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A Bivariate Divisibility Problem (Posted on 2006-06-08) Difficulty: 4 of 5
Determine all possible pairs (p,q) of positive integers such that :
pq2+ q + 23 divides p2q + p + q .

See The Solution Submitted by K Sengupta    
Rating: 1.6667 (6 votes)

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  1. Prove that no solution exists for p< q; so p> = q.
  2. Substitute s = ((p^2)*q + p + q) / (p*(q^2) + q + 23)
  3. This would  give a simple parametric solution for x and y in terms of s in a particular situation. In all other available  situations q must be less than or equal to 4.(Why?)
  4. It is now relatively easy to determine all possible solutions to the problem.

Edited on June 8, 2006, 11:27 pm
  Posted by K Sengupta on 2006-06-08 23:26:37

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