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