Modular equation solver

# -*- coding: utf-8 -*-

#
# ax ≡ b (mod c)
# Looking for two smallest positive numbers (x) satisfying equation ax - b ≡ 0 (mod c)
# Having two solutions m & n is possible to determine formula for all solutions
#
# n - m is a difference between next solutions; m is smaller then n;
# Hence if ax = b (mod c), x = (n - m)t + m
#

def getEquation(a, b, c):
    if a is c and b is not 0 and b is not c:
        return "There is no solution"

    congruent = u"\u2261".encode('utf-8')
    answer = "{0}x {1} {2} (mod {3}) \n".format(a, congruent, b, c) # ax ≡ b (mod c)
    if a %c is not a or b % c is not b:
        answer += "{0}x {1} {2} (mod {3}) \n".format(a % c, congruent, b % c, c)

    solutions, i = [], 0
    a %= c
    b %= c

    while len(solutions) is not 2:
        if i > 2 * c: return "There is no solution"
        if (a * i - b) % c is 0: solutions.append(i)
        i += 1

    answer += "x {0} {1} (mod {2}) \n".format(congruent, solutions[0], solutions[1] - solutions[0]) # x ≡ m (mod n - m)
    answer += "x = {0}t + {1}".format(solutions[1] - solutions[0], solutions[0]) # x = (n - m)*t + m
    return answer

print(getEquation(411, 3207, 911))