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

Since (q-p)³=q³-p³-3pq²+3p²q, (q-p)³+3pq(q-p) = 27pq + 2009.
Let x = q-p and y=pq. So x³ + 3xy = 27y + 2009.
Equivalently, 3y = (x³-2009)/(9-x) = -x² - 9x - 81 + 1280/(x-9).
Since 1280 has 18 positive factors, there are only 36 values of x to consider. They are 10,11,13,14,17,19,25,29,41,49,73,89,137,169,
265,329,649,1289,8,7,5,4,1,-1,-7,-11,-23,-31,-55,-71,-119,-151,-247,
-311,-631, and -1271. Since 3y is divisible by 3, half of the above values of x are eliminated leaving 11,14,17,29,41,89,137,329,1289,
8,5,-1,-7,-31,-55,-151,-247, and -631. Finally, solving x=q-p and
y=pq simultaneously yields 2p + x = ±√(x²+4y)  so x²+ 4y must
be a perfect square. This forces x = 14 and y = -49 only, which in
turn yields p=-7 and q=7 only.

Posted by Dennis on 2006-09-13 16:24:09