All Questions
Tagged with factorization nt.number-theory
14 questions
118
votes
5
answers
33k
views
How did Cole factor $2^{67}-1$ in 1903?
I just heard a This American Life episode which recounted the famous anecdote about Frank Nelson Cole factoring $N:=2^{67}-1$ as $193{,}707{,}721\times 761{,}838{,}257{,}287$. There doesn't seem to be ...
- 156k
44
votes
1
answer
17k
views
Conjecturally unsafe RSA primes $p=27a^2+27a+7$
We got strong numerical evidence that primes of the form $p=27a^2+27a+7$
are unsafe for cryptographic purposes since they can be found in the factorization.
Consider the following generic factoring ...
- 25.4k
6
votes
2
answers
804
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 ...
- 25.4k
5
votes
1
answer
326
views
Generalizing Kasteleyn's formula even more?
Inspired and intrigued by this question, I decided just for fun to throw in another integer into the factors and look what happens. So for $k\in\mathbb Z$, let us define $$K_r(n,k):=\prod_{\ell_1=1}^...
- 13.4k
4
votes
3
answers
3k
views
$2^n$-1 consisting only of small factors
I've checked the factorization of $2^N - 1$ up through N = 120 for the largest prime factor, and it looks like the largest value of N where $2^N-1$ has a largest prime factor under 2500 is N = 60 (...
- 663
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 ...
- 25.4k
10
votes
3
answers
3k
views
Reduction from factoring to solving Pell equation
The paper Polynomial-Time Quantum Algorithms for Pell's Equation and the Principal Ideal Problem claims
There are reductions from factoring to solving Pell’s equation, and from solving Pell’s
...
- 25.4k
9
votes
1
answer
698
views
Hensel's lemma, Bezout's identity, and the integers
Factorization in the ring $\mathbb{Z}[x]/(x^2+1)\mathbb{Z}[x]\cong \mathbb{Z}[i]$ is well known. For instance, $5$ and $13$ (and any prime $\equiv 1\pmod{4}$) are no longer prime.
The factorization ...
- 18.7k
9
votes
3
answers
980
views
$\omega(p^n - 1)$ as $n \rightarrow \infty$
Although I am also interested in the number of distinct prime factors (not counting
multiplicity), today I use $\omega(m)$ to denote the number of (positive) prime
factors (with multiplicity) of the ...
- 2,158
7
votes
0
answers
859
views
Optimal Gear Trains
Suppose you need to slow down a turning motor so that a gear turns at
an angular velocity $\frac{a}{b}$ of that of the motor shaft, where $a$ and
$b$ are natural numbers. For example, this set of ...
- 151k
3
votes
1
answer
520
views
How much space between these smooth numbers?
In looking at OEIS sequence A063539, $1,8,12,16,18,24,27,30,32,36,40,45,...$ I noticed that the first 1000 members were less than 4000, and thought there were no large gaps between them. What (if ...
- 13k
2
votes
1
answer
360
views
Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$
This is a follow-up question to Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$
Let \begin{equation}
P(x,n)= 1+x+x^2+ \cdots + x^n, \end{equation}
\begin{...
- 319
2
votes
1
answer
382
views
Irreducibility of Faulhaber-like Polynomials over $\mathbb Q[x]$
Motivation: Inspired by the famous Faulhaber polynomials $F_k(N)=\displaystyle\sum_{n=0}^Nn^k,$ I decided to study their alternating versions, $\Phi_k(M)=\displaystyle\sum_{n=0}^M(-1)^nn^k$.
For $k\...
- 655
2
votes
2
answers
405
views
When is a prime factor of Mersenne number Wieferich prime?
Wieferich prime is a prime number $p$ such that $p^2$ divides $2^{p - 1} - 1$.
There are only two Wieferich primes known and it is an open problem if
there are infinitely many non-Wieferich primes.
...
- 25.4k