Are there any non zero real triplet(s) (p, q, r) that satisfy the following system of equations?

p + 1/p = q, q + 1/q = r, and r + 1/r = p

If such triplet(s) exist, find at least one of these. Otherwise, prove that no such triplet can conform to the given conditions.

(In reply to alternate solution by Dennis)

"1/p^2 + 1/q^2 + 1/r^2 + 6 = 0"

....... and  since each of p, q and r must be non zero real numbers,  the lhs of the above equation must be > 6, which leads to a contradiction.

Very nice methodology especially since the said process obviates the necessity to consider separately the positive case and the negative case.

Consequently, I confirm having hyperlinked your methodology in my solution.

Edited on May 5, 2008, 2:50 pm

Posted by K Sengupta on 2008-05-05 14:45:40