Trying to prove $c^3a^2+(9c^2-b^2)a+(27c-10b)=0$ has no positive integer solutions

Question

I'm trying to prove (or, I suppose, disprove) the following claim, in either version.

Conjecture (Strong Version): There are no positive integers $a,b,c$ such that $$c^3a^2+(9c^2-b^2)a+(27c-10b)=0.$$

Conjecture (Weak Version): There are no positive integers $a,b,c$, with $b$ odd, $\gcd(b,c)=1$, and $ab \equiv ac \equiv 0\!\pmod{3}$ such that $$c^3a^2+(9c^2-b^2)a+(27c-10b)=0.$$

I believe the Strong Version (and, hence, the Weak Version) to be true for the following reasons:

1. It is derived from another Diophantine problem, the solutions of which suggest the equation should have no positive integer solutions.

2. I have been doing increasingly large brute-force computer verification: no solutions found with $1 \le a,b,c \le 500$.

A proof by infinite descent would be my preference — so I've been trying to use Vieta jumping, but can't seem to get the proof across the goal line. But any elementary proof would suffice for my purposes.

Any help or hints would be appreciated.

EDIT: Solving quadratically in $a$ gives $$ a = \frac{b^2-9c^2 \pm \sqrt{b^4-18b^2c^2+40bc^3-27c^4}}{2c^3}. \tag{$\star$} $$ A brute-force search (admittedly low-limit) finds no solution to the radical except $(b,c)=(27k,10k)$ for integer $k=1,2,\dots$.

From my Vieta-jumping work, I can prove that $$ a > \frac{b^2-9c^2}{c^3}. $$ Therefore, only the $+$ sign in ($\star$) is valid. Is there anything I can do with that to unearth a contradiction?

Answer 1

If $x = ab+5$ and $y =3+ac$, this becomes $$ x^2 + 2 =y^3 $$ By Siegel's theorem, this has only finitely many integer solutions. I wouldn't be surprised if explicit bounds were available, although I don't know them. I suspect the only integer solutions are $y = 3$, $x = \pm 5$, corresponding to $a=0$, or $b=c=0$, or $ab=-10$, $c=0$. Thus the original equation should have no positive integer solutions.

In fact, it is believed that the only positive integer solution to $x^p + 2 = y^q$ with $p,q>1$ is $5^2 + 2 = 3^3$: see OEIS sequence A076427.

EDIT: At http://tnt.math.se.tmu.ac.jp/simath/MORDELL/MORDELL- we find:

E_-00002: r = 1   t = 1   #III =  1
      E(Q) = <(3, 5)>
      R =   1.3495768357
       2 integral points
        1. (3, 5) = 1 * (3, 5)
        2. (3, -5) = -(3, 5)

` which, if I understand it correctly, confirms that $y=3$, $x = \pm 5$ are the only integer solutions.