Determine all possible triplet(s) (p, q, r) of positive integers, with p ≤ q ≤ r, that satisfy the equation: pq+qr+rp - pqr = 2.
a) If p = 1, then (by substitution)
q + qr + r - qr = 2
q + r = 2
only solution is (1,1,1)
b) If p = 2, then (by substitution)
2q + qr + 2r - 2qr = 2
solving for r,
r = (2q-2)/(q-2) = 2 + (2/(q-2))
r is integral only if q = 3 or 4,
both of which lead to the ordered solution of (2,3,4)
c) If p >= 3, then
pq+qr+ rp <= 3qr (since r and q are the biggest values)
3qr <= pqr (since 3 <= p)
so pq + qr + rp <= pqr
pq + qr + rp - pqr <= 0
so there are no solutions if p >= 3
Therefore, the only solutions are (1,1,1) and (2,3,4)
Edited on December 8, 2009, 3:38 pm
Edited on December 8, 2009, 5:51 pm
Analytical solution (spoiler)