A diophantine problem with big numbers!

Question

Find all pairs of positive integers $(a, b)$ such that

$a^2 b^2 +300 \mid a^2(300 b^2 -a)$

and $300 b^2 -a>0$.

I've tried so many different ways, I only concluded that $a<300$.

Answer 1

Not a complete answer, but this approach might be helpful :

$$a^2b^2+300|a^2(300b^2-a)$$

with $300b^2-a>0$

implies $$a^2b^2+300|300a^2b^2-a^3-300(a^2b^2+300)=-a^3-90000$$

So, we have $$a^2b^2+300|a^3+90000$$

$$a^2(b^2k-a)=90000-300k=300(300-k)$$

If $k>300$, then the right side is negative, but the left side is postitive because of $b^2k-a>300b^2-a>0$

If $k=300$, then the right side is $0$, but the left side is $300b^2-a>0$.

So, we have $k<300$, but certainly $k>0$ because the left side must be positive.