<s>[INST] To find relatively prime ordered pairs of positive integers $(a,b)$ such that $ \dfrac ab +\dfrac {14b}{9a}$ is an integer How many ordered pairs $(a,b)$ of positive integers are there such that g.c.d.$(a,b)=1$ , and 
$ \dfrac ab +\dfrac {14b}{9a}$ is an integer ? 
 [/INST] I first find out what the values of a/b are which make the expression an integer, then check which of the solutions are rational numbers. Let t = a/b, then t + 14/(9t) should be an integer. 
Solve t + 14/(9t) = n: t^2 + 14/9 = nt, t^2 - nt + n^2/4 = n^2/4 - 14/9.
t - n/2 = +/- sqrt (n^2/4 - 14/9). 
t should be rational, so n^2/4 - 14/9 should be the square of a rational number. Multiply by 36, and 9n^2 - 56 should be the square of a rational number, which means it must be the square of an integer. 
When n >= 10, (3n-1)^2 < 9n^2 - 56 < (3n)^2, so 9n^2 - 56 cannot be the square of an integer. n = 1 to n = 9 can be checked manually, and we find that n = 3 makes 9n^2 - 56 = 25 = 5^2, n = 3 makes 9n^2 - 56 = 169 = 13^2. 
n = 3: t - 3/2 = +/- sqrt (9/4 - 14/9) = +/- sqrt (25/36) = +/- 5/6; t = 14/6 or t = 4/6. 
n = 5: t - 5/2 = +/- sqrt (25/4 - 14/9) = +/- sqrt (169/36) = +/- 13/6, t = 28/6 or t = 2/6. 
Together: a/b is one of 7/3, 2/3, 14/3 or 1/3
Checking: 
7/3 + (14*3 / 9*7) = 7/3 + 2/3 = 3
2/3 + (14*3 / 9*2) = 2/3 + 7/3 = 3
14/3 + (14*3 / 9*14) = 14/3 + 1/3 = 5
1/3 + (14*3 / 9*1) = 1/3 + 14/3 = 5. 
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