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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions

2 votes
0 answers
43 views

Wieferich primes and identities for the Euler quotients of $2^n+1$ and $\frac{2^n+1}{3}$

Let $n>1$ be odd integer. Define the Euler quotient $a(n)=\frac{2^{\varphi(n)}-1 \bmod n^2}{n}$. Number $n$ with $a(n)=0$ is Wieferich number and if it is prime it is Wieferich prime. It is open probl …
1 vote
1 answer
139 views

On a probabilistic integer factorization algorithm given bounds for one prime factor

We got a probabilistic integer factorization algorithm and experimental evidence with large integers given bounds for one factor. Let $D \ge 2$ be real number and let $p,q$ be primes and $N=pq$. Assum …
8 votes
1 answer
299 views

Identity?: $\frac{\varphi(2^n-1)}{n}=\frac{2^{\varphi(2^n-1)}-1 \bmod (2^n-1)^2}{2^n-1}$

The computer found this. Let $n$ be a positive integer. Up to $n=200$ we have: $$\frac{\varphi(2^n-1)}{n}=\frac{2^{\varphi(2^n-1)}-1 \bmod (2^n-1)^2}{2^n-1}. \tag{1}\label{483144_1}$$ Q1 Is \eqref{48 …
1 vote
0 answers
49 views

Conjecture about Euler quotients related to non-Wieferich numbers $W(n)=\frac{2^n+1}{3}$

For odd natural $n$ define the Euler quotient: $$ a(n)=\frac{(2^{\phi(n)}-1) \bmod n^2}{n}=\frac{2^{\phi(n)}-1}{n} \bmod n$$ $a(n)=0$ is $n$ being Wieferich number (not necessarily prime). For odd $n$ …
1 vote
0 answers
88 views

Can we find curves with many rational points using linear algebra?

Probably this is impossible, but let us try. Working over $\mathbb{Q}[x_1,...,x_n]$. Let $T_i$ be $n$ sets of rationals with cardinality $B$. Assume we are given $n-2$ linear equations $f_i$ which are …
1 vote
0 answers
54 views

System of linear diophantine equations with many small solutions?

Let $n$ be positive integer, $k$,$B$ fixed positive integers. Let $f_i(x_1,x_2...x_n)$ be a system of $n-k$ linearly independent linear equations over the integers. Let $S(f_i,k,B)$ be the set of solu …
1 vote
0 answers
58 views

On the parity of $(2^{\varphi(n)}-1) \bmod{n^2}$

For odd integer $n$ define the function $$ J(n)=(2^{\varphi(n)}-1) \bmod{n^2}$$ $J(n)$ is integer in $[0,n^2-1]$ and it is divisible by $n$. Integer $n$ is Wieferich number iff $J(n)=0$ and if $n$ is …
5 votes
1 answer
163 views

On vanishing of $p$-adic logarithms

Might be related to Wieferich primes. Let $p$ be odd prime and define the Fermat quotient $$F(n)=\frac{(2^{n-1} -1)}{n} \mod n=\frac{(2^{n-1} \bmod n^2 )-1}{n}$$ For integer $b$ let $L_p(b)$ be the $p …
6 votes
2 answers
793 views

Must Mersenne numbers be divisible by arbitrary large primes with exponent one?

Let $M_n$ denote the Mersenne numbers $M_n=2^n-1$. As $n$ varies, must $M_n$ be divisible by arbitrary large prime $p$ with exponent one, i.e. $p \mid M_n, p^2 \nmid M_n$? In other words, must the exp …
0 votes
0 answers
61 views

On base $b$ digits of $n\#$ (primorial)

Related to normal numbers. Let $n\#$ denote the primorial, the product of the first $n$ primes. Q1 For all bases $b>1$, do the base $b$ digits of $n\#$ occur with equal asymptotic frequency $\frac1b$ …
10 votes
4 answers
1k views

The smallest solution to $2^{2k}-1=\text{powerful}$

Integer is powerful if all the exponents in its factorization are at least $2$. Every powerful integer can be written in the form $a^2 b^3$. For odd $k$, define $F(k)=2^{2k}-1=(2^k-1)(2^k+1)$. This pa …
0 votes
0 answers
62 views

Parametrization of elliptic curve with differential equation $(x,y)=(f(x),f'(x))$ involving ...

For non-zero complex $A$, define the curve over the complex numbers $C: x^2 y^2-A x-y=0$. $C$ is an elliptic curve. $C$ has the differential equation parametrization $(x,y)=(f(x),f'(x))$ where $$ f(x) …
6 votes
1 answer
232 views

Congruence obstructions for three consecutive powerful numbers

A powerful number is an integer $m$ such that if $p$ is prime and $p \mid m$ then $p^2 \mid m$. Powerful numbers can be represented in the form $m=u^2 v^3$. Erdos conjectured that three consecutive po …
4 votes

Genus 0 curves on surfaces and the abc conjecture

I wasted some electricity on a very similar problem until I realized that if we fix all variables except one, then the abc theorem for polynomials may imply non-existence after parametrization of all …
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4 votes
2 answers
383 views

Non-linear recurrence for rational sequences with generating function with radicals?

Let $a(n)$ be a sequence of rational numbers with generating function $F(x)$. Assume $F(x)$ is composition of rational functions and radicals (roots). Is the following conjecture true: Conjecture 1: …

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