how many pairs of integers (a,b)are there with a = or < b that 1/6=1/a+1/b?

Answer 1

Solve for a,
a = 1/(1/6 - 1/b) = 6b/(b-6) = 6 + 36/(b-6)
As long as b-6 divides 36, you have a potential answer.
b-6 = ±1, ±2, ±3, ±4, ±6, ±9, ±12. ±18, ±36
Can you count the number of answers that a≤b?

Answer 2

First note that there is an oversight in the work by "sahsjing." He states that b-6 should divide 36. The condition is obviously necessary, but it is not sufficient. For example, 2 is a divisor of 36, but with b - 6 = 2, we have b = 8 and a = 24. This violates the condition that a <= b. More significantly, suppose b - 6 = -3; then b = 3 and a = -9, but 1/(-9) + 1/3 is not 1/6, even though a< b is satisfied.

From the given equation, we get ab = 6a + 6b. Let b = a + k, with k >= 0. This leads to the quadratic
(1) a^2 + (k - 12)*a - 6k = 0.
The discriminant is k^2 + 144. Since we are only interested in integer solutions, we need
(2) k^2 + 144 = x^2, for some integer x.
From the theory of Pythagorean Triples, we know that the only primitive solutions of (2) occur when sqrt(144) = 12 = 2rs, for 0 < r <= s, where r and s are relatively prime and of opposite parity. This tells us that either r = 1 and s = 6, or r = 2 and s = 3. Then k = s^2 - r^2 = 35 or 5. In turn, in (2) we get x = 37 or 13 for x.

With k = 35 in (1) we find a = 7 or -30, and then b = 42 or 5. This gives us the solutions (a,b) = (7,42) and (-30,5). With k = 5 we find a = -3 or 10, so we get solutions (-3,2) and (10,15).