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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.

Anonymous - a group
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    This would imply that either x or $x\pm1$ is a multiple of $9$. – Lucian Dec 12 '13 at 16:25
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    This is the same as asking whether $$\exists x,y\in\Bbb Z^+,\ 6{x\choose 3}=18{y\choose 3}\\implies{x\choose 3}=3{y\choose 3}$$ – abiessu Dec 12 '13 at 16:33
  • I don't know if it helps but $y=5k,5k+3,5k+4$. Check the last digit of $x(x+1)(x+2)$. – hhsaffar Dec 12 '13 at 20:51
  • A related question – 2'5 9'2 Dec 15 '13 at 19:27
  • In the comments of the question I linked to, I lay out an argument why such a thing would be extremely rare. Here the corresponding argument would be based on having to have two different sets of three consecutive integers with very similar prime factorizations integer by integer. – 2'5 9'2 Dec 15 '13 at 19:40
  • WolframAlpha shows a nice plot with interesting stuff happening for negative numbers, a set of integer solutions with no both positive values, and $y$ in terms of $x$ for three roots. – Mark Hurd Dec 17 '13 at 00:14

3 Answers3

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The equation $x(x+1)(x+2) = 3y(y+1)(y+2)$ is equivalent to $\left(\frac{24}{3y-x+2}\right)^2 = \left(\frac{3y-9x-6}{3y-x+2}\right)^3-27\left(\frac{3y-9x-6}{3y-x+2}\right)+90$.

This is an elliptic curve of conductor $3888$. Cremona's table says its group of rational points is of rank $2$, and is generated by the obvious solutions $(x,y) \in \{-2;-1;0\}^2$

I am not sure how one would go about proving an upper bound for the integral solutions of the original equation. There are papers on the subject (for example, Stroeker and de Weger's "Solving elliptic diophantine equations: The general cubic case." seems to be applicable here)

Also, see How to compute rational or integer points on elliptic curves

Community
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mercio
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It seems unlikely. The search you have done seems to get us out of the law of small numbers. Somebody's law says that when the reciprocal powers in an equation like this sum to less than $1$ you should expect finitely many solutions. Here it is $\frac 23$. The equation can be written $(x+1)((x+1)^2-1)=3(y+1)((y+1)^2-1)$ so you need $\frac {x+1}{y+1}$ to be very close to $\sqrt[3]3$ You could use the continued fraction to find convergents

Ross Millikan
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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.

Yong Hao Ng
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    Can you get a bound on the height precise enough to prove that you have found all the integer solutions? – David E Speyer Dec 20 '13 at 16:39
  • @DavidSpeyer Let $(x,y)=R=[a]P\oplus [b]Q\not\in {P,Q}, a,b\geq 0$. I considered $R\oplus P=(f(x,y)/d(x,y), g(x,y)/d(x,y))$ and saw that numerator of $Y$-coordinates increases. Similarly for $R\oplus Q$. Hence by induction the numerator of $Y$-coordinates is increasing and cannot correspond to an integral solution in the original question (since that requires numerator to be 24). It suffices to test $[a]P\oplus [b]Q, (a,b)\in [0,4]\times[0,4]$, from there on the numerator exceeds $24$. It is a bit tricky to bound the cancellation and I need to check for mistakes before I update my solution. – Yong Hao Ng Dec 21 '13 at 11:14
  • For example: Let $R=(x,y)\in E(\mathbb Q)$. Then $(x,y)=(m/e^2,n/e^3)$ for some $(m,e)=(n,e)=1$. From my workings for $R\oplus P$, the cancellation in $g(x,y)/d(x,y)$ after clearing denominators is bounded by $8 (5 e^2 + m)^2$. Then it is possible to show that the numerator is greater than before (which is $n$). – Yong Hao Ng Dec 21 '13 at 11:28
  • Mistake: $8(5e^2+m)^3$. – Yong Hao Ng Dec 21 '13 at 11:53