Solving quadratic or higher degree congruence with very large modulus.

Question

Is there any general way to solve a polynomial congruence with a very large modulus?

An example could be

$$ x^2-377x+1\equiv 0 \pmod {8683317618811886495518194401279999999 } $$

or $$ x^2-29478x\equiv 13^{67} \pmod {17^{2001}} $$

the following briefly describe general way (and I think number theory books do too): http://www.cs.xu.edu/math/math302/08f/16_QuadraticCongruences.pdf and http://www.johndcook.com/blog/quadratic_congruences/ and http://math.stackexchange.com/questions/160385/how-to-solve-this-quadratic-congruence-equation — Mirko, May 06 '15 at 00:07

Robert Israel · Accepted Answer · 2015-05-05T23:30:09.920

1

If you can factor the modulus into coprime factors, you can solve the congruence with respect to each factor and then use the Chinese Remainder Theorem. On the other hand, being able to solve $x^2 = a \mod m$ would give you the ability to factor $m$.

EDIT: For lots more on these questions, see "Basic algorithms in number theory" by Joe Buhler and Stan Wagon

edited May 05 '15 at 23:30

answered May 05 '15 at 23:15

Robert Israel

448,999

And what would you do if you had a very large prime number as modulust (first example)? Is there any way to simplify that congruence? – YoTengoUnLCD May 05 '15 at 23:17
The modulus, in this particular case, is a large prime. – Lucian May 06 '15 at 01:23
Not all that large... Maple's msolve quickly returns NULL, because $5685$ is not a square mod $8683317618811886495518194401279999999$. – Robert Israel May 06 '15 at 03:00

score -1 · Answer 2 · answered May 06 '15 at 14:56

-1

Algorithms for this are included in Mathematica. E.g.:

Solve[x^2 - 29478*x == 13^67, x, Modulus -> 17^2001]

yields the two solutions in an eyeblink.

answered May 06 '15 at 14:56

stan wagon

1,653

Solving quadratic or higher degree congruence with very large modulus.

2 Answers2