Pythagorean Triples

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include <conio.h>

typedef enum {false, true} bool;

bool isPerfectSquare(long int n)
{
    if (n % 7 == 3 || n % 7 == 5 || n % 11 == 8) {return false;} else
    {
        if (fabs(fmod(sqrt(n), 1.)) < 0.0001) {return true;}
        return false;
    }
}

//Writing this stuff on my own is beyond my knowledge of pointers.
int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

void main(void)
{
    int x;
    int y;
    int k;
    int c = 0;
    bool inter;
    long int z;
    int sl[40];
    int dl[40][3];
    int ol[40][3];
    int t = (int)clock();
    for(x = 1; x < 1024; x++)
    {
        for(y = 2; y < x; y++)
        {
            z = x*x + y*y;
            if (isPerfectSquare(z))
            {
                z = (int)sqrt(z);
                //printf("DING x=%d y=%d\n", x, y);
                inter = true;
                for (k = 2; k < (int)sqrt(z) + 1; k++)
                {
                    if (x % k == 0 && y % k == 0 && z % k == 0) {inter = false; break;}
                }
                //if (!inter) {continue;} else {printf("DING DING x=%d y=&d");}
                if (inter)
                {
                    //printf("DING DING x=%d y=%d z=%d n=%d\n", y, x, z, c);
                    dl[c][0] = y;
                    dl[c][1] = x;
                    dl[c][2] = z;
                    sl[c] = y;
                    c++;
                    if (c > 39) {goto done;}
                }
            }
        }
    }
    done:;
    // let's get some quicksort all up in 'dis
    qsort(sl, 40, sizeof(int), compare);
    // messily match and sort y and z vals; we're scrambled like eggs now
    c = 0;
    do
    {
        x = sl[c];
        for(y = 0; y < 40; y++)
        {
            if (x == dl[y][0])
            {
                // I'm sure this is better with pointers.
                ol[c][0] = dl[y][0];
                ol[c][1] = dl[y][1];
                ol[c][2] = dl[y][2];
                c++;
            }
        }
    }
    while(c < 40);
    // time to display the hopefully correct output
    t = (int)clock()-t;
    for(c = 0; c < 20; c++)
    {
        printf("> x=%d  y=%d  z=%d\n", ol[c][0], ol[c][1], ol[c][2]);
    }
    printf("Execution time: %d ms\n", t);
    _getch();
}