Find all integers p and q that satisfy p³+27pq+2009=q³.

I have made a further restriction to the solution(s)

the only solution found so far is p=-7 q=7 and thus is of the form
p=-|p| q=|q|  I shall now prove that it is the only solution of this form and thus any other solutions would have to be of a different form

if p=-|p| and q=|q| then we have
-|p|³-27|p||q|+2009=|q|³
2009-27|p||q|=|q|³+|p|³
now since the right hand side is always positive then so also must the left hand side thus
2009-27|p||q|>0
27|p||q|<2009
|p||q|<2009/27
since p,q are integers we can simplify like so
|p||q|≤74
since p,q are integers then we have
|p|≤74  |q|≤74 and from my previous work with mathematica I have already shown that the only solution under these constraints is p=-7 and q=7.  Thus there are no further solutions with
p=-|p| and q=|q|

I am attempting to arrive at proofs that no solutions exist for the other 3 sign combinations for p,q but they are not so simple as this one and thus far ellude me :-)

Posted by Daniel on 2006-08-27 22:14:38