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7 votes
1 answer
290 views

On $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i+j}(2i+j)$ and $\prod^{(p-1)/2}_{i,j=1\atop p\nmid 2i-j}(2i-j)$ modulo a prime $p>3$

QUESTION: Is my following conjecture true? Conjecture. Let $p>3$ be a prime and let $h(-p)$ be the class number of the imaginary quadratic field $\mathbb Q(\sqrt{-p})$. Then $$\frac{p-1}2!!\prod^{...
Zhi-Wei Sun's user avatar
  • 15.6k
8 votes
1 answer
407 views

Cardinality of the image of a polynomial modulo $p^n$

Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial and let $p$ be a prime number. I'm looking for results about $$N_f(p^k) := \#\{(f(n) \bmod p^k) : n \in \mathbb{Z}\},$$ as $k \to +\infty$, where $...
OhhiMark's user avatar
7 votes
1 answer
185 views

2-adic valuation of $L(0,\chi)$ for a Dirichlet character

Let $\chi : (\mathbb Z/f\mathbb Z)^\times \to K = \mathbb Q(\mu_{\phi(f)})$ be a primitive Dirichlet character. Assume moreover that it is not quadratic, that is, $\chi^2$ is not the trivial character....
Asvin's user avatar
  • 7,746
2 votes
0 answers
110 views

"Close" roots of polynomials and Pillai property

A sequence of integers $a(n)_{n \geq 0}$ has the Pillai property if there exists an integer $G \geq 2$ such that for all integers $k \geq G$ there exists an integer $n \geq 0$ such that none of the $k$...
MacNamara's user avatar
13 votes
1 answer
1k views

On cubic reciprocity for $x^3+y^3+z^3 = 996$?

I. The Diophantine equation, $$x^3+y^3+z^3 = 3w^3\tag1$$ with $x\geq y \geq z$ and $w=1$ has only two known solutions, namely $1,1,1$ and $4,4,-5$. Are there larger ones? As Noam Elkies points out ...
Tito Piezas III's user avatar