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I am no mathematician but have studied mathematics some 20 years ago. So I know basics of number theory but have lost the skills to solve problems.

I was wondering if the equation $3ax^2 + (3a^2+6ac)x-c^3=0$ in which $a$ and $c$ are arbitrary positive integers and $c$ is divisible by 6 has an integer solution.

So we need to show there is either no $a$, $c$ and $x$ all positive that satisfy the equation, or one counter example exist.

Rational root theorem didn't help me.

Thanks.

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  • $\begingroup$ If the polynomial has integer roots, then its discriminant must be a perfect square: $3 (3 a^4 + 12 a^3 c + 12 a^2 c^2 + 4 a c^3)$. It doesn't look like it is always a perfect square. If we set $c=6d$, then the discriminant is $9 a (a^3 + 24 a^2 d + 144 a d^2 + 288 d^3)$ which doesn't like any better. $\endgroup$ – lhf Apr 30 '20 at 21:45
  • $\begingroup$ Sorry, I mean $a$ and $c$ are positive. $\endgroup$ – user781723 Apr 30 '20 at 21:55
  • $\begingroup$ If $a=1, c=6$ the quadratic has complex roots. $\endgroup$ – saulspatz Apr 30 '20 at 22:10
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    $\begingroup$ Are there ever integer solutions (other than $a=0, c=0$)? $\endgroup$ – Robert Israel Apr 30 '20 at 23:12
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Not true. Take $a=13$ and $c=-6$. Then the equation reduces to $13 x^2 + 13 x + 72 = 0$, which does not have an integer root. It does not even have real roots.

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  • $\begingroup$ @saulspatz, now they are. $\endgroup$ – lhf Apr 30 '20 at 21:53
  • $\begingroup$ If it doesn't have an integer solution for a set of parameters can we say it doesn't have at all? Here $a$ and $c$ can be any positive integer and $c$ is divisible by 6. $\endgroup$ – user781723 Apr 30 '20 at 21:53
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    $\begingroup$ @YetiYota, I suggest you clarify your question then. $\endgroup$ – lhf Apr 30 '20 at 21:55
  • $\begingroup$ Sure. I corrected the question. $\endgroup$ – user781723 Apr 30 '20 at 22:02
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The main condition is that $c^3$ be divisible by $a.$ If, for instance, $a = p$ is chosen to be prime, the requirement for any possibility of a solution is that $p | c.$ For that matter, if $a$ is squarefree, it is still necessary to have $a|c$ in order to have an integer solution.

Let's see, $3ax^2 + (3a^2+6ac)x-c^3=0$ with $6|c,$ so we take $$ c = 6 w $$ for $$ 3ax^2 + (3a^2 + 36 aw)x = 216 w^3 \; . $$ $$ ax^2 + (a^2 + 12aw)x = 72 w^3. $$ $$ a (x^2 + a + 12wx) = 72 w^3. $$

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  • $\begingroup$ Unfortunately I do not see how it can be proved by this. Can you elaborate a bit? $\endgroup$ – user781723 May 1 '20 at 18:32
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Writing $x=aX, c=aY$, the equation becomes (if $a \ne 0$) the elliptic curve $-Y^3 + 3 X^2 + 6 X Y + 3 X = 0$, and we're looking for rational solutions $(X,Y)$. This has Weierstrass form $-s^3 + 27/4 + t^2 = 0$, with $X = -s/3 + 1/2 - t/9$ and $Y = s/3 - 1$. Sage tells me this has rank $0$ with torsion elements $(s,t) = (3,-9/2)$ and $(3,9/2)$, corresponding to $(X,Y) = (0,0)$ and $(-1,0)$. Thus, if I'm interpreting Sage correctly, the only integer solutions are the trivial ones with $c=0$ and either $a=0$ or $x=0$ or $x=-a$.

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  • $\begingroup$ Well advanced and out of my league :). Is there any layman's explanation for this? $\endgroup$ – user781723 May 1 '20 at 18:31
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$3 a x^2 + (3 a^2 + 6 a c) x - c^3 = 0\overset{c\to 6d}{\implies}72 d^3 - a^2 x - 12 a d x - a x^2 = 0$

$a\mid 72d^3\implies$ let $72d^3 = ab$

Apply Resultant over $a$:

$Res_a(72 d^3 - a^2 x - 12 a d x - a x^2, 72 d^3 - a b)=72 d^3 (b^2 - 12 b d x - 72 d^3 x - b x^2)$,

i.e. $b^2 - 12 b d x - 72 d^3 x - b x^2 = 0$ and $x\mid b^2$.

Let $b^2 = xy\implies$

$Res_x(b^2 - 12 b d x - 72 d^3 x - b x^2, b^2 - x y) = -b^2 (b^3 + 12 b d y + 72 d^3 y - y^2)\implies$

$b^3 + 12 b d y + 72 d^3 y - y^2 = 0 \;\;\;\;\;\;\;\;(1)$

Let $b^3 = yz\implies$

$Res_y(b^3 + 12 b d y + 72 d^3 y - y^2, b^3 - y z) = -b^3 (b^3 - 12 b d z - 72 d^3 z - z^2)\implies$

$b^3 - 12 b d z - 72 d^3 z - z^2 = 0$

Let $b^3 = zt\implies$

$Res_z(b^3 - 12 b d z - 72 d^3 z - z^2, b^3 - z t) = -b^3 (b^3 + 12 b d t + 72 d^3 t - t^2)\implies$

$b^3 + 12 b d t + 72 d^3 t - t^2 = 0 \;\;\;\;\;\;\;\;(2)$

$(1)$ like $(2)$, then we have infinite descent and source equation have no solutions.

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  • $\begingroup$ This is a quite smart move. It took me some time to understand Resultant and I am not sure I understand it correctly, but it seems that if I study it I should be able to fully understand your solution. Can this be expressed in a simpler way? $\endgroup$ – user781723 May 1 '20 at 18:09
  • $\begingroup$ Dmitry! Can you still help here? I think your solution does not work. Having $b^3 =yz$ and $b^3 =zt$, there is $y=t$ and thus (1) and (2) are exactly the same. So it is not infinite descent. $\endgroup$ – user781723 May 2 '20 at 21:28
  • $\begingroup$ Yes, here me mistake, it is not infinite descent. Sorry! $\endgroup$ – Dmitry Ezhov May 3 '20 at 4:56
  • $\begingroup$ Other idea, consider Weierstrass form $(12 a c)^3 + 36 a^2 (12 a c)^2 + 432 a^4 (12 a c) + 1296 a^6 = (36 a^2 (2 x + a + 2 c))^2$ $\endgroup$ – Dmitry Ezhov May 3 '20 at 6:45

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