Re: 61. - THE SILVER CUBES

In article <da38f2628fe7919ad1a2f3495eeb2483@www.novabbs.com>,
IlanMayer <ilan_no_spew@hotmail.com> wrote:

The C# program below finds the solution in under a minute:

Here is a C program.  You can choose whether to use floating point
(which should be big and accurate enough) but it you don't you will
need to check whether you have the flsll function.

/* Find rationals a/d and b/d the sum of whose cubes is 17 */

#include <stdio.h>
#include <stdlib.h>
#include <strings.h>
#include <inttypes.h>
#include <math.h>

/* integer cube root, return -1 if not a cube */
long long int cuberoot(long long c)
{
#ifdef USE_FLOAT
    long long a = cbrt(c);
#else
    int n = flsll(c);
    long long a = 1ll << ((n+2)/3); /* a good first approximation */

    while(c/a/a < a)
a = (2*a+c/a/a) / 3;
#endif
    return a*a*a == c ? a : -1;
}

int main(int argc, char **argv)
{
    int n = (argc > 1 ? atoi(argv[1]) : 17);
    long long denom, denom3n, num1, num2, num13, num23;

    for(denom=1; denom<1000000; denom++)
    {
if(denom % 1000 == 0)
    printf("(%lld)\n", denom);

denom3n = denom * denom * denom * n;

for(num1=1; ; num1++)
{
    num13 = num1 * num1 * num1;
    num23 = denom3n - num13;

    if(num23 < num13)
break;

    num2 = cuberoot(num23);

    if(num2 != -1)
    {
printf("%lld^3 + %lld^3 = %d x %lld^3\n",
       num1, num2, n, denom);
return 0;
    }
}
    }
}