3
$\begingroup$

From pari's implementation of Coppersmith method

zncoppersmith(P, N, X, {B=N}): finds all integers $x$ with $|x| \le X$ such that $\gcd(N, P(x)) \ge B$. $X$ should be smaller than

$$\exp((\log B)^2 / (\deg(P) \log N)) \qquad (1) $$

Observe that this might find non-trivial factor and pari's documentation gives example of this.

Linear $P(x)$ is allowed and looks like the content (the gcd of coefficients) of $P(x)$ need not be $1$.

I believe this algorithm (if successful) might find non-trivial factor for general $N$.

Let $v$ be positive integer and set $P(x)=v(x-1), N=nv$ where $n$ is integer to be factored.

For a divisor $d$ of $n$, $\gcd(P(d+1),N)=dv$.

With few guesses, one can find $v,X,B$ such that (1) holds.

Experimentally, for very small $n$, this indeed factors $n$.

Examining the source, I believe this is explained by small $N$ handled specially.

For larger $N$, the algorithm, doesn't return in reasonable time and debugging info shows signs of infinite loop, possibly caused by C double.

Q1 When Coppersmith's algorithm is polynomial and this approach factors $n$? Is the content the only obstacle?

$\endgroup$
3
  • $\begingroup$ Coppersmith mentions integer factorization in his second paper on the topic: link.springer.com/chapter/10.1007%2F3-540-68339-9_16 $\endgroup$ Commented Feb 19, 2016 at 16:31
  • $\begingroup$ @MaxAlekseyev Thanks. Your link is bivariate and doesn't mention the bound AFAICT. In some cases it requires coprime content, which may answer the question. $\endgroup$
    – joro
    Commented Feb 20, 2016 at 8:34
  • $\begingroup$ @joro Where did you get the bound? $\endgroup$
    – Turbo
    Commented Dec 31, 2018 at 18:10

1 Answer 1

3
$\begingroup$

We asked on the pari-dev mailing list and the developers replied that the documentation was incorrect if the leading coefficient is not coprime to $N$.

This is fixed in pari-master.

Discussion: http://pari.math.u-bordeaux1.fr/archives/pari-dev-1912/msg00002.html

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .