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15 votes
2 answers
1k views

Saying things rapidly about integer factorisations

Let $N$ be a positive integer. Thanks to the Miller-Rabin test and the work of Agrawal, Kayal and Saxena, these days people have much much faster algorithms for testing whether $N$ is prime or ...
James Cranch's user avatar
  • 3,064
11 votes
2 answers
1k views

Why going to number fields in number field sieve help beat quadratic sieve?

To factor an $n$ bit integer number field sieve roughly takes $$e^{c{(\ln\ln n)^{\frac23}}({\ln n})^{\frac13}}$$ time while quadratic sieve takes $$e^{c{(\ln\ln n)^{\frac12}}({\ln n})^{\frac12}}$$ ...
user avatar
5 votes
0 answers
153 views

On factorization algorithms for $\mathcal{O}[x]$

We know that $\mathsf{LLL}$ algorithm provides factorization procedure that runs in poly time for polynomials in $\Bbb Z[x]$ that are primitive. What other rings $\mathcal{O}$ can we use instead of $\...
user avatar
4 votes
1 answer
288 views

Is total degree version and $x,y$ degree version of Coppersmith's theorem correct?

The notes here https://web.eecs.umich.edu/~cpeikert/lic13/lec04.pdf have the note 'Small decryption exponent $d$: so far the best known attack recovers $d$ if it is less than $N^{.292}$. This uses a ...
Turbo's user avatar
  • 13.9k
3 votes
1 answer
261 views

Could efficient solutions of $x^2+n y^2=A$ be related to integer factorization?

Let $n$ be positive integer with unknown factorization and $A$ integer with known factorization. According to pari/gp developers pari can efficiently find all solutions of: $$x^2+n y^2=A \qquad (1)$$ ...
joro's user avatar
  • 25.4k
3 votes
1 answer
137 views

Subexponential algorithms that apply only one of factoring and discrete logarithm?

Shor (quantum polynomial), Number Field Sieve (subexponential), Pollard rho (square root) all have both factoring and discrete logarithm over $\mathbb F_p^*$ variants. What are the subexponential ...
Turbo's user avatar
  • 13.9k
2 votes
1 answer
536 views

Modular square roots problem which is $NP$ hard

It is well known extracting modular square roots modulo a composite number factors the modulus. On other hand given $u,v>0$ and an integer $n$, deciding if there is a factor of $n$ in $[u,v]$ is $...
Turbo's user avatar
  • 13.9k
2 votes
1 answer
514 views

Eisenstein polynomial of totally ramified extension over $p$-adic field

Let $p\geq 3$ be a prime number, $K$ be a finite extension of $\mathbb{Q}_p$ with no non-trivial unramified subextension, $f(x)$ be an irreducible monic polynomial in $\mathcal{O}_K[x]$, making $L=K[x]...
Yijun Yuan's user avatar
2 votes
0 answers
110 views

Evidence of optimality of sieve algorithms

Sieve techniques apply to integer factoring and discrete logarithm to provide $2^{O(((\log n)(\log\log n)^2)^{1/3})}$ complexity for $n$ bit factoring and $n$ bit prime discrete logarithm. The state ...
Turbo's user avatar
  • 13.9k
2 votes
0 answers
127 views

Coppersmith's method to quadrivariate degree $2$ polynomials that behave as bivariate?

We have a polynomial $f(x_1,x_2,x_3,x_4)\in\mathbb Z[x_1,x_2,x_3,x_4]$ where the only monomials are either from set $$\{x_1,x_1x_2,x_2,x_3,x_3x_4,x_4\}$$ and we seek solutions $(x_1,x_2,x_3,x_4)\in\...
Turbo's user avatar
  • 13.9k
1 vote
1 answer
144 views

Factoring with partial information on gaps

If $N=PQ$ is a semi-prime with $P=N^{\frac12 +\delta}$ and $Q=N^{\frac12-\delta}$ then if we know $\delta\in(0,\frac12)$ to a reasonable precision we can factor $N$ quickly. What precision (number of ...
Turbo's user avatar
  • 13.9k
1 vote
0 answers
70 views

Is this factorization problem in EXP?

Factorization is not known to have a polynomial time algorithm. Traditionally the input length is number of bits in representation of the integer to be factored. However now consider integers of form $...
Turbo's user avatar
  • 13.9k
1 vote
0 answers
96 views

Smooth number pairs satisfying a congruence

Let $\mathcal P=\{p_1,\dots,p_{2t}\}$ be $2t$ primes between $2^\ell$ and $2^{\ell+1}$ and fix an exponent bound $a\in\mathbb Z_{\geq2}$. Fix $N\in\mathbb N$ whose prime factors $p$ satisfy $p>2^{\...
Turbo's user avatar
  • 13.9k
0 votes
0 answers
78 views

Factoring totient of a prime

Is it any easy to factor $p-1$ when $p$ is a prime compared to general factorization problem? What about when $2p+1$ is also a prime?
Turbo's user avatar
  • 13.9k