"""
Euler discovered the remarkable quadratic formula: n^2 + n + 41

It turns out that the formula will produce 40 primes for the consecutive integer values 0<=n<=39.
However, when n = 40, 40^2+40+41=40*(40+1)+41 is divisible by 41, and certainly when n = 41,
41^2+41+41 is clearly divisible by 41.

The incredible formula n^2-79n+1601 was discovered, which produces 80 primes for the consecutive values
0<=n<=79 . The product of the coefficients, -79 and 1601, is -126479 .

Considering quadratics of the form:
n^2+an+b , where |a|<1000 and |b|<=1000
where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |-4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that produces
the maximum number of primes for consecutive values of n, starting with n=0.
"""

from typing import Callable


def quadratic_primes(a: int, b: int) -> Callable[[int], int]:
    return lambda n: n**2 + a * n + b


def is_prime(n: int) -> bool:
    if n < 2:
        return False
    if n == 2:
        return True
    if n % 2 == 0:
        return False
    for i in range(3, int(n**0.5) + 1, 2):
        if n % i == 0:
            return False
    return True


def max_prime_quadratic(a_max: int, b_max: int) -> tuple[int, int, int]:
    max_primes = 0
    max_a = 0
    max_b = 0
    for a in range(-a_max, a_max + 1):
        for b in range(-b_max, b_max + 1):
            primes = 0
            n = 0
            while is_prime(quadratic_primes(a, b)(n)):
                primes += 1
                n += 1
            if primes > max_primes:
                max_primes = primes
                max_a = a
                max_b = b
    return max_a, max_b, max_a * max_b


def main():
    a_max = 1000
    b_max = 1000
    max_a, max_b, product = max_prime_quadratic(a_max, b_max)
    print(f"{max_a} x {max_b} = {product}.")


if __name__ == "__main__":
    main()