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.

(In reply to what mathematica has to say by Daniel)

Mathematica is awesome!  However, its price is astronomical unless you have some "in" which I don't.  So I had to do it by hand.  I used the change of variables

3P-2Q=x, 3P+2Q=y or P=(x+y)/6, Q=(y-x)/4

in terms of which the equation simplfies to x^2=y so that

P=x(x+1)/6, Q=x(x-1)/4

and we then need to analyze the cases where integers are obtained.  One case is where x+1 is divisible by 6 and x-1 is divisible by 4 which gives P=5-22k+24k^2, Q=5-27k+36k^2.  Another case is where x is divisble by 12 which gives P=2j+24j^2, Q=36j^2-3j. Your other two cases must be lurking in there somewhere, too, but were not obvious to me.

Posted by Richard on 2006-08-03 19:14:50