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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
17
votes
Accepted
Is there always, for a given prime $p$, a prime $\ell<p$ that is not a quadratic residue mod...
Of course. Take a quadratic nonresidue $1\leq n\leq p-1$, then some prime divisor $\ell$ of $n$ will be a quadratic nonresidue.
See this MO question for what is known about number fields.
3
votes
A difficult proof about integers
There are infinitely many solutions with the prime $p:=5$ (in which case $2p+1=11$ is also prime).
It is known that the Pell equation $u^2-55v^2=1$ has infinitely many solutions in positive integers. …
7
votes
Squares of the form $2^j\cdot 3^k+1$
There are finitely many pairs $(j,k)$, and this follows from results on $S$-unit equations. Moreover, the solutions can be effectively determined. Here is a quick treatment going back to the classical …
3
votes
When is the sum of two quadratic residues modulo a prime again a quadratic residue?
To complement the answers so far let me show using Gauss sums that the number of solutions of $ ax^2+by^2=c $ in $\mathbb{F}_p$ equals $p-\left(\frac{-ab}{p}\right)$ for any $a,b,c\in\mathbb{F}_p^\tim …
5
votes
Accepted
Are there lower bounds on the quality of a rational approximation?
Here are some standard facts that you can find in many textbooks, e.g. in Cassels: An introduction to diophantine approximation.
There exists an $A>0$ such that $\left|\frac{p}{q}-\xi\right|<\frac{1 …
16
votes
Accepted
Is there a constant $c>0$, such that every natural number $n>1$ is the sum of primes, each w...
The usual proof of Vinogradov's result can be modified to show that every sufficiently large odd $n$ has $\asymp n^2/(\log n)^3$ representations as a sum of three primes with each prime exceeding $cn$ …
11
votes
Accepted
Irreducible cubics modulo primes
No. It follows from Chebotarev's density theorem that for any polynomials $p_1,\dots,p_k\in\mathbb{Z}[x]$ there are infinitely many primes that split all these polynomials. Simply, apply this theorem …
27
votes
Accepted
p^2 dividing n^8-n^4+1
Your conjecture is true in the light of the following statements.
Proposition 1. A prime $p$ has the form $x^2+24y^2$ if and only if $p\equiv 1\pmod{24}$.
Proposition 2. A prime square $p^2$ divides …
3
votes
Accepted
Diophantine approximations and quadratic polynomials
I think the answer is no. For example, the Oppenheim conjecture (proved by Margulis in 1987) states that if an indefinite nondegenerate quadratic form has at least 3 variables and it is not proportion …
3
votes
Accepted
Numbers of a certain form not expressible as squares
I recommend this survey about the solution of Catalan's conjecture. We learn from here that the special case you are considering, namely $x^p-y^q=1$ for $p=2$, was solved by Chao Ko in 1964. In 1976 C …
8
votes
Proving the Irrationality of this Number
To resonate with Henry Cohn's comment, Schanuel's conjecture implies that the natural logarithms of the primes are algebraically independent over $\mathbb{Q}$. In particular, the statement in the orig …
3
votes
looking for integer pairs $(a,c)$ such that $4a^2 + 8c^2 - 4c + 1$ is a perfect square
With the substitution $x:=2a$, $y:=4c-1$, $z:=2n+1$ your framed equation becomes
$$ 2x^2+y^2-2z^2=-1. $$
Equations of this type are studied thoroughly in Section 13.6 "Representation by Anisotropic Te …
5
votes
Divisibility in a set
The set $A$ has positive lower density, hence the result follows from the main theorem in this paper of Paul Erdős (the proof takes one page).
5
votes
Accepted
Possible ratios of Pythagorean fractions
This is a sketch how to decide the question for $\frac{4}{9}$.
The question is if there are positive integers $a,b,d,e$ such that $\frac{4}{9}=\frac{a/b}{d/e}$ with $a^2+b^2$ and $d^2+e^2$ squares. D …
5
votes
Function that gives 1 only when an integer is divisible by another integer
In analytic number theory, this function is usually written as
$$ f(a,b):=\frac{1}{b}\sum_{n=1}^b e^{2\pi i na/b}.$$
This works in practice.
Added. To respond to Andrej Bauer's comment, "in practice" …