Problem:
How many pairs of positive integers (a,b) are there such that a and b have no common factors greater than 1 and
ba​+9a14b​
is an integer?
Answer Choices:
A. 4
B. 6
C. 9
D. 12
E. infinitely many
Solution:
Let u=a/b. Then the problem is equivalent to finding all positive rational numbers u such that
u+9u14​=k
for some integer k. This equation is equivalent to 9u2−9uk+14=0, whose solutions are
u=189k±81k2−504​​=2k​±61​9k2−56​
Hence u is rational if and only if 9k2−56​ is rational, which is true if and only if 9k2−56 is a perfect square. Suppose that 9k2−56=s2 for some positive integer s. Then (3k−s)(3k+s)=56. The only factors of 56 are 1,2,4,7, 8,14,28, and 56, so (3k−s,3k+s) is one of the ordered pairs (1,56),(2,28), (4,14), or (7,8). The cases (1,56) and (7,8) yield no integer solutions. The cases (2,28) and (4,14) yield k=5 and k=3, respectively. If k=5, then u=1/3 or u=14/3. If k=3, then u=2/3 or u=7/3. Therefore there are (A)4​ pairs (a,b) that satisfy the given conditions, namely (1,3),(2,3),(7,3), and (14,3).
OR
Rewrite the equation
ba​+9a14b​=k
in two different forms. First, multiply both sides by b and subtract a to obtain
9a14b2​=bk−a
Because a,b, and k are integers, 14b2 must be a multiple of a, and because a and b have no common factors greater than 1, it follows that 14 is divisible by a. Next, multiply both sides of the original equation by 9a and subtract 14b to obtain
b9a2​=9ak−14b
This shows that 9a2 is a multiple of b, so 9 must be divisible by b. Thus if (a,b) is a solution, then b=1,3, or 9, and a=1,2,7, or 14. This gives a total of twelve possible solutions (a,b), each of which can be checked quickly. The only such pairs for which
ba​+9a14b​
is an integer are when (a,b) is (1,3),(2,3),(7,3), or (14,3).