TITLE: Is it possible that $(x+2)(x+1)x=3(y+2)(y+1)y$ for positive integers $x$ and $y$?
QUESTION [18 upvotes]: Let $x$ and $y$ be positive integers. Is it possible that $(x+2)(x+1)x=3(y+2)(y+1)y$?
I ran a program for $1\le{x,y}\le1\text{ }000\text{ }000$ and found no solution, so I believe there are none.

REPLY [7 votes]: Completing mercio's answer: There are no other integer solutions.  
Mercio's portion:
Let $(x,y)$ be an integer solution to $x(x+1)(x+2)=3y(y+1)(y+2)$.
We can rewrite it as
$$\left(\dfrac{24}{3y-x+2}\right)^2=\left(\dfrac{3y-9x-6}{3y-x+2}\right)^3-27\left(\dfrac{3y-9x-6}{3y-x+2}\right)+90$$
or simply
$$E:Y^2=X^3-27X+90.$$
$E$ is an Elliptic curve since $\Delta=-139968\neq 0$.
Since $x,y$ are integers, they give rise to a rational solution in $E(\mathbb Q)$, via
$$(x,y)\mapsto \left(\dfrac{3y-9x-6}{3y-x+2},\dfrac{24}{3y-x+2}\right)=(X,Y).$$
However this only makes sense if $3y-x+2\neq 0$. Suppose otherwise, $x=3y+2$, then
$$24^2(3y-x+2)=(3y-9x-6)^3-27(3y-9x-6)(3y-x+2)^2+90(3y-x+2)^3$$
$$0=3y-9x-6\Leftrightarrow 0=y-3x-2$$
Together with $x=3y+2$, we have
$$ y-9y-6-2=0\implies y=-1$$
So it fails when $(x,y) = (-1,-1)$.
Ignoring this, then each $(x,y)$ integral solution must give rise to a rational point $(X,Y)$ in $E$. 
We can use the following commands in Sage:
E = EllipticCurve([0,0,0,-27,90]);
E.rank()
E.gens()
Which will tell us that $E(\mathbb Q)$ is torsion free and of rank 2 and generated by $(-5, 10), (-3, 12)$.
Therefore all the rational points on $E$ can be written as:
$$P\in E(\mathbb Q)\Longleftrightarrow P = [a](-5,10)\oplus [b](-3,12)$$
for some $a,b\in\Bbb Z$.
Solving for $(x,y)$ using $E(\mathbb Q)$:
If any $P\in E(\mathbb Q)$ can be mapped from an integral solution $(x,y)$, then we have relations
$$\dfrac{3y-9x-6}{3y-x+2}=X,\dfrac{24}{3y-x+2}=Y,$$
and we can check validity by solving the simultaneous equations.
Note that for $Y$, the numerator will always be small since $x$ and $y$ are integers.
Therefore all we have to do is to generate all $(a,b)\in \mathbb Z^2$, then form $P=(X,Y)$ and check if it satisfies the simultaneous equations.  
The problem is $\{(a,b)\}$ is an infinite set.
However, a (Height) property of Elliptic curve tells us that:
(1) As we add more points, we expect $X$ and $Y$ to be not integers.
(2) Since they are not torsion points, we also expect numerator/denominator to grow exponentially.
(3) Numerator and denominator will have approximately same number of digits as they grow.  
Combining all (3), this says that gradually the numerator of $Y$ must increase.
Edit2: Not quite enough. This is true for almost all points but we need to show that we did not miss any. Temporary solution in comments, will add proper solution after checking my workings.
End_Edit_2
As an example, let $P=(-5,10)$ and consider $\varphi(P)=\{P,[2]P,\dots\}$.
We will get:
$$\left\{(-5,10),\left(\dfrac{394}{25}, \dfrac{-7478}{125} \right), \left(\dfrac{148795}{269361},\dfrac{1212735770}{139798359} \right), \left(\dfrac{25189696321}{5592048400},\dfrac{-3233187530793631}{418173379352000}\right),\left(\dfrac{41697179388698395}{10487471993072881}, \dfrac{7244632674771290918438950}{1074004796869053110612279}\right),\dots\right\}$$
We then observe that for our solution in $(x,y)$, we require $Y=24/(3y-x+2)$.
In particular, this says that the numerator must be rather small.
Therefore it suffices to check for $P=[a](-5,10)\oplus [b](-3,12)$ for small values of $(a,b)$.
(For example, say $-5\leq a,b\leq 5$.)  
We will then find that the only solutions are the trivial negative ones.