"Euler discovered the remarkable quadratic formula:\n",
"\n",
"$n^2 + n + 41$\n",
"\n",
"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.\n",
"\n",
"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 $−126479$.\n",
"\n",
"Considering quadratics of the form:\n",
"\n",
"$n^2 + an +b$, where |a|<1000 and |b|≤1000\n",
"\n",
"where |n| is the modulus/absolute value of n e.g. |11|=11 and |−4|=4.\n",
"\n",
"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",
"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",
"Where $(-2p + 1) = a$ and $p^2 -p + 41 = b$. We can now calulate the bounds for a:\n",
"\n",
"$-2p + 1 > -1000 \\rightarrow p < 500.5$\n",
"\n",
"$-2 p + 1 < 1000 \\rightarrow p > -499.5$\n",
"\n",
"And b:\n",
"\n",
"$p^2 - p + 41 <= 1000$\n",
"\n",
"$-30.472 < p < 31.472$\n",
"\n",
"$p^2 - p + 41 >= 1000$ True for $\\forall p \\in \\mathbb{N}$\n",
"\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:\n",