I asked the following question here, but I did not get a full answer, so I put it here that may be some help.

Let $n$ be a positive integer. The Diophantine equation $$ n^2=c(4ab-a-b)-b,\qquad (a,b,c\in\mathbb{Z}^+) $$ is solvable for $n\equiv\pm1\pmod3$, but I stuck for $n\equiv0\pmod3$.

Is there any method to solve it?

thanks!

P.S. The method I used for the cases $n\equiv\pm1\pmod3$ is as follows: $$ n^2+b=c(4ab-a-b) $$ Assume for a moment the left-hand side is prime. Since $4ab-a-b>1$ for all $a,b>0$, therefore $c=1$. Now let $n=3k\pm1$ then $$ n^2=9k^2\pm6k+1=3(3k^2\pm2k+1)-2=(4ab-a-b)-b=(4b-1)a-2b $$ If we let $b=1$, we have $a=3k^2\pm2k+1$.