/*
 * 4nbg - solve alternate formulation of 4/n problem.
 * David Eppstein, UC Irvine, 26 Oct 1998
 *
 * 4/n = 1/x + 1/y + 1/z can be solved if we can find b and g s.t.
 * b+g = 0 mod (-n mod 4bg): From any solution b, g,
 * let a = (b+g)/(-n mod 4bg), f = (an+b)/g, c = (f+1)/4ab.
 * Then 4/n = 1/abcn + 1/bcg + 1/acgn.
 *
 * Not all solutions to 4/n = 1/x + 1/y + 1/z look like this,
 * even when n is prime, but this seems to work for many n.
 *
 * There never exist solutions b and g when n is square (apparently this
 * is related to a 1956 paper by A. Schinzel cited in the section of
 * Unsolved Problems in Number Theory on the 4/n problem).  But they
 * also do not exist for certain other special values of n: 288, 336,
 * and 4545 (and no others less than a million).  What are these numbers?
 */

#include <stdio.h>
#include <stdlib.h>

int euclid(long long a, long long b) {
  if (a > b) return euclid(b,a);
  else if (a == 0) return b;
  else return euclid(b % a, a);
}

void out4n(long long n, long long b, long long g)
{
  long long a,c,f,k,m;
  m = (-n) % (4*b*g);
  if (m <= 0) m += 4*b*g;
  a = (b+g)/m;
  if (euclid(a,b) != 1) return; /* filter duplicate solutions */
  f = (a*n+b)/g;
  c = (f+1)/(4*a*b);
  k = a*b*c;
  printf("4/%lld = 1/%lld + 1/%lld + 1/%lld\n", n, b*c*g, k*n, a*c*g*n);
}

void findBG(long long n)
{
  long long b,g;
  for (b = 1; b*b < n; b++)
    for (g = b; 4*g*b < 2*n; g++) {
      long long m = (-n) % (4*b*g);
      if (m <= 0) m += 4*b*g;
      if ((b+g) % m == 0) out4n(n,b,g);
    }
}

static int nout = 0;
static long long sqroot = 1;

void testBG(long long n)
{
  long long b,g;
  if (sqroot * sqroot == n) {
    sqroot++;
    return;
  }
  for (b = 1; b*b < n; b++)
    for (g = b; 4*g*b <= 2*n; g++) {
      long long m = (-n) % (4*b*g);
      if (m <= 0) m += 4*b*g;
      if ((b+g) % m == 0) return;
    }
  printf("%lld", n);
  if (++nout == 5) {
    printf("\n"); nout = 0;
  }
  else printf(" ");
  fflush(stdout);
}

main(int ac, char **av) {
  long long i,n;

  if (ac != 2) {
    printf("usage: 4nbg n expands 4/n; 4nbg -n finds exceptional values up to n\n");
    exit(0);
  }
  n = atoll(av[1]);
  if (n > 0) findBG(n);
  else {
    for (i = 1; i <= -n; i++) testBG(i);
    if (nout > 0) printf("\n");
  }
}