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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 $$\cot\... • 251 14 votes ### Is -\det\big[\big(\frac{i^2+j^2}p\big)\big]_{1\le i,j\le (p-1)/2} always a square for each prime p\equiv 3\pmod 4? It can be seen that S_p is not divisible by p. The argument about the decomposition of the matrix as \frac{2}{i\sqrt{p}}A^2 suggested above implies that -S_p is a square in \mathbb{Q}[\zeta_p]... • 1,241 14 votes ### A series of conjectures on \sum_{x=0}^{(p-1)/2}(\frac{x^5+cx^3+dx}p) (III) Here is a proof of (i): Since the relevant primes p are \equiv 1 \bmod 4, we have S_p(c,d) = \frac{1}{2} T_p(c,d), where$$ T_p(c,d) = \sum_{x=0}^{p-1} \left(\frac{x^5+cx^3+dx}{p}\right) \,. $$... • 11.2k 14 votes ### A new formula for the class number of the quadratic field \mathbb Q(\sqrt{(-1)^{(p-1)/2}p})? If by "the class number h(p^*) of the quadratic field \mathbb{Q}(\sqrt{p^*})" you mean "the minus class number h^{-} of \mathbf{Q}(\zeta_p)" and if by " a possible new formula for the ... • 165 13 votes ### A new formula for the class number of the quadratic field \mathbb Q(\sqrt{(-1)^{(p-1)/2}p})? The conjecture is not true, as some examples show. Let D(p) denote your number. For primes p \equiv 1 \textrm{ mod } 4, we have D(29)=8, D(37)=37, D(41)=121 while h(29)=h(37)=h(41)=1. ... • 20.5k 13 votes Accepted ### On sums of quadratic residues By standard formulas for values of L functions at negative integers, for p\equiv1\pmod4 one has$$A_p=(p^2-1)/16+aL(\chi_p,-1)\;,$$with a=3/4 if p\equiv1\pmod8 and a=5/4 if p\equiv5\pmod8 ... • 11.3k 12 votes ### Does the expression x^4 +y^4 take on all values in \mathbb{Z}/p\mathbb{Z}? Expanding on a comment, the curve X^4+Y^4=aZ^4 (for a\ne0) has genus 3. So the Hasse-Weil bound says$$ N_p(a) := \#\bigl\{ [X,Y,Z]\in\mathbb P^2(\mathbb F_p) : X^4+Y^4=aZ^4 \bigr\} $$satisfies ... • 45.6k 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) ... • 96.9k 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. • 102k 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{... • 30.1k 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. ... • 102k 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 ... • 4,745 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\...
• 102k
Accepted

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 ...
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### 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 $(\ln m)^A$ ...
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### 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 $A$ ...
• 51

### 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 $R_p(x)=\prod_r(x-\zeta_p^r),\;\;\; N_p(x)=\prod_n(x-\zeta_p^n).$ ...
• 487

### 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^...
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Accepted

• 1,616
### 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 ...