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Curious Real Additive Relationship (Posted on 2008-04-12) Difficulty: 2 of 5
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.

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

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Solution solution | Comment 1 of 5


s = r + 1/r
  = q + 1/q + 1/(q + 1/q)
  = p + 1/p + 1/(p + 1/p) + 1/(p + 1/p + 1/(p + 1/p))

When p is positive, s > p because of all the positive terms added to p on the right side. Thus s cannot equal p.

When p is zero s is undefined and again cannot equal p.

When p is negative, the terms added to p are all negative so s < p, and again, s cannot equal p.

Since s = r + 1/r is what is asked to equal p, this can never occur.


  Posted by Charlie on 2008-04-12 14:26:33
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