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 $0 \le n \le 39$. 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 $0 \le n \le 79$. The product of the coefficients, $−79$ and $1601$, is $−126479$.

Considering quadratics of the form:

$n^2 + an + b$, where $|a|