# Project Euler 27 Solution: Quadratic primes

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.

$n^2 + an + b,$ where $$|a| < 1000$$ and $$|b| \le 1000$$ and 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 $$n = 0$$.

## Solution

Based on the problem statement, a very simple brute force algorithm can be stated, using the isPrime function from Problem 7:

function solution() {

var maxC = 0;
var maxAB = 0;
for (var a = -999; a <= 999; a++) {
for (var b = -1000; b <= 1000; b++) {

var c = countConsecutivePrimes(a, b);
if (c > maxC) {
maxC = c;
maxAB = a * b;
}
}
}
return maxAB;
}

function countConsecutivePrimes(a, b) {

for (var n = 0; ; n++) {
var t = n * n + a * n + b;
if (!isPrime(t)) {
return n;
}
}
}


This sledgehammer method is quite fast already, but far from satisfying. Since $$n^2 + an + b$$ must generate a series of primes, we can easily find optimizations to algorithm above. Starting with $$n_0:= 0$$, the first element becomes

$n_0^2+an_0+b = b$

Since the quadratic equation must be prime for every element, it follows that $$b$$ must be prime as well. For the next element, a similar step can be performed with $$n_1=1$$:

$n_1^2+an_1+b = 1 + a + b$

Since $$b$$ is prime, and all primes but 2 are odd, it follows that $$a$$ must be odd for all $$b\neq 2$$ and even for $$b= 2$$. This lets us improve the algorithm as follows:

var primes = [
2,   3,   5,   7,  11,  13,  17,  19,  23,  29,  31,  37,  41,
43,  47,  53,  59,  61,  67,  71,  73,  79,  83,  89,  97, 101,
103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313,
317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467,
479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569,
571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643,
647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823,
827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911,
919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997];

function solution() {

var maxC = 0;
var maxAB = 0;
for (var a = -999; a <= 1001; a+= 2) {
for (var i = 0; i < primes.length; i++) {

var b = primes[i];
var c = countConsecutivePrimes(a - (i == 0 ? 1 : 0), b);
if (c > maxC) {
maxC = c;
maxAB = a * b;
}
}
}
return maxAB;
}


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