# Integer solution to $3ax^2 + (3a^2+6ac)x-c^3=0$

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.

• 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. – lhf Apr 30 '20 at 21:45
• Sorry, I mean $a$ and $c$ are positive. – user781723 Apr 30 '20 at 21:55
• If $a=1, c=6$ the quadratic has complex roots. – saulspatz Apr 30 '20 at 22:10
• Are there ever integer solutions (other than $a=0, c=0$)? – Robert Israel Apr 30 '20 at 23:12

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.

• @saulspatz, now they are. – lhf Apr 30 '20 at 21:53
• 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. – user781723 Apr 30 '20 at 21:53
• @YetiYota, I suggest you clarify your question then. – lhf Apr 30 '20 at 21:55
• Sure. I corrected the question. – user781723 Apr 30 '20 at 22:02

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 $$3ax^2 + (3a^2 + 36 aw)x = 216 w^3 \; .$$ $$ax^2 + (a^2 + 12aw)x = 72 w^3.$$ $$a (x^2 + a + 12wx) = 72 w^3.$$

• Unfortunately I do not see how it can be proved by this. Can you elaborate a bit? – user781723 May 1 '20 at 18:32

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 $$x=-a$$.

• Well advanced and out of my league :). Is there any layman's explanation for this? – user781723 May 1 '20 at 18:31

$$3 a x^2 + (3 a^2 + 6 a c) x - c^3 = 0\overset{c\to 6d}{\implies}72 d^3 - a^2 x - 12 a d x - a x^2 = 0$$

$$a\mid 72d^3\implies$$ 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.

• 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? – user781723 May 1 '20 at 18:09
• 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. – user781723 May 2 '20 at 21:28
• Yes, here me mistake, it is not infinite descent. Sorry! – Dmitry Ezhov May 3 '20 at 4:56
• Other idea, consider Weierstrass form $(12 a c)^3 + 36 a^2 (12 a c)^2 + 432 a^4 (12 a c) + 1296 a^6 = (36 a^2 (2 x + a + 2 c))^2$ – Dmitry Ezhov May 3 '20 at 6:45