14
votes

### A conjectural formula for the class number of the field $\mathbb Q(\sqrt{-p})$ with $p\equiv3\pmod8$

Here is a possible approach based on a formula of Zhang (see Page 432 of Wenpeng Zhang, On the mean values of Dedekind sums. J. Théor. Nombres Bordeaux 8 (1996), no. 2, 429–442.)
Recalling that
$$\cot\...

14
votes

### Is $-\det\big[\big(\frac{i^2+j^2}p\big)\big]_{1\le i,j\le (p-1)/2}$ always a square for each prime $p\equiv 3\pmod 4$?

It can be seen that $S_p$ is not divisible by $p$. The argument about the decomposition of the matrix as $\frac{2}{i\sqrt{p}}A^2$ suggested above implies that $-S_p$ is a square in $\mathbb{Q}[\zeta_p]...

14
votes

### A series of conjectures on $\sum_{x=0}^{(p-1)/2}(\frac{x^5+cx^3+dx}p)$ (III)

Here is a proof of (i):
Since the relevant primes $p$ are $\equiv 1 \bmod 4$, we have
$S_p(c,d) = \frac{1}{2} T_p(c,d)$, where
$$ T_p(c,d) = \sum_{x=0}^{p-1} \left(\frac{x^5+cx^3+dx}{p}\right) \,. $$
...

14
votes

### A new formula for the class number of the quadratic field $\mathbb Q(\sqrt{(-1)^{(p-1)/2}p})$?

If by
"the class number $h(p^*)$ of the quadratic field $\mathbb{Q}(\sqrt{p^*})$"
you mean
"the minus class number $h^{-}$ of $\mathbf{Q}(\zeta_p)$"
and if by
" a possible new formula for the ...

13
votes

### A new formula for the class number of the quadratic field $\mathbb Q(\sqrt{(-1)^{(p-1)/2}p})$?

The conjecture is not true, as some examples show.
Let $D(p)$ denote your number. For primes $p \equiv 1 \textrm{ mod } 4$, we have $D(29)=8$, $D(37)=37$, $D(41)=121$ while $h(29)=h(37)=h(41)=1$.
...

13
votes

Accepted

### On sums of quadratic residues

By standard formulas for values of L functions at negative integers, for
$p\equiv1\pmod4$ one has
$$A_p=(p^2-1)/16+aL(\chi_p,-1)\;,$$
with $a=3/4$ if $p\equiv1\pmod8$ and $a=5/4$ if $p\equiv5\pmod8$
...

12
votes

### Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$?

Expanding on a comment, the curve $X^4+Y^4=aZ^4$ (for $a\ne0$) has genus $3$. So the Hasse-Weil bound says
$$ N_p(a) := \#\bigl\{ [X,Y,Z]\in\mathbb P^2(\mathbb F_p) : X^4+Y^4=aZ^4 \bigr\} $$
satisfies
...

11
votes

Accepted

### Is the permanent of the matrix $[(\frac{i+j}{2n+1})]_{0\le i,j\le n}$ always positive?

This is the sequence A322898 in OEIS.
I used a program in PARI and calculated the values of a(26) to a(34).
...

9
votes

Accepted

### Quadratic non-residue problem

This question is closely related to Linnik's problem on the least quadratic nonresidue for a given prime modulus.
Let us consider the quadratic Dirichlet character $\chi(m):=\left(\frac{-n}{m}\right)$ ...

8
votes

Accepted

### Distribution of quadratic residues in an interval

Yes. The points $(\frac{a}p,\frac{a^2\pmod p}p)$ are asymptotically equidistributed in $[0,1]^2$ by Weyl's criterion.

8
votes

### Is $-\det\big[\big(\frac{i^2+j^2}p\big)\big]_{1\le i,j\le (p-1)/2}$ always a square for each prime $p\equiv 3\pmod 4$?

This is not an answer, but a reduction to supposedly simpler problem. (edited with more details)
Using quadratic Gauss sums, we can express Legendre symbol $\left(\frac{i^2+j^2}p\right)$
as
\begin{...

7
votes

### A new determinant question for primes $p\equiv3\pmod4$

The conjectures are true. If you negate the first row of $A_p^{-}$ you get a cofactor of the matrix in "Chapman's evil determinant". In particular you can get the answer from the same matrix ...

7
votes

### Is $|\{(j,k):\ 1\le j<k\le\frac{p-1}2:\ \&\ (j^{16}\ \text{mod}\ p)>(k^{16}\ \text{mod}\ p)\}|$ even for each prime $p\equiv1\pmod {16}$?

Start with Conjecture 1. Two other look similar, but possibly require additional ideas (UPDATE: they do not actually).
Write $(x)_p\in \{0,\ldots,p-1\}$ for the remainder of integer $x$ modulo $p$. ...

6
votes

Accepted

### Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$?

emtom has found the right reference, but there is a more explicit result in that book (Ireland and Rosen, A Classical Introduction to Modern Number Theory). In fact, Theorem 5 of Chapter 8 (on page ...

6
votes

### 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 $p>3$

Here is a respectively short way to write down what we came up with Dmitry Krachun tonight.
Denote $p=2m+1$.
The idea is very simple: calculate the product $$\prod_{j\in\{s,s+1\}, 1\leqslant i\...

6
votes

Accepted

### Quadratic Nonresidue

I don't think this is true in general, for instance, if there exists $1\leq i <j<k<r$ such that $n_in_j = n_k$, then $(n_i/p)=(n_j/p)=-1\implies (n_k/p)=1$ (where $(a/p)$ is the Legendre ...

6
votes

### Counting squares modulo $p$ that are also prime in an interval

Here is the paper by P. Pollack on the distribution of non-residues and residues. Theorem 1.3 states that for any $\varepsilon>0$, $A<\infty$ and large enough $m$ there are at least $(\ln m)^A$ ...

6
votes

### Set of all primes $p$ that split in $\mathbb{Q}\left(\sqrt{-k}\right)$

The following answer concerns the original version of the question, while my comment below addresses the updated version.
For $p>3$ the Chinese Remainder Theorem shows that there are integers $k\...

5
votes

Accepted

### Question: How to find the smallest value $x$ satisfying the equation: $x^2 = a \pmod c$ (known is $a$ and $c$, $c$ is not the prime)?

This is an NP-hard problem. That is, Manders and Adleman [1] proved that given $a$, $b$, and $c$, it is NP-complete to determine if there exists $x\le b$ such that $x^2\equiv a\pmod c$, and that it ...

5
votes

### On triangular numbers modulo primes

Differences
A general useful fact (compare with my answer to your previous question) is that whenever we have $A=\{a_1,\dots,a_n\}\subset \{0,1,\dots,p-1\}$ such that $n=|A|$ is odd, the sign of a ...

5
votes

Accepted

### Is it possible to find a (nonsquare) integer which is a quadratic residues modulo a given infinite list of primes?

It depends on the given list of primes. A simpler but necessary condition is that there be a $d$ so that all the primes of the list (greater than $d$) are concentrated in a few congruence classes $\...

5
votes

### Does the expression $x^4 +y^4$ take on all values in $\mathbb{Z}/p\mathbb{Z}$?

Not exactly an answer, but in exercise $18$ in page $106$ of Ireland and Rosen's A Classical Introduction to Modern Number theory it states: Let $p\equiv 1\mod 4$ and let $p=A^2+B^2$ where we fix $A$ ...

5
votes

### A conjecture on primitive tenth roots of unity

Not a complete solution.
Let $p$ be (1 mod 4) and $r,n$ runs over quadratic residues/non residues mod $p$ in $[1,p-1]$ and let
$R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$
...

5
votes

### Pythagorean triples and quadratic residues modulo primes

The Conjecture 1 is true. We are looking for integers $m, n$ such that for sufficiently large prime $p$ we have
$$
x_{1}^2\equiv 2mn\pmod{p},\quad x_{2}^2\equiv m^2-n^2\pmod{p}, \quad x_{1}^2\equiv m^...

4
votes

Accepted

### A cubic equation, and integers of the form $a^2+32b^2$

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version.
The equation is unsolvable. The proof requires two theorems
Theorem 1. Let $p = ...

3
votes

### Does $\det[\lfloor(i^2+j^2)/p\rfloor]_{1\le i,j\le(p-1)/2}$ vanish for each prime $p>7$ with $p\equiv3\pmod4$?

Let $p$ be large enough. Then there are two pairs of consecutive squares $a$, $a+1$ and $b$, $b+1$ modulo $p$ (otherwise the parities of sqiares modulo $p$ cannot alternate more than constant times, ...

3
votes

### Question: How to find the smallest value $x$ satisfying the equation: $x^2 = a \pmod c$ (known is $a$ and $c$, $c$ is not the prime)?

In general, finding the smallest integer solution to a system of modular (in)equalities seems to be very hard. To be precise: the question is how to find the smallest $x$ such that e.g. $x\mod{3}\in\{...

3
votes

### Does $n^2$ divide $\det\left[\left(\frac{i^2+2ij+3j^2}n\right)\right]_{1\le i,j\le n-1}$ for each odd integer $n>3$?

This is not an answer, either. Just some attempts to attack the problem.
Let $r(n)$ be the square-free part of $(2, 3)_n$, i.e. $r(n)$ is square-free and we have $(2, 3)_n = r(n) B^2$ for some ...

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