# Euler Problem 27¶

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.

Okay, bruteforcing this complete thing is definitely hard. The interesting thing is that euler provided two examples. If we calculate the primes for both terms we see that there is a certain overlap. This indicates that there is a relation between the two. Sure enough if we put $n-40$ into the first term we get the following.

$(n - 40)^2 + (n - 40) + 41 = n^2 - 80n + 1600 + n - 40 + 41 = n^2 - 79n +1601$

Let's assume that all incredible formulas can be derived by inserting $(n - p)$ into the formula. Of course, what ever value we choose for p the resulting terms must not exceed the boundaries for a or b. We calculate the boundaries.

$(n-p)^2 + (n-p) + 41 = n^2 + n(-2p + 1) + p^2 - p + 41$

Where $(-2p + 1) = a$ and $p^2 -p + 41 = b$. We can now calulate the bounds for a:

$-2p + 1 > -1000 \rightarrow p < 500.5$

$-2 p + 1 < 1000 \rightarrow p > -499.5$

And b:

$p^2 - p + 41 <= 1000$

$-30.472 < p < 31.472$

$p^2 - p + 41 >= 1000$ True for $\forall p \in \mathbb{N}$

So now we only have to check for the values p in range(-30, 32). Alternatively, for the example $p = 40$ was used, maybe the next smaller value $p = 31$ yields the correct solution:

$s = a\times b = (-2 * 31 + 1) * (31^2 - 31 + 41) = -61 \times 971 = -59231$

In [1]:

    from functools import lru_cache

    @lru_cache(maxsize=1000)
    def is_prime(n):
        if n < 2:
            return False
        from math import sqrt
        for s in range(2, int(sqrt(n) + 1)):
            if n % s == 0:
                return False
            if s * s > n:
                return True
        return True

    assert(is_prime(41) == True)

    def get_quadratic(n, p):
        return n*n + n * (-2* p + 1) + p*p - p + 41

    n_max = 0
    p_max = 0

    for p in range(-30, 32):
        for n in range(0, 10000):
            if not is_prime(get_quadratic(n, p)) and n > n_max:
                n_max = n
                p_max = p
                break

    p = p_max
    s = (-2 * p + 1) * (p*p - p + 41)
    assert(s == -59231)
    s

Out[1]:

    -59231