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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 $a=3k^2\pm2k+1$.

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    $\begingroup$ The problem looks comparable in difficulty to the Erdos-Straus conjecture en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Straus_conjecture which remains open (as with this problem, one can handle various congruence classes by explicit algebraic solutions, but these do not cover all cases). $\endgroup$
    – Terry Tao
    Commented Jul 17, 2019 at 20:08
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    $\begingroup$ $9k^2=c(4ab-a-b)-b\overset{a\to u+v\\b\to u-v}{\implies}\\ (8 c u - 2 c - 1)^2 - (8 c v - 1)^2 - (2 c + 36 k^2 + 1)^2 + (36 k^2 + 1)^2 = 0 \implies\\(8 c u - 2 c - 1)^2 - (8 c v - 1)^2 - (2 c + 36 k^2 + 1)^2 + (36 k^2 - 1)^2 + (12 k)^2 = 0$ $\endgroup$
    – Dmitry Ezhov
    Commented May 13, 2020 at 7:18
  • $\begingroup$ @TerryTao, indeed, this $\ 4\!\cdot\! a\!\cdot\! b - a- b\ $ associates immediately with Erdős-Straus. $\endgroup$
    – Wlod AA
    Commented Jan 8, 2021 at 9:02

1 Answer 1

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https://math.stackexchange.com/questions/3214034/integer-solutions-to-a-two-sheeted-hyperboloid/3214271#3214271

$$z^2=axy+bx+cy+d$$ Use another equation. $$q=\frac{A^2-d}{b}$$

And we use solutions to the Pell equation. $k,t -$ any number.

$$p^2-akts^2=1$$

Decisions then write down so.

$$z=Ap^2-((aq+c)t+bk)ps+aAkts^2$$

$$x=qp^2-2kAps+(k((aq+c)t+bk)-aqkt)s^2$$

$$y=ts(((aq+c)t+bk)s-2Ap)$$

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