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**13**

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**1**answer

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### Evidence for integer factorization is in $P$

Peter Sarnak believes that integer factorization is in $P$. It is a well-known open problem in TCS to identify the real complexity class of integer factorization. Take a look at this link for Peter ...

**8**

votes

**2**answers

5k views

### Fast trace of inverse of a square matrix

Which would be the most efficient way (in computational time) to compute tr(inv(H)), where H is a (dense) square matrix?
In my particular problem I also have a LU decomposition of H already ...

**12**

votes

**6**answers

1k views

### Seeking Noetherian normal domain with vanishing Picard group but not a UFD

Once again, the question says it all.
My motivation is the article on factorization I am writing. I want to explain (as well as to understand!) why for normal Noetherian domains of dimension greater ...

**6**

votes

**3**answers

2k 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
...

**7**

votes

**3**answers

799 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 ...

**6**

votes

**2**answers

262 views

### Cubic graphs whose 2-factors all have the same cycle type

Let $G$ be a bridgeless cubic graph. I am interested in such graphs where all 2-factors are isomorphic (as graphs), i.e. have the same partition as cycle type. We'll say that this partition is ...

**5**

votes

**1**answer

203 views

### Large gaps between consecutive irreducible polynomials with small heights

For a prime gap of length at least $n$, a trivial upper bound for its first occurrence is $N=n!$ or $N=lcm(2,\dots,n)$. A bit better is $N=p_1\cdots p_n$ where $p_k$ is the $k$th prime, as then ...

**0**

votes

**0**answers

68 views

### Is this factoring algorithm sufficiently efficient for some integers of special kind?

Basically the question is if $m$ is factored over the integers,
can it be relatively efficiently factored over $\mathbb{Z}[\sqrt{n}]$
where $n$ is not factored, but might be of special form?
Suppose ...