Determine all pairs (a, b) of positive integers such that ab2 + b + 7 divides a2b + a + b.
Answer (a, b) = (11, 1), (49, 1) or (7k2, 7k).
Solution
If a < b, then b ≥ a + 1, so ab2 + b + 7 > ab2 + b ≥ (a + 1)(ab + 1) = a2b + a + ab ≥ a2b + a + b. So there can be no solutions with a < b. Assume then that a ≥ b.
Let k = the integer (a2b + a + b)/(ab2 + b + 7). We have (a/b + 1/b)(ab2 + b + 7) = ab2 + a + ab + 7a/b + 7/b + 1 > ab2 + a + b. So k < a/b + 1/b. Now if b ≥ 3, then (b - 7/b) > 0 and hence (a/b - 1/b)(ab2 + b + 7) = ab2 + a - a(b - 7/b) - 1 - 7/b < ab2 + a < ab2 + a + b. Hence either b = 1 or 2 or k > a/b - 1/b.
If a/b - 1/b < k < a/b + 1/b, then a - 1 < kb < a + 1. Hence a = kb. This gives the solution (a, b) = (7k2, 7k).
It remains to consider b = 1 and 2. If b = 1, then a + 8 divides a2 + a + 1 and hence also a(a + 8) - (a2 + a + 1) = 7a - 1, and hence also 7(a + 8) - (7a - 1) = 57. The only factors bigger than 8 are 19 and 57, so a = 11 or 49. It is easy to check that (a, b) = (11, 1) and (49, 1) are indeed solutions.
If b = 2, then 4a + 9 divides 2a2 + a + 2, and hence also a(4a + 9) - 2(2a2 + a + 2) = 7a - 4, and hence also 7(4a + 9) - 4(7a - 4) = 79. The only factor greater than 9 is 79, but that gives a = 35/2 which is not integral. Hence there are no solutions for b = 2.
A variant on this from Johannes Tang Lek Huo is as follows:
We have ab2 + b + 7 divides a(ab2 + b + 7) - b(a2b + a + b) = 7a - b2 . If 7a = b2, then b must be a multiple of 7, so b = 7k for some k. Then a = 7k2, and it is easy to check that this is a solution. We cannot have 7a < b2 for then 0 < b2 - 7a < ab2 < ab2 + b + 7. If 7a > b, then we must have 7a - b ≤ ab2 + b + 7 > ab2, so 7 > b2, so b = 1 or 2.
We can then continue as above.
(C) John Scholes
jscholes@kalva.demon.co.uk
26 Oct 1998
Last corrected/updated 20 Aug 03