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 Add to Product Inverse (Posted on 2009-03-06)
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.

 See The Solution Submitted by K Sengupta Rating: 4.0000 (2 votes)

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 computer exploration | Comment 1 of 5

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

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