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 0
elif p % 4 == 3:
return pow(a, (p + 1) >> 2, 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) >> 1, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in range(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) >> 1, p)
return -1 if ls == p - 1 else ls