Quadratic primes Problem 27 Project Euler

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.

BRUTE FORCE:

#include<stdio.h>
#include<math.h>
#include<limits.h>
int pcheck(int n)
{
    int i;
    for(i=2;i<=sqrt(n);i++)
    {
        if(n%i==0){return 0;}
    }
    if(n<=1){return 0;}
    return 1;
}
int max(int a,int b)
{
    if(a>b)return a;
    return b;
}
main()
{
int a,b,i;
int max=INT_MIN;
int amax,bmax;
for(a=-999;a<1000;a++)
{
    i=0;
    for(b=-999;b<1000;b++)
    {
         if(((a+b)%2)){continue;}
    int j=0;
    while(1)
    {
        int res=(j*j)+(a*j)+b;
        if(pcheck(res)){j++;}
        else break;
    }
    if(j>max){max=j;amax=a;bmax=b;}
    }
}
int t=amax*bmax;
printf("%d",t);

}

TIME TAKEN:

.591s

ANSWER

-59231