Determine all possible integer pairs (p,q) such that p+q²+s³=pqs, where s=gcd(p,q) and gcd denotes the

greatest common divisor.
In all cases, p = (q^2 + s^3)/(qs - 1)

If p and q are relatively prime (s = 1), then

p = (q^2 + 1)/(q - 1) = (q - 1) + 2q/(q - 1)

P is therefore integral if q- 1 divides 2 (i.e, q = 3)

or if q - 1 divides q (i.e, q = 2 or 0)

This leads to three solutions:

(5, 3), (5,2) and (-1,0)

I checked the definitiion, and the gcd of -1 and 0 is in fact 1!

Next step: s > 1

*Edited on ***June 21, 2007, 7:01 pm**