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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...

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  • $\begingroup$ Regarding the addendum: For $n = 4$, there are no non-trivial ($y \neq 0$) integer solutions, by the Erdos-Selfridge result that I quoted. $\endgroup$
    – Vesselin Dimitrov
    Commented Jul 21, 2015 at 18:19
  • $\begingroup$ @VesselinDimitrov Oops, forgot that $4$ was a square :), thanks! But still, can you parse what the sage result actually means? $\endgroup$
    – Igor Rivin
    Commented Jul 21, 2015 at 18:24
  • $\begingroup$ I do not know if I can read the conclusion of no other integral solutions without actually computing all seven non-trivial points of the transformed equation, as well as the change of coordinates map. In any case, we know that four of these seven non-identity rational points were given as $y = 0$ and $x \in \{-1,\ldots,-4\}$ in the original coordinates. $\endgroup$
    – Vesselin Dimitrov
    Commented Jul 21, 2015 at 18:44
  • $\begingroup$ @VesselinDimitrov Yes, exactly. But what about those four other mysterious points. $\endgroup$
    – Igor Rivin
    Commented Jul 21, 2015 at 18:45
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    $\begingroup$ For a hyperelliptic curve $y^{2} = f(x)$, where $f(x)$ has even degree, there are two points at infinity on the non-singular model. In addition, there are the points $(-5/2,3/2)$ and $(-5/2,-3/2)$. This explains the four points other than those with $y = 0$. $\endgroup$
    – Jeremy Rouse
    Commented Jul 21, 2015 at 20:28

4 Answers 4

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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 $n$.

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    $\begingroup$ Actually, Erd\H{o}s and Selfridge proved something stronger, that implies that the equation $$\prod_{i=1}^n (x+i)=b y^2$$ has no solutions with $n \geq 4$, $x > n^2$ and $b$ an integer with greatest prime factor $< n$. This was extended by Saradha and Shorey to $n=3$ and to the greatest prime factor of $b$ being $\leq n$. In the $n=3$ case, one gets a single solution corresponding to $(x,b,y)=(47,6,140)$. $\endgroup$
    – Mike Bennett
    Commented Jul 22, 2015 at 2:01
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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.

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  • $\begingroup$ Ah, so this is actually the first nontrivial example! $\endgroup$
    – Igor Rivin
    Commented Jul 21, 2015 at 21:26
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    $\begingroup$ Maybe easier for $n=6$, because there the curve is bielliptic [involution $(x,y) \leftrightarrow (7-x,y)$] and it's easier to compute ranks of two elliptic curves than one genus-2 Jacobian. There's also an elliptic quotient for $n=8$. $\endgroup$
    – Noam D. Elkies
    Commented Jul 22, 2015 at 8:33
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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 $-1/2*(-24*u+208)/(-3*u+44)$

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  • $\begingroup$ How do you find the rational points? Does Maple do that too? $\endgroup$
    – Igor Rivin
    Commented Jul 21, 2015 at 15:42
  • $\begingroup$ @IgorRivin Don't know how to to do this in Maple, but it is possible in sage. If the rank is zero, all rational points are given by: E.torsion_points() $\endgroup$
    – joro
    Commented Jul 22, 2015 at 6:00
  • $\begingroup$ Thanks! I figured it out in sage, but how did you get the transform to the Weierstrass form? From your typesetting, I assume that this is something Sage did, but how? $\endgroup$
    – Igor Rivin
    Commented Jul 22, 2015 at 13:59
  • $\begingroup$ @IgorRivin maybe sage can did the first form, but the second Weierstrass form was from Maple. I have sage code quartic with rational point to Weierstrass form, but didn't use it. $\endgroup$
    – joro
    Commented Jul 22, 2015 at 14:21
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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 $x\in\{-8,-7,\ldots,-1\}.$

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  • $\begingroup$ That's a nice idea, and the computation you mention give a boost to the idea that there is a high likelihood of only trivial solutions. $\endgroup$
    – Igor Rivin
    Commented Aug 3, 2015 at 11:18

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