from fractions import Fraction
from lib_misc import proper_divisors
from math import ceil


def get_distinct_solutions(n):
    """ Get the number of distinct solutions for 1/x + 1/y = 1/n. The idea is
    that 1/n*2 + 1/n*2 = 1/n is always a solution. So what we do is
    brute-force all 1/x for x in [2, n * 2]. We have a solution if the
    difference between 1/n and 1/x has a numerator of 1. """
    n_inv = Fraction(1, n)
    n_distinct = 0
    for x in range(2, n * 2 + 1):
        x_inv = Fraction(1, x)
        y_inv = n_inv - x_inv
        if y_inv.numerator == 1:
            n_distinct += 1
    return n_distinct


def get_distinct_solutions2(n):
    ds = proper_divisors(n * n)
    return ceil((len(ds) + 1) / 2)


def euler_108():
    d_max, n_prev = 0, 0
    # I arrived at the starting values empirically by observing the deltas.
    for n in range(1260, 1000000, 420):
        d = get_distinct_solutions2(n)
        if d > d_max:
            # print("n={} d={} delta={}".format(n, d, n - n_prev))
            n_prev = n
            d_max = d
        if d > 1000:
            return n


if __name__ == "__main__":
    solution = euler_108()
    print("e108.py: " + str(solution))
    assert(solution == 180180)