1. Program

Let's try to find the solutions of Diophantine equation n=(x+y+z)(1/x+1/y+1/z) where -100 ≤ n ≤ 100.
Assume 1 ≤ |x| ≤ |y| ≤ |z|.

When d=gcd(x,y,z)>1, let x=dx', y=dy', z=dz', then,

  (x+y+z)(1/x+1/y+1/z)
= (dx'+dy'+dz')(1/dx'+1/dy'+1/dz')
= (x'+y'+z')(1/x'+1/y'+1/z')

so, we can assume gcd(x,y,z)=1.

And let l=lcm(x,y,z), x'=l/x, y'=l/y, z'=l/z, then,

  (x+y+z)(1/x+1/y+1/z)
= (lx'+ly'+lz')(1/lx'+1/ly'+1/lz')
= (x'+y'+z')(1/x'+1/y'+1/z')

so, (x',y',z') is also the solution of above equation. This is called as "dual".

If we consider the case d= -1, we should search the following four cases,

( x, y, z)
(-x, y, z)
( x,-y, z)
( x, y,-z)

Program

 10   ' xyz.ub
 20   M=1000
 30   for X=1 to M
 40     for Y=X to M:D=gcd(X,Y)
 50       for Z=Y to M:D=gcd(D,Z):if D>1 then 110
 60         if and{X=Y,Y=Z} then 110
 70         R=fnSub(X,Y,Z)
 80         R=fnSub(-X,Y,Z)
 90         R=fnSub(X,-Y,Z)
100         R=fnSub(X,Y,-Z)
110       next Z
120     next Y
130   next X
140   end
150   fnSub(X,Y,Z)
160   local N
170   N=(X+Y+Z)*(1//X+1//Y+1//Z)
180   if or{N=0,N=1} then 220
190   if den(N)>1 then 220
200   if abs(N)>100 then 220
210   print N;":";X;",";Y;",";Z
220   return(0)

Start from the range 1 ≤ |x| ≤ |y| ≤ |z| ≤ 1000.
Computation results are next page.

The order of this program is O(n³), so we should find out the other methods for further computation.