RSA Chinese Remainder Theorem

#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Author: john
# @Date:   2016-08-25 14:33:10
# @Last Modified by:   john
# @Last Modified time: 2017-02-28 18:52:57

from Crypto.Util.number import *
import binascii
import sympy


e = 3

bitsize = 512

n1 = n2 =n3 = 0

while ( ( sympy.gcd(n3,n2) !=  1 ) and ( sympy.gcd(n1,n2) !=  1 ) and ( sympy.gcd(n1,n2) !=  1 ) ):

    n1 = getPrime( bitsize )
    n2 = getPrime( bitsize )
    n3 = getPrime( bitsize )
    # n4 = getPrime( bitsize )


# print n1, n2, n3
print "n1 = ", n1
print "n2 = ", n2
print "n3 = ", n3
# print "n4 = ", n4
# print sympy.gcd(n1,n2), sympy.gcd(n1,n3), sympy.gcd(n3,n2)

m = 'USCGA{the_chinese_remainder_theorem_is_so_cool}'
m = binascii.hexlify(m)
m = int(m, 16)

c1 = pow(m, e, n1)
c2 = pow(m, e, n2)
c3 = pow(m, e, n3)
# c4 = pow(m, e, n4)
print ""

print "c1 = ", c1
print "c2 = ", c2
print "c3 = ", c3

def chinese_remainder(n, a):
    sum = 0
    prod = reduce(lambda a, b: a*b, n)

    for n_i, a_i in zip(n, a):
        p = prod / n_i
        print "n_i", n_i
        print "a_i", a_i
        print "p", p
        #sum += a_i * mul_inv(p, n_i) * p
        sum += a_i * inverse(p, n_i) * p
        print "sum", sum
    return sum % prod

x = chinese_remainder([n1, n2, n3], [c1, c2, c3])
# print x

# 1. Simple attack
# m = sympy.root(x, e)

m = pow(sympy.E,  sympy.ln(x) / e )


print "Trey's one sum", c1 * n2 * n3 * inverse( n2*n3, n1 )
# # THIS IS THE CHINESE REMAINDER THEOREM IN ONE EQUATION ... FROM TREY
# m = ( c1 * n2 * n3 * inverse( n2*n3, n1 ) + c2 * n1 * n3 * inverse( n1*n3, n2 ) + c3 * n1 * n2 * inverse( n1*n2, n3 ) ) % ( n1 * n2 * n3 )
# print m == x

# print pow(10, sympy.ln(pow(m,e)))


print ""

# print

try:
    print hex(int(m))[2:-1].decode('hex')
except:
    print "Supposedly odd length string..."
    print str("0"+ hex( int(m))[2:-1]).decode('hex')