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
if s = 1 (one more solution!)
