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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?

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    $\begingroup$ As was pointed out by M.Bennett in this post, the radical is an integer for infinitely many solutions with $\gcd(b,c)=1$ such as $b,c = 5383,1710$, etc. $\endgroup$ – Tito Piezas III Mar 23 '16 at 2:20
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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.

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  • $\begingroup$ It's possible that this particular instance of Mordell's equation can be solved by elementary methods. I'd have a look at Uspensky & Heaslet, where a number of Mordell equations are solved by elementary means, or Mordell's book, Diophantine Equations. $\endgroup$ – Gerry Myerson Mar 22 '16 at 6:14
  • $\begingroup$ @GerryMyerson: That Mordell equation has no known elementary solution. $\endgroup$ – Kieren MacMillan Mar 22 '16 at 13:20
  • $\begingroup$ Since I'm searching for an elementary solution, returning to a Mordell equation without a [known] elementary solution isn't really an option. But thanks (I upvoted the answer)! $\endgroup$ – Kieren MacMillan Mar 22 '16 at 13:21
  • $\begingroup$ The title of your question, Kieren, is "Trying to prove ... has no positive integer solutions." The body begins, "I'm trying to prove ..." and then gives two conjectures. Robert has settled the conjectures, and thus he has answered the question. Not only that, he has related your equation to a much simpler and well-studied one of which you were unaware (I'm assuming you didn't know about the reduction to Mordell, since, if you had, you would have been morally obligated to say so in the body of the question). The upvote is a nice gesture, but I think acceptance is called for. $\endgroup$ – Gerry Myerson Mar 22 '16 at 22:09
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    $\begingroup$ Anyone stopping by here might be interested in an earlier discussion of $x^2+2=y^3$ on MathOverflow, mathoverflow.net/questions/142220/fermats-proof-for-x3-y2-2 $\endgroup$ – Gerry Myerson Mar 22 '16 at 22:15

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