Consider the equation:

(3P–2Q)² = 24PQ/(3P+2Q-1)

where P and Q are positive integers. It can be verified that (P=5, Q=5) and (P=26, Q=33) are two solutions.

(A) Can you give at least three other solutions to the above equation?

(B) Determine whether or not the equation admits of an infinite number of solutions.

I used the reduce function in mathematica to come up with the following general solutions

these work for all values of k both positive and negative
p=5+22k+24k^2 q=5+27k+36k^2
p=12+34k+24k^2 q=14+45k+36k^2
p=15+38k+24k^2 q=18+51k+36k^2
p=2k+24k^2 q=36k^2-3k

Posted by Daniel on 2006-08-03 18:03:12