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 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