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A 2009 problem (Posted on 2006-08-27) Difficulty: 3 of 5
Find all integers p and q that satisfy p³+27pq+2009=q³.

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

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Factors and restrictions | Comment 4 of 5 |

The prime factorization of 2009 is 7x7x41, and none of these are factors of 27. This means that they may be the only factors that p and q share. because otherwise, you can factor it out from all three terms except 2009 which would be impossible to solve.

Looking at the cases p=|q|, it is clear that p doesn't equal q. if p=-q, then p³-27p²+2009=-p³ or -p³-27p²+2009=p³. The first form equals p²(2p-27)=-2009 and the second form equals p²(2p-27)=2009. It is clear that (7,-7) is the only possible solution there.

Edited on August 28, 2006, 12:02 am
  Posted by Gamer on 2006-08-28 00:00:41

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