From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: Perfect squares among the binomial coefficients
Date: 10 Dec 1999 07:52:32 GMT
Newsgroups: sci.math

In article <82846e$2gc$1@wanadoo.fr>,
Jean-Pierre.EHRMANN <Jean-Pierre.EHRMANN@wanadoo.fr> wrote:
>It is easy to find that C(n, 2) is a perfect square iff n is one of the u[k]
>where
> u[0] = 1; u[1] = 2; u[k + 1] = 6 u[k] - u[k-1] -2.
>Is it possible to find all the perfect squares among the
>C(n, k) with 3 <= k <= n/2 ? - as, for exemple, C(50, 3) -

For fixed  k, the equation  y^2 = C(n,k)  describes a curve on which you
seek integer points. Of course  C(n,0)  is a square for all n;  C(n,1)  is
a square precisely when  n  is.  C(n,2)  is square precisely when 
(2n-1) ^2 - 8 y^2 = 1  for some  n; there are infinitely many such  n
but they grow exponentially. 

For  k = 3  and  4  the curve is elliptic and so can have only finitely many
integer points. Simath reports that the only integer solutions when k=3
are n=0,1,2,3,4,50 . I didn't go through the labor of computing the
integral points when  k=4  but a quick search found none up to 10000 except
for n=0,1,2,3,4 (of course) so I would imagine one could easily prove
there are no more. There are infinitely many rational points however -- in
both cases k=3 and k=4, since these turn out to be isogenous curves.

For k > 4, the curve  y^2 = C(n,k)  has genus greater than  1, and so there
there are only finitely many rational solutions, and so a fortiori only
finitely many integer ones. Conceivably, of course, we could find finitely
many for _each_ k, and thus infinitely many. But in fact a quick search 
found none at all with  n, k less than  10000, for example. There may
well be none at all, but I don't know how you would directly prove that
for _all_  k  and  m.

However, there is a very useful result which applies here, proved by 
Erdos and Selfridge: "The product of consecutive integers is never a power",
Illinois J. Math. 19 (1975), 292--301. MR 51 #12692 . They prove, and
I quote from the review,

   Let $k,l,n$ be
   integers such that $k\geq 3$, $l\geq 2$ and $n+k\geq p\sp {(k)}$,
   where $p\sp {(k)}$ is the least prime $p$ satisfying $p\geq k$. Then
   there is a prime $p\geq k$ for which the power of $p$ dividing
   $(n+1)\cdots(n+k)$ is not divisible by $l$.

Take  l = 2; then there is a prime appearing with odd exponent in  C(n+k, k)
if  n+k  is sufficiently large, except perhaps if  k  is prime. Here 
n = k is already "sufficiently large".

So that about wraps up most of the possibilities. I haven't gone to look
at the article but I suppose it's possible you could adjust the argument to
cover the case  k  is prime, too. (Indeed, the review points to an
earlier papers of Erdos and Rigge which cover the case  l=2, which might
more easily be adapted to the question at hand.)

There are a few other papers of Erdos and Selfridge which study related
questions about the prime factorizations of products of consecutive integers.

dave