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Say I have a polynomial $F$ of degree $n$ with coefficients in $Z_m$ and I wish to find $x$ such that $F(x)=0$ (mod $m$). For instance if $F(x)=x^{2}-a$ the solution would be the modulo $m$ squareroot of $a$ (if there is a solution).

I'm primarily interested in solving the general case.

Without loss of generality we may assume $F$ is irreducible since otherwise we can just factor $F=f_0\cdots f_1$ with $f_i$ irreducible and the solutions of $F$ will be the union of the solutions for all of the $f_i$.

One method similar to Hensel lifting I've already considered roughly would involve factoring $m=p_0^{a_0}\cdots p_k^{a_k}$ and brute forcing $x$ (mod $p_i$) for each $i$ and lifting them to solutions mod $p_i^{a_i}$. This would be problematic if $m$ was a large prime though.

So any of the following would be very useful:

  • Fast solutions for when $m$ is prime
  • Methods to prove there are no solutions
  • Any relevant research
QutieParrot
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  • For primes, and $x^m \equiv a$, there are fast methods for determining whether there is a solution. And there can be fast deterministic methods for some classes of primes and some $m$, and usually fast probabilistic methods for the others. – André Nicolas Jul 19 '12 at 06:30
  • If $m$ is not a prime then a product can have more zeroes than the union of those of its factors. – WimC Jul 19 '12 at 06:42

1 Answers1

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There is a nice randomized algorithm for finding the roots when $m=p$ is prime, it goes roughly as follows: first compute $$G(x) = \operatorname{gcd}\bigl( x^p-x,F(x)\bigr)$$ then $G(x)$ is the product of all the roots of $F(x)$ that are in $\mathbb Z/p$. Now for several random values of $a$ compute $$ \operatorname{gcd}\bigl( (x+a)^{(p-1)/2}-1,G(x)\bigr)$$ this will probably result on a non trivial factor of $G$, because it separates roots $\alpha$ for which $x+a$ is a quadratic residue from those which are not. You can repeat with the resulting factors until all the factors have degree one. This algorithm is very fast. It is described here.

Esteban Crespi
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