from lib_prime import prime_nth


def varray(i):
    return i * 2 + 1


def rem(i):
    n = varray(i)
    p = prime_nth(n)
    return (pow(p - 1, n) + pow(p + 1, n)) % (p * p)


def bin_search(lo, hi, target):
    if hi - lo == 1:
        return varray(hi)
    i = lo + ((hi - lo) // 2)
    r = rem(i)

    # print(f"{i=:<6} {r=}")
    if r < target:
        return bin_search(i, hi, target)
    elif r > target:
        return bin_search(lo, i, target)
    assert False


def euler_123():
    """
    I found out that when n is even, the remainder is always 2. When the
    remainder is odd, the series grows monotonically. That means we can use a
    binary search on odd values to find the solution.
    """
    return bin_search(1000, 30000, 10**10)


if __name__ == "__main__":
    solution = euler_123()
    print("e123.py: " + str(solution))
    assert(solution == 21035)