Integral points on a particular family of curves This is a follow-up to this question (and comments thereon). Namely, it follows from Felipe Voloch's comment that for any $n>2$ there is a finite set of integral $(x, y),$ such that
$$
\prod_{i=1}^n(x+i) = n y^2.$$
The question is: is there any $n$ for which the explicit list is known (and if so, how it was generated)? I would assume that $n=3, 4$ should be tractable...
ADDENDUM In joro's answer he points out that for $n=3$ there are no nontrivial solutions. For $n=4,$ sage computes the Weierstrass form of the curve as
$$y^2 = x^3 - 208/3*x + 4480/27.$$
It then finds that the rank of the MW group is $0,$ and that the torsion subgroup is of the form $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/4\mathbb{Z}.$ The generators are claimed to be $(-4/3: 16:1)$ and $(8/3:0:1).$ I assume this is for the transformed $x, y,$ so I am not sure what this means for the integral solutions - I am sure someone here does, though...
 A: Erdos and Selfridge have proved that the product of two or more consecutive non-zero integers is never a power (Illinois J. Math, vol. 19, no. 2, 1975). This implies in particular that for $n > 1$ a perfect square, the only integer solutions are the ones given by $y =0$ and $x \in \{-1,\ldots,-n\}$. The proof of the Erdos-Selfridge theorem is elementary and very clever, based on Sylvester's theorem: For $n > k$, the product $(n+1) \cdots (n+k)$ has at least one prime factor exceeding $k$. It is possible that the method might extend to certain other values of $n$. A related but simpler problem was the determination of all perfect power binomial coefficients, solved similarly by Erdos in 1951 and included in the collection Proofs from the Book.
In the general case it may be asking too much to have a complete description of the triples $(x,y,n)$. But for any hyperelliptic equation $dy^2 = f(x)$, the determination of all integer solutions $x,y \in \mathbb{Z}$ is at least in theory possible by Alan Baker's estimates for logarithmic linear forms, and a finite search. Siegel had explained how to reduce the hyperelliptic to a finite set of two-variable $S$-unit equations, see section 5.3.4 from Bombieri and Gubler (Heights in Diophantine Geometry) for the structure of this argument. Then Baker's theorem applies straightforwardly. This allows in principle allowing to determine all solutions $x,y \in \mathbb{Z}$ for any given value $n$.
A: The $n = 5$ case gives the hyperelliptic curve $C : y^{2} = 5x^{5} - 25x^{3} + 20x$. The Jacobian $J$ has $J(\mathbb{Q}) \cong \mathbb{Z} \times (\mathbb{Z}/2\mathbb{Z})^{4}$ and a generator of the infinite part is $\infty - (4,60)$. Using this, the method of Chabauty proves that the only rational points on $C' : y^{2} = 5(x+1)(x+2)(x+3)(x+4)(x+5)$ are the point at infinity, those with $y = 0$ and also $x = 1$, $y = \pm 60$.
The same sort of argument works for the $n = 6$ case.
A: The case $n=3$ leads to the elliptic curve in Weierstrass form $u^3  - 9u = v^2$.
The map is $(x,y) =( u/3 - 2, v/9) $.
It has finite number of rational points and none of them leads to solution other than $y=0$.
One way to find the Weirstrass form is in Maple:
with(algcurves);Weierstrassform( (x+1)*(x+2)*(x+3)-3*y^2,x,y,u,v);


Added Extending Igor Rivin's answer about $n=4$.
The Weierstrass form is rank zero, so we must map the torsion points
to the original curve and check for integrality.
The $x$ coordinate on the original curve is $-1/2*(-24*u+208)/(-3*u+44)$
A: The following idea may work for some not too large fixed $n.$ E.g. let $n=7.$ The equation has the following form $$(x+4)(x^2+8x+7)(x^2+8x+12)(x^2+8x+15)=7y^2.$$ It remains to check the possible $\gcd$ values. As far as I see one obtains that 
\begin{eqnarray*}
(x+4)(x^2+8x+7)&=&7\delta\square,\\
(x^2+8x+12)(x^2+8x+15)&=&\delta\square,
\end{eqnarray*}
where $\delta\mid 160$ or
\begin{eqnarray*}
(x+4)(x^2+8x+7)&=&\delta\square,\\
(x^2+8x+12)(x^2+8x+15)&=&7\delta\square,
\end{eqnarray*}
where $\delta\mid 160.$
That is the problem is reduced to determine integral points on certain elliptic curves.
As Elkies already noted the case $n=8$ leads to elliptic curves. One obtains that $$8(X+8)(X+14)(X+18)(X+20)=y^2,$$ where $X=x^2+9x.$ Using the MAGMA procedure 
IntegralQuarticPoints([8,480,10464,97408,322560],[-8,0]);
the list of integral points is as follows:
[
    [ -26, 288 ],
    [ -38, -1440 ],
    [ -14, 0 ],
    [ -20,0],[ -8,0],                                                                         [ -18, 0],                                                                             [ -16, -32]]. It remains to solve the equations $x^2+9x=x_0,$ where $x_0$ is an $x$-coordinate of an integral point. There are only trivial solutions, that is $x\in\{-8,-7,\ldots,-1\}.$
