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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: 3.0000 (2 votes)

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Solution alternate solution | Comment 3 of 5 |

Square both sides of all three equations to get:

p^2 + 1/p^2 + 2 = q^2, q^2 + 1/q^2 + 2 = r^2, r^2 + 1/r^2 + 2 = p^2

Now add all three equations to get:

p^2+q^2+r^2+1/p^2+1/q^2+1/r^2+6 = p^2+q^2+r^2 -->

1/p^2 + 1/q^2 + 1/r^2 + 6 = 0 which is impossible.

So there are no such triples. 


  Posted by Dennis on 2008-04-12 20:51:13
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