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 supposedly is simple to modify from square roots to cube roots, but this eludes me. I’ve scoured the web and math libraries and a few books to no avail. Code would be wonderful, so would an algorithm explained in plain English.
Here is the Python (2.6?) code for finding square roots:
def modular_sqrt(a, p):
""" Find a quadratic residue (mod p) of 'a'. p
must be an odd prime.
Solve the congruence of the form:
x^2 = a (mod p)
And returns x. Note that p - x is also a root.
0 is returned is no square root exists for
these a and p.
The Tonelli-Shanks algorithm is used (except
for some simple cases in which the solution
is known from an identity). This algorithm
runs in polynomial time (unless the
generalized Riemann hypothesis is false).
"""
# Simple cases
#
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return n
elif p % 4 == 3:
return pow(a, (p + 1) / 4, p)
# Partition p-1 to s * 2^e for an odd s (i.e.
# reduce all the powers of 2 from p-1)
#
s = p - 1
e = 0
while s % 2 == 0:
s /= 2
e += 1
# Find some 'n' with a legendre symbol n|p = -1.
# Shouldn't take long.
#
n = 2
while legendre_symbol(n, p) != -1:
n += 1
# Here be dragons!
# Read the paper "Square roots from 1; 24, 51,
# 10 to Dan Shanks" by Ezra Brown for more
# information
#
# x is a guess of the square root that gets better
# with each iteration.
# b is the "fudge factor" - by how much we're off
# with the guess. The invariant x^2 = ab (mod p)
# is maintained throughout the loop.
# g is used for successive powers of n to update
# both a and b
# r is the exponent - decreases with each update
#
x = pow(a, (s + 1) / 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in xrange(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
def legendre_symbol(a, p):
""" Compute the Legendre symbol a|p using
Euler's criterion. p is a prime, a is
relatively prime to p (if p divides
a, then a|p = 0)
Returns 1 if a has a square root modulo
p, -1 otherwise.
"""
ls = pow(a, (p - 1) / 2, p)
return -1 if ls == p - 1 else ls
Answers:
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"
Here is a complete code in pure python. By considering special cases first, it is almost as fast as the Peralta algoritm.
#assumes p prime, it returns all cube roots of a mod p
def cuberoots(a, p):
#Non-trivial solutions of x**r=1
def onemod(p,r):
sols=set()
t=p-2
while len(sols)<r:
g=pow(t,(p-1)//r,p)
while g==1: t-=1; g=pow(t,(p-1)//r,p)
sols.update({g%p,pow(g,2,p),pow(g,3,p)})
t-=1
return sols
def solutions(p,r,root,a):
todo=onemod(p,r)
return sorted({(h*root)%p for h in todo if pow(h*root,3,p)==a})
#---MAIN---
a=a%p
if p in [2,3] or a==0: return [a]
if p%3 == 2: return [pow(a,(2*p - 1)//3, p)] #One solution
#There are three or no solutions
#No solution
if pow(a,(p-1)//3,p)>1: return []
if p%9 == 7: #[7, 43, 61, 79, 97, 151]
root = pow(a,(p + 2)//9, p)
if pow(root,3,p) == a: return solutions(p,3,root,a)
else: return []
if p%9 == 4: #[13, 31, 67, 103, 139]
root = pow(a,(2*p + 1)//9, p)
print(root)
if pow(root,3,p) == a: return solutions(p,3,root,a)
else: return []
if p%27 == 19: #[19, 73, 127, 181]
root = pow(a,(p + 8)//27, p)
return solutions(p,9,root,a)
if p%27 == 10: #[37, 199, 307]
root = pow(a,(2*p +7)//27, p)
return solutions(p,9,root,a)
#We need a solution for the remaining cases
return tonelli3(a,p,True)
An extension of Tonelli-Shank algorithm.
def tonelli3(a,p,many=False):
def solution(p,root):
g=p-2
while pow(g,(p-1)//3,p)==1: g-=1 #Non-trivial solution of x**3=1
g=pow(g,(p-1)//3,p)
return sorted([root%p,(root*g)%p,(root*g**2)%p])
#---MAIN---
a=a%p
if p in [2,3] or a==0: return [a]
if p%3 == 2: return [pow(a,(2*p - 1)//3, p)] #One solution
#No solution
if pow(a,(p-1)//3,p)>1: return []
#p-1=3**s*t
s=0
t=p-1
while t%3==0: s+=1; t//=3
#Cubic nonresidu b
b=p-2
while pow(b,(p-1)//3,p)==1: b-=1
c,r=pow(b,t,p),pow(a,t,p)
c1,h=pow(c,3**(s-1),p),1
c=pow(c,p-2,p) #c=inverse modulo p
for i in range(1,s):
d=pow(r,3**(s-i-1),p)
if d==c1: h,r=h*c,r*pow(c,3,p)
elif d!=1: h,r=h*pow(c,2,p),r*pow(c,6,p)
c=pow(c,3,p)
if (t-1)%3==0: k=(t-1)//3
else: k=(t+1)//3
r=pow(a,k,p)*h
if (t-1)%3==0: r=pow(r,p-2,p) #r=inverse modulo p
if pow(r,3,p)==a:
if many:
return solution(p,r)
else: return [r]
else: return []
You can test it using:
test=[(17,1459),(17,1000003),(17,10000019),(17,1839598566765178548164758165715596714561757494507845814465617175875455789047)]
for a,p in test:
print "y^3=%s modulo %s"%(a,p)
sol=cuberoots(a,p)
print "p%s3=%s"%("%",p%3),sol,"--->",map(lambda t: t^3%p,sol)
which should yield (fast):
y^3=17 modulo 1459
p%3=1 [483, 329, 647] —> [17, 17, 17]
y^3=17 modulo 1000003
p%3=1 [785686, 765339, 448981] —> [17, 17, 17]
y^3=17 modulo 10000019
p%3=2 [5188997] —> [17]
y^3=17 modulo 1839598566765178548164758165715596714561757494507845814465617175875455789047
p%3=1 [753801617033579226225229608063663938352746555486783903392457865386777137044, 655108821219252496141403783945148550782812009720868259303598196387356108990, 430688128512346825798124773706784225426198929300193651769561114101322543013] —> [17, 17, 17]
I converted the code by Rolandb above into python3. If you put this into a file, you can import it and run it in python3, and if you run it standalone it will validate that it works.
#! /usr/bin/python3
def ts_cubic_modular_roots (a, p):
""" python3 version of cubic modular root code posted
by Rolandb on stackoverflow. With new formatting.
https://stackoverflow.com/questions/6752374/cube-root-modulo-p-how-do-i-do-this
"""
#Non-trivial solution of x**r = 1
def onemod (p, r):
t = p - 2
while pow (t, (p - 1) // r, p) == 1:
t -= 1
return pow (t, (p - 1) // r, p)
def solution(p, root):
g = onemod (p, 3)
return [root % p, (root * g) % p, (root * (g ** 2)) % p]
#---MAIN---
a = a % p
if p in [2, 3] or a == 0:
return [a]
if p % 3 == 2:
return [pow (a, ((2 * p) - 1) // 3, p)] #Eén oplossing
#There are 3 or no solutions
#No solution
if pow (a, (p-1) // 3, p) > 1:
return []
if p % 9 == 4: #[13, 31, 67]
root = pow (a, ((2 * p) + 1) // 9, p)
if pow (root, 3, p) == a:
return solution (p, root)
else:
return []
if p % 9 == 7: #[7, 43, 61, 79, 97
root = pow (a, (p + 2) // 9, p)
if pow (root, 3, p) == a:
return solution (p, root)
else:
return []
if p % 27 == 10: #[37, 199]
root = pow (a, ((2 * p) + 7) // 27, p)
h = onemod (p, 9)
for i in range (0,9):
if pow (root, 3, p) == a:
return solution (p, root)
root *= h
return []
if p % 27 == 19: #[19, 73, 127, 181]
root = pow (a, (p + 8)//27, p)
h = onemod (p, 9)
for i in range (0, 9):
if pow (root, 3, p) == a:
return solution (p, root)
root *= h
return []
#We need an algorithm for the remaining cases
return tonelli3 (a, p, True)
def tonelli3 (a, p, many = False):
#Non-trivial solution of x**r = 1
def onemod (p, r):
t = p - 2
while pow (t, (p - 1) // r, p) == 1:
t -= 1
return pow (t, (p - 1) // r, p)
def solution (p, root):
g = onemod (p, 3)
return [root % p, (root * g) % p, (root * (g**2)) % p]
#---MAIN---
a = a % p
if p in [2, 3] or a == 0:
return [a]
if p % 3 == 2:
return [pow (a, ((2 * p) - 1) // 3, p)] #Eén oplossing
#No solution
if pow (a, (p - 1) // 3, p) > 1:
return []
#p-1 = 3^s*t
s = 0
t = p - 1
while t % 3 == 0:
s += 1
t //= 3
#Cubic nonresidu b
b = p - 2
while pow (b, (p - 1) // 3, p) == 1:
b -= 1
c, r = pow (b, t, p), pow (a, t, p)
c1, h = pow (c, 3 ^ (s - 1), p), 1
c = pow (c, p - 2, p) #c=inverse_mod(Integer(c), p)
for i in range (1, s):
d = pow (r, 3 ^ (s - i - 1), p)
if d == c1:
h, r = h * c, r * pow (c, 3, p)
elif d != 1:
h, r = h * pow (c, 2, p), r * pow (c, 6, p)
c = pow (c, 3, p)
if (t - 1) % 3 == 0:
k = (t - 1) // 3
else:
k = (t + 1) // 3
r = pow (a, k, p) * h
if (t - 1) % 3 == 0:
r = pow (r, p - 2, p) #r=inverse_mod(Integer(r), p)
if pow (r, 3, p) == a:
if many:
return solution(p, r)
else: return [r]
else: return []
if '__name__' == '__main__':
import ts_cubic_modular_roots
tscr = ts_cubic_modular_roots.ts_cubic_modular_roots
test=[(17,1459),(17,1000003),(17,10000019),(17,1839598566765178548164758165715596714561757494507845814465617175875455789047)]
for a,p in test:
print ("y**3=%s modulo %s"%(a,p))
sol = tscr (a,p)
print ("p%s3=%s"%("%",p % 3), sol, [pow (t,3,p) for t in sol])
# results of the above
#y**3=17 modulo 1459
#p%3=1 [] []
#y**3=17 modulo 1000003
#p%3=1 [785686, 765339, 448981] [17, 17, 17]
#y**3=17 modulo 10000019
#p%3=2 [5188997] [17]
#y**3=17 modulo 1839598566765178548164758165715596714561757494507845814465617175875455789047
#p%3=1 [753801617033579226225229608063663938352746555486783903392457865386777137044, 655108821219252496141403783945148550782812009720868259303598196387356108990, 430688128512346825798124773706784225426198929300193651769561114101322543013] [17, 17, 17]
Sympy has a nice implementation for arbitrary integer modulo and arbitrary power: https://docs.sympy.org/latest/modules/ntheory.html#sympy.ntheory.residue_ntheory.nthroot_mod
from sympy.ntheory.residue_ntheory import nthroot_mod
a = 17
n = 3
modulo = 10000019
roots = nthroot_mod(a, n, modulo)
print(roots)
# 5188997
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 supposedly is simple to modify from square roots to cube roots, but this eludes me. I’ve scoured the web and math libraries and a few books to no avail. Code would be wonderful, so would an algorithm explained in plain English.
Here is the Python (2.6?) code for finding square roots:
def modular_sqrt(a, p):
""" Find a quadratic residue (mod p) of 'a'. p
must be an odd prime.
Solve the congruence of the form:
x^2 = a (mod p)
And returns x. Note that p - x is also a root.
0 is returned is no square root exists for
these a and p.
The Tonelli-Shanks algorithm is used (except
for some simple cases in which the solution
is known from an identity). This algorithm
runs in polynomial time (unless the
generalized Riemann hypothesis is false).
"""
# Simple cases
#
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return n
elif p % 4 == 3:
return pow(a, (p + 1) / 4, p)
# Partition p-1 to s * 2^e for an odd s (i.e.
# reduce all the powers of 2 from p-1)
#
s = p - 1
e = 0
while s % 2 == 0:
s /= 2
e += 1
# Find some 'n' with a legendre symbol n|p = -1.
# Shouldn't take long.
#
n = 2
while legendre_symbol(n, p) != -1:
n += 1
# Here be dragons!
# Read the paper "Square roots from 1; 24, 51,
# 10 to Dan Shanks" by Ezra Brown for more
# information
#
# x is a guess of the square root that gets better
# with each iteration.
# b is the "fudge factor" - by how much we're off
# with the guess. The invariant x^2 = ab (mod p)
# is maintained throughout the loop.
# g is used for successive powers of n to update
# both a and b
# r is the exponent - decreases with each update
#
x = pow(a, (s + 1) / 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in xrange(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
def legendre_symbol(a, p):
""" Compute the Legendre symbol a|p using
Euler's criterion. p is a prime, a is
relatively prime to p (if p divides
a, then a|p = 0)
Returns 1 if a has a square root modulo
p, -1 otherwise.
"""
ls = pow(a, (p - 1) / 2, p)
return -1 if ls == p - 1 else ls
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"
Here is a complete code in pure python. By considering special cases first, it is almost as fast as the Peralta algoritm.
#assumes p prime, it returns all cube roots of a mod p
def cuberoots(a, p):
#Non-trivial solutions of x**r=1
def onemod(p,r):
sols=set()
t=p-2
while len(sols)<r:
g=pow(t,(p-1)//r,p)
while g==1: t-=1; g=pow(t,(p-1)//r,p)
sols.update({g%p,pow(g,2,p),pow(g,3,p)})
t-=1
return sols
def solutions(p,r,root,a):
todo=onemod(p,r)
return sorted({(h*root)%p for h in todo if pow(h*root,3,p)==a})
#---MAIN---
a=a%p
if p in [2,3] or a==0: return [a]
if p%3 == 2: return [pow(a,(2*p - 1)//3, p)] #One solution
#There are three or no solutions
#No solution
if pow(a,(p-1)//3,p)>1: return []
if p%9 == 7: #[7, 43, 61, 79, 97, 151]
root = pow(a,(p + 2)//9, p)
if pow(root,3,p) == a: return solutions(p,3,root,a)
else: return []
if p%9 == 4: #[13, 31, 67, 103, 139]
root = pow(a,(2*p + 1)//9, p)
print(root)
if pow(root,3,p) == a: return solutions(p,3,root,a)
else: return []
if p%27 == 19: #[19, 73, 127, 181]
root = pow(a,(p + 8)//27, p)
return solutions(p,9,root,a)
if p%27 == 10: #[37, 199, 307]
root = pow(a,(2*p +7)//27, p)
return solutions(p,9,root,a)
#We need a solution for the remaining cases
return tonelli3(a,p,True)
An extension of Tonelli-Shank algorithm.
def tonelli3(a,p,many=False):
def solution(p,root):
g=p-2
while pow(g,(p-1)//3,p)==1: g-=1 #Non-trivial solution of x**3=1
g=pow(g,(p-1)//3,p)
return sorted([root%p,(root*g)%p,(root*g**2)%p])
#---MAIN---
a=a%p
if p in [2,3] or a==0: return [a]
if p%3 == 2: return [pow(a,(2*p - 1)//3, p)] #One solution
#No solution
if pow(a,(p-1)//3,p)>1: return []
#p-1=3**s*t
s=0
t=p-1
while t%3==0: s+=1; t//=3
#Cubic nonresidu b
b=p-2
while pow(b,(p-1)//3,p)==1: b-=1
c,r=pow(b,t,p),pow(a,t,p)
c1,h=pow(c,3**(s-1),p),1
c=pow(c,p-2,p) #c=inverse modulo p
for i in range(1,s):
d=pow(r,3**(s-i-1),p)
if d==c1: h,r=h*c,r*pow(c,3,p)
elif d!=1: h,r=h*pow(c,2,p),r*pow(c,6,p)
c=pow(c,3,p)
if (t-1)%3==0: k=(t-1)//3
else: k=(t+1)//3
r=pow(a,k,p)*h
if (t-1)%3==0: r=pow(r,p-2,p) #r=inverse modulo p
if pow(r,3,p)==a:
if many:
return solution(p,r)
else: return [r]
else: return []
You can test it using:
test=[(17,1459),(17,1000003),(17,10000019),(17,1839598566765178548164758165715596714561757494507845814465617175875455789047)]
for a,p in test:
print "y^3=%s modulo %s"%(a,p)
sol=cuberoots(a,p)
print "p%s3=%s"%("%",p%3),sol,"--->",map(lambda t: t^3%p,sol)
which should yield (fast):
y^3=17 modulo 1459
p%3=1 [483, 329, 647] —> [17, 17, 17]
y^3=17 modulo 1000003
p%3=1 [785686, 765339, 448981] —> [17, 17, 17]
y^3=17 modulo 10000019
p%3=2 [5188997] —> [17]
y^3=17 modulo 1839598566765178548164758165715596714561757494507845814465617175875455789047
p%3=1 [753801617033579226225229608063663938352746555486783903392457865386777137044, 655108821219252496141403783945148550782812009720868259303598196387356108990, 430688128512346825798124773706784225426198929300193651769561114101322543013] —> [17, 17, 17]
I converted the code by Rolandb above into python3. If you put this into a file, you can import it and run it in python3, and if you run it standalone it will validate that it works.
#! /usr/bin/python3
def ts_cubic_modular_roots (a, p):
""" python3 version of cubic modular root code posted
by Rolandb on stackoverflow. With new formatting.
https://stackoverflow.com/questions/6752374/cube-root-modulo-p-how-do-i-do-this
"""
#Non-trivial solution of x**r = 1
def onemod (p, r):
t = p - 2
while pow (t, (p - 1) // r, p) == 1:
t -= 1
return pow (t, (p - 1) // r, p)
def solution(p, root):
g = onemod (p, 3)
return [root % p, (root * g) % p, (root * (g ** 2)) % p]
#---MAIN---
a = a % p
if p in [2, 3] or a == 0:
return [a]
if p % 3 == 2:
return [pow (a, ((2 * p) - 1) // 3, p)] #Eén oplossing
#There are 3 or no solutions
#No solution
if pow (a, (p-1) // 3, p) > 1:
return []
if p % 9 == 4: #[13, 31, 67]
root = pow (a, ((2 * p) + 1) // 9, p)
if pow (root, 3, p) == a:
return solution (p, root)
else:
return []
if p % 9 == 7: #[7, 43, 61, 79, 97
root = pow (a, (p + 2) // 9, p)
if pow (root, 3, p) == a:
return solution (p, root)
else:
return []
if p % 27 == 10: #[37, 199]
root = pow (a, ((2 * p) + 7) // 27, p)
h = onemod (p, 9)
for i in range (0,9):
if pow (root, 3, p) == a:
return solution (p, root)
root *= h
return []
if p % 27 == 19: #[19, 73, 127, 181]
root = pow (a, (p + 8)//27, p)
h = onemod (p, 9)
for i in range (0, 9):
if pow (root, 3, p) == a:
return solution (p, root)
root *= h
return []
#We need an algorithm for the remaining cases
return tonelli3 (a, p, True)
def tonelli3 (a, p, many = False):
#Non-trivial solution of x**r = 1
def onemod (p, r):
t = p - 2
while pow (t, (p - 1) // r, p) == 1:
t -= 1
return pow (t, (p - 1) // r, p)
def solution (p, root):
g = onemod (p, 3)
return [root % p, (root * g) % p, (root * (g**2)) % p]
#---MAIN---
a = a % p
if p in [2, 3] or a == 0:
return [a]
if p % 3 == 2:
return [pow (a, ((2 * p) - 1) // 3, p)] #Eén oplossing
#No solution
if pow (a, (p - 1) // 3, p) > 1:
return []
#p-1 = 3^s*t
s = 0
t = p - 1
while t % 3 == 0:
s += 1
t //= 3
#Cubic nonresidu b
b = p - 2
while pow (b, (p - 1) // 3, p) == 1:
b -= 1
c, r = pow (b, t, p), pow (a, t, p)
c1, h = pow (c, 3 ^ (s - 1), p), 1
c = pow (c, p - 2, p) #c=inverse_mod(Integer(c), p)
for i in range (1, s):
d = pow (r, 3 ^ (s - i - 1), p)
if d == c1:
h, r = h * c, r * pow (c, 3, p)
elif d != 1:
h, r = h * pow (c, 2, p), r * pow (c, 6, p)
c = pow (c, 3, p)
if (t - 1) % 3 == 0:
k = (t - 1) // 3
else:
k = (t + 1) // 3
r = pow (a, k, p) * h
if (t - 1) % 3 == 0:
r = pow (r, p - 2, p) #r=inverse_mod(Integer(r), p)
if pow (r, 3, p) == a:
if many:
return solution(p, r)
else: return [r]
else: return []
if '__name__' == '__main__':
import ts_cubic_modular_roots
tscr = ts_cubic_modular_roots.ts_cubic_modular_roots
test=[(17,1459),(17,1000003),(17,10000019),(17,1839598566765178548164758165715596714561757494507845814465617175875455789047)]
for a,p in test:
print ("y**3=%s modulo %s"%(a,p))
sol = tscr (a,p)
print ("p%s3=%s"%("%",p % 3), sol, [pow (t,3,p) for t in sol])
# results of the above
#y**3=17 modulo 1459
#p%3=1 [] []
#y**3=17 modulo 1000003
#p%3=1 [785686, 765339, 448981] [17, 17, 17]
#y**3=17 modulo 10000019
#p%3=2 [5188997] [17]
#y**3=17 modulo 1839598566765178548164758165715596714561757494507845814465617175875455789047
#p%3=1 [753801617033579226225229608063663938352746555486783903392457865386777137044, 655108821219252496141403783945148550782812009720868259303598196387356108990, 430688128512346825798124773706784225426198929300193651769561114101322543013] [17, 17, 17]
Sympy has a nice implementation for arbitrary integer modulo and arbitrary power: https://docs.sympy.org/latest/modules/ntheory.html#sympy.ntheory.residue_ntheory.nthroot_mod
from sympy.ntheory.residue_ntheory import nthroot_mod
a = 17
n = 3
modulo = 10000019
roots = nthroot_mod(a, n, modulo)
print(roots)
# 5188997