Cube root modulo P — how do I do this?

Question

I am trying to calculate the cube root of a many-hundred digit number modulo P in Python, and failing miserably.

I found code for the Tonelli-Shanks algorithm which supp

Answer 1

Note added later: In the Tonelli-Shanks algorithm and here it is assumed that p is prime. If we could compute modular square roots to composite moduli quickly in general we could factor numbers quickly. I apologize for assuming that you knew that p was prime.

See here or here. Note that the numbers modulo p are the finite field with p elements.

Edit: See this also (this is the grandfather of those papers.)

The easy part is when p = 2 mod 3, then everything is a cube and athe cube root of a is just a**((2*p-1)/3) %p

Added: Here is code to do all but the primes 1 mod 9. I'll try to get to it this weekend. If no one else gets to it first

#assumes p prime returns cube root of a mod p
def cuberoot(a, p):
    if p == 2:
        return a
    if p == 3:
        return a
    if (p%3) == 2:
        return pow(a,(2*p - 1)/3, p)
    if (p%9) == 4:
        root = pow(a,(2*p + 1)/9, p)
        if pow(root,3,p) == a%p:
            return root
        else:
            return None
    if (p%9) == 7:
        root = pow(a,(p + 2)/9, p)
        if pow(root,3,p) == a%p:
            return root
        else:
            return None
    else:
        print "Not implemented yet. See the second paper"