Explanation:
Putting a – 1 = p, b - 1 = q and c –1 = r, the equation may be written in form
pqr = 2(p + q + r) + 4,
where p, q, r are integer such that 0 ≤ p ≤ q ≤ r. Observe that p = 0 is not possible, for then 0 = 2(p + q) + 4 which is impossible in nonnegative integers. Thus we may write this in the form
2(1/pq + 1/qr + 1/rp) + 4/pqr = 1.
If p ≥ 3 then q ≥ 3 and r ≥ 3. Then left side is bounded by 6/9 + 4/27 which is less then 1. We conclude that p = 1 or 2.
Case 1. Suppose p = 1. Then we have qr = 2(q + r) + 6 or (q - 2)(r – 2) = 10 . This gives q – 2 = 1, r – 2 = 10, or q – 2 = 2 and r – 2 = 5(recall q r). This implies (p, q, r) = (1, 3,12),(1, 4, 7).
Case 2. If p = 2 the equation reduce to 2qr = 2(2 + q + r) + 4 or qr = q + r + 4. This reduce to (q – 1) (r – 1) = 5. Hence q – 1 = 1 and r – 1 = 5. Is the only solution. This gives (p, q, r) = (2, 2, 6)
Reverting back to a,b,c we get three triples(a,b,c) =(2,4,13), (2,5,8),(3,3,7).
