Shorten the intervals

If with take a closer look at the equation, we should find useful information. For example, with some peculiar values of \( n \), we can narrow down the ranges of \( a \) and \( b \).

We are searching \( p \) prime such that:

\[ p = n^2 + an + b \]

For \( n = 0\), the equation is \( p = b \). That means that \( b \) must be a prime.

For \( n = b \), the equation is \( p = b^2 + ab + b = b(b + a + 1) \) which is divisible by \( b \). \( b \) is in fact the limit of consecutive primes that can be found using this equation.

Rearranging the equation we have:

\[ p = n^2 + an + b \Leftrightarrow p - b = n(n + a) \]

Since both \( p \) and \( b \) are prime, \( p - b \) is an even number.

If \( n \) is even then \( n(n + a) \) will also be even.
If \( n \) is odd, then \( n(n + a) \) will be odd only if \( a \) is odd.

For \( n = 1 \), \( n^2 + an + b \) will never be a prime if \( a \) is even.

All this information can help quite a bit:

• \( a \) must be an odd number between.
• b must be a prime.
• b is the upper limit of consecutive primes.

def quadratic_primes():
    primes_b = list((primerange(0, 1000)))[::-1]  # b is prime.
    res = 0
    max_primes = 0
    for a in range(-999, 1000, 2):  # a is odd.
        for b in primes_b:
            if b < max_primes:  # b is the limit for consecutive prime.
                continue

            n = 0
            while isprime(n ** 2 + a * n + b):
                n += 1

            if n > max_primes:
                max_primes = n
                res = a * b

    return res