Project Euler

Problem 27: Quadratic Primes

Euler discovered the remarkable quadratic formula: n² + n + 41. It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40² + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² - 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, -79 and 1601, is -126479.

Considering quadratics of the form: n² + 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.

The Catch

How to speed up the search from |a| < 1000 and |b| < 1000. That is looping from -999 to 999 for a and b, totaling 1999 * 1999 = 3,996,001 iterations!

The Light

By doing a little mathematical analysis, it is obvious that for n = 0, n² + an + b = 0² + 0 + b = b. Since this function must return a prime number, b must be prime. Thus, we can skip all non-prime values for b and keep track of the streak. It can be shown that this simple tweak lowers the number of iterations to 355,822. Use the prime checking method discussed in Problem 3 to check for prime number.

The Code

public class Problem27
{
  public static void main(String[] args)
  {
    int maxA = 0, maxB = 0 , maxN = 0;
    int count = 0;

    for(int a = -999; a < 1000; a++)
    {
      for(int b = -999; b < 1000; b++)
      {
        if(!isPrime(b))
          continue;

        int n = 0;
        while(isPrime( n*n + a*n + b )) 		
        {
          n++;
        }

        if(n > maxN)
        {
          maxN = n;
          maxA = a;
          maxB = b;
        }
      }
    }
    System.out.println("a * b = " + (maxA * maxB));
  }

  public static boolean isPrime(int n)
  {
    if( n < 2 )
      return false;

    if(n % 2 == 0 && n != 2)
      return false;

    for(int i = 3; i * i < n; i++)
    {
      if(n % i == 0)
        return false;
    }
    return true;
  }
}