13

(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 ...


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 ...


11

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.


5

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 ...


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 ...


3

Rewrite the inequality as $$f_{p,a}(b):=\frac b{(a+b)^p}>a,\tag{1}$$ where $p:=1-\epsilon\in(0,1)$. We have $$f'_{p,a}(b)=\frac{a+(1-p)b}{(a+b)^{p+1}}>0$$ for $a,b>0$, so that the function $f_{p,a}$ is continuous and strictly increasing from $0$ to $\infty$ on $(0,\infty)$. So, (1) can be rewritten as $$b>f_{p,a}^{-1}(a),\tag{1}$$ where $f_{p,a}^{...


3

Looking more carefully in Pierce - Associative algebras, I found the answer I was looking for, which I'm going to describe here for future reference. The algebra $D$ in the question is a realisation of a special type of crossed product algebra, which gives precisely the central division algebras of degree 3 over $\mathbb{Q}$. The first step is to use the ...


3

The same group of authors work out the answer to your question in theorem 2 of this paper, so you can simply cite their result. As mentioned in the comments, the answer gets more complicated because it doesn't just involve the divisor function, but also the fourier coefficients of newforms of weight 4 on $\Gamma_0(6)$ and $\Gamma_0(12)$. You can check a ...


2

A very similar question was asked even earlier in Atkin–Lehner operator for GL(3)?. I was looking for the same generalisations of the Atkin-Lehner operators for $GL(n)$ and I managed to find something useful recently. Though the question is old, I think it could help other people like me to give an answer here. In short, the operator you are looking for is ...


1

We have $$\prod_{k\geq 2} (1+x^k) = \frac{\prod_{k\geq 1} (1+x^k)}{1+x} \equiv \sum_{j\geq 0} \frac{x^{j(3j+1)/2} + x^{(j+1)(3j+2)/2}}{1+x}$$ $$\equiv \sum_{j\geq 0} x^{j(3j+1)/2}\frac{1 - x^{2j+1}}{1-x}\equiv \sum_{j\geq 0} x^{j(3j+1)/2}(1+x+\dots+x^{2j}) \pmod{2}.$$


1

Factoring as Optimization could be a useful link in this direction as well.


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