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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
8
votes
Accepted
Is the given expression, monotonically increasing or decreasing with increasing x?
I'm not sure I should bother answering this question, because it seems like the original poster may not have asked the right question. However, it is a nice exercise in basic asymptotics.
For $x$ …
3
votes
prime ideal factorization in an extension field
Pete's answer is very good. Some self-promotion:
An expository post on the relationship between factoring polynomials and factoring primes.
The previous answer regarding the polynomial $t^3+t^2−2t−1 …
14
votes
Accepted
Algorithms for Diophantine Systems
No. Given any set of diophantine equations $f_1(z_1, \ldots, z_n) = \ldots = f_m(z_1, \ldots, z_n)=0$, we can rewrite in terms of linear equations and quadratics. Create a new variable $w_{k_1 \cdots …
6
votes
Laurent polynomials with constant term of their $(p-1)$-st power congruent to one modulo $p$
I should explain how to think about these things in terms of Frobenius splitting. This perspective will let you write down lots more examples like the ones you already have. For example, if I haven't …
29
votes
Accepted
A possibly surprising appearance of Fibonacci numbers
The statement is true. Write $F_n$ for the $n$-th Fibonacci (my indexing starts at $(F_0, F_1, F_2, F_3, \dots) = (0,1,1,2,\dots)$). We are being asked to show that
$$\frac{1}{F_{j+1}} > \sum_{m=F_{j} …
6
votes
Smallest solution to $x^2 \equiv x\pmod{n}$
Building on Fedor Petrov's answer, if $n$ has $k$-distinct factors, we can find a solution below $(n-1)/k$. Let $n=q_1 q_2 \cdots q_k$, with $q_i$ prime powers. Let $e_i$ be $1$ modulo $q_i$ and $0$ m …
5
votes
Accepted
sums of fractional parts of linear functions of n
I think I can show that
$$\sum_{1 \leq h,k \leq N} \frac{GCD(h,k)^2}{hk}$$
grows linearly. But I get the constant is
$$\sum_{ GCD(i,j)=1} \frac{1}{\max(i,j) i j}$$
This constant is incredibly close t …
12
votes
Accepted
is $2$ a $2^n$-th power mod $p$ ?
There is no $N$ such that $p \equiv 1 \mod N$ implies that $2$ is a fourth power modulo $p$.
Proof: Suppose otherwise. Without loss of generality, suppose that $4$ divides $N$.
Let $K$ be the field …
6
votes
Smallest solution to $x^2 \equiv x\pmod{n}$
If there are infinitely many twin primes, then we can't do much better than $n/2$. If $q=2k-1$ and $p = 2k+1$ are primes, then the solutions to $x^2\equiv x \bmod pq$ are $0$, $1$, $k(2k-1) \approx (p …
8
votes
Accepted
Distribution of moduli of quadratic residues
For $D=-1$, Landau proved that
$$\# S(-1, x) \sim K \frac{x}{\sqrt{\log x}}$$
where $K = \frac{1}{\sqrt{2}} \prod_{p \equiv 3 \bmod 4} \frac{1}{\sqrt{1-p^{-2}}}$.
This shows that, for fixed $0<a<b$,
…
4
votes
Composition of a transcendental function with a rational function
Yes. If there is a polynomial relation $P(x, f(\psi(x))=0$, then we also have $P(\psi^{-1}(y), f(y))=0$ (passing to an open set where $\psi^{-1}$ is defined and analytic.) But $\psi^{-1}(y)$ is algebr …
5
votes
homogeneous forms as norms
A necessary condition: Let $f: \mathbb{Q}^d \to \mathbb{Q}$ be a homogenous map of degree $d$. If $f$ is a norm, then the corresponding map $\mathbb{C}^d \to \mathbb{C}$ will be a product of linear fo …
7
votes
Where does the error term of the Prime Number Theorem touch the predicted asymptotic behavior
The Riemann Hypothesis is also equivalent to $|\pi(x) - Li(x)| = O(x^{1/2 + \epsilon})$, so let's look at that instead. In other words, $\log$ of the error should be about $(1/2) \log x$.
The sequenc …
2
votes
On the divisibility of $(x+y)^k - 1$ by $xy$
Another family of answers which I don't think is of the other forms: For $k=3$, consider $x=t^2$, $y=1-t^3$. More generally, if $(t^j)^k-1$ has a factor of the form $\phi(t) = 1 + \sum_{m \geq j} c_m …
6
votes
Accepted
A polynomial in multiple variables with nice properties
This is a CW answer to remove this question from the unanswered list (once someone upvotes it). This is the determinant of the Moore matrix $\left( x_i^{q^{j-1}} \right)_{1 \leq i,j \leq n}$. This det …