Determine all possible triplet(s) (x,y,z) of positive rational numbers, with x ≥ y ≥ z, such that each of x + (yz)^-1, y + (zx)^-1 and, z + (xy)^-1 is an integer.

    5   dim Hist(50,3)
   10   for T=6 to 1000000
   20   for A=1 to T-5
   30   for B=1 to T-A-4
   40   for C=1 to T-A-B-3
   50   for D=1 to T-A-B-C-2
   60   for E=1 to T-A-B-C-D-1
   70    F=T-A-B-C-D-E
   80    X=A//B:Y=C//D:Z=E//F
   90    V=X+1//(Y*Z)
  100    if V=int(V) then
  110     :V=Y+1//(X*Z)
  120     :if V=int(V) then
  130      :V=Z+1//(X*Y)
  140      :if V=int(V) then
  150          :if X<Y then Hold=X:X=Y:Y=Hold:endif
  151          :if Y<Z then Hold=Y:Y=Z:Z=Hold:endif
  152          :if X<Y then Hold=X:X=Y:Y=Hold:endif
  160          :Good=1
  165          :for I=1 to SCt
  170            :if X=Hist(I,1) and Y=Hist(I,2) and Z=Hist(I,3) then Good=0:cancel for:goto 180:endif
  175          :next
  180          :if Good then
  190            :inc SCt
  195            :Hist(SCt,1)=X
  196            :Hist(SCt,2)=Y
  197            :Hist(SCt,3)=Z
  198            :print X;Y;Z,A;B;C;D;E;F
  200   next
  210   next
  220   next
  230   next
  240   next
  250   next

With the total of the numerators and denominators having gone to 71, only 3 sets have been found:

(1,1,1); (2,1,1/2) and (4,1/2,1/2)

(1,1,1);
1 + 1/(1*1) = 2

(2,1,1/2):
2+1/(1/2) = 4
1+1/1 = 2
1/2 + 1/2 = 1

(4,1/2,1/2):
4+1/(1/4) = 8
1/2 + 1/2 = 1

Edited on March 6, 2009, 1:00 pm

Posted by Charlie on 2009-03-06 13:00:00