision in Mathematica is not set high enough to handle these numbers when doing real calcualtions. In ...

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