What are the pairs $(A,N)$ where $A,N$ are integers such that the following equation is satisfied:
$\large A=\frac{-6+\sqrt{144-12N^2}}{6}$
I know that we should have:
$k^2=144-12N^2$ for some integer $k$.
How do I proceed from here? Is there any other way?
First of all, note that $$6|\sqrt{144-12N^2}\Rightarrow 144-12N^2=36k^2$$ for some integer $k$ . Then it follows $$N^2=3(4-k^2)\Rightarrow N=3m$$ for some integer $m$. So ultimately we get $$4=3m^2+k^2$$ which implies the only solutions as $ k={}_{-}^{+}1,\ m=0,{}_{-}^{+}1 $. So $$N={}_{-}^+3,0$$ and consequently $$A= \left\{ \begin{array}{lr} 0,-2 & \mbox{if} N={}_{-}^+3\\ 1,-3 & \mbox{if} N=0 \end{array} \right. $$