Is it possible for $(900q^2+ap^2)/(3q^2+b^2p^2)$ to be an integer? The original problem is: "Find all possible pairs of positive integers $(a, b)$ 
$$k = \dfrac{a^3+300^2}{a^2b^2+300}\tag1$$
such that $k$ is an integer."
I've tried so many different ways. Now this question comes up from one of them. Let, 
$$\left(\dfrac{a}{10} \right)^2=\dfrac{3(300-k)}{kb^2-a}=\dfrac{p^2}{q^2}\tag2$$
If you calculate $k$ in terms of $a, b, p, q,$ then the following question comes up:
"Suppose $a$, $b$, $p$ and $q$ are natural numbers such that $a<300$ and $\gcd(p, q)=1$. Is it possible for
$$k=\frac{900q^2+ap^2}{3q^2+b^2 p^2}\tag3$$
to become an integer?" 
I think this is easier than original problem, but I don't know how to proceed from here.

While your $(3)$ can be derived from $(2)$, if it is an integer does not guarantee that $(1)$ will be an integer as well. Note that it has the four variables $a,b,p,q$ that can "integerize" your expression, while you're stuck with only $a,b$ for $(1)$. 
(For example, $a,b,p,q =3,2,9,5$ makes $(3)$ an integer, but $a,b = 3,2$ does not for $(1)$.)
Also, your original problem should explicitly specify the constraint that $a \neq 300b^2$. If not, then $a = 300b^2$ yields an infinite family which always has $k=300$.
A computer search with $a<10000$ and $b<100$ yields the infinite family and,
$$a,b,k = 10,2,130$$  
$$a,b,k = 20,1,140$$  
$$a,b,k = 30,2,30$$  
which seem to be the only other solutions.
