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.

First of all, there is the trivial solution P=Q=0. This is part of the family of solutions P= 2j+24j^2, Q=36j^2-3j, j any integer. This family also includes P=26, Q=33.

There is also another family of solutions that includes P=Q=5, namely P=5-22k+24k^2, Q=5-27k+36k^2, k any integer.

The key to solving was .... (later, maybe).  Let's see if some other  solvers can get off their duffs.

Posted by Richard on 2006-08-03 16:41:51