Find $a$ for $\sqrt{\frac {9a + 4}{a - 6}} = n$, with $a \in \Bbb Z$ and $n \in \Bbb Q$.

Question

The problem:

Find all values of $a$ such that $\sqrt{\frac {9a + 4}{a - 6}} = n$, with $a \in \Bbb Z$ and $n \in \Bbb Q$.

What i tried:

I arrived to this: $$ n^2 = \frac {58}{a - 6} + 9 $$

i tried setting some form of boundaries for $n$, and then brute-force my way to the solutions, but i tried some of the $n$ values and i didn't find anything. After that, i went on and tested most $a$'s for which $a - 6$ and $9a + 4$ are perfect squares, and i saw that $n \to \sqrt {9}$, but for that value of $n$, the equation has no value.

I don't actually know whether this has a solution or not, if it doesn't, then can i get a proof of it?

Answer 1

There are two cases.

1. $9a+4=km^2$ and $a-6=kn^2$, where $k$, $m$ and $n$ are natural numbers;

2)$9a+4=-km^2$ and $a-6=-kn^2$, where $k$, $m$ and $n$ are natural numbers.

In the first case we obtain: $$k(m-3n)(m+3n)=58,$$ which gives not so many cases.

Can you end it now?

For example, the second case for $k=2$ gives $a=-44.$