sol1.py

"""
Problem 120 Square remainders: https://projecteuler.net/problem=120

Description:

Let r be the remainder when (a−1)^n + (a+1)^n is divided by a^2.
For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 ≡ 42 mod 49.
And as n varies, so too will r, but for a = 7 it turns out that r_max = 42.
For 3 ≤ a ≤ 1000, find ∑ r_max.

Solution:

On expanding the terms, we get 2 if n is even and 2an if n is odd.
For maximizing the value, 2an < a*a => n <= (a - 1)/2 (integer division)
"""


def solution(n: int = 1000) -> int:
    """
    Returns ∑ r_max for 3 <= a <= n as explained above
    >>> solution(10)
    300
    >>> solution(100)
    330750
    >>> solution(1000)
    333082500
    """
    return sum(2 * a * ((a - 1) // 2) for a in range(3, n + 1))


if __name__ == "__main__":
    print(solution())