Divisibility of a binomial coefficient by $p^2$ -- current status - MathOverflow most recent 30 from http://mathoverflow.net2013-06-19T06:38:03Zhttp://mathoverflow.net/feeds/question/45923http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/45923/divisibility-of-a-binomial-coefficient-by-p2-current-statusDivisibility of a binomial coefficient by $p^2$ -- current statussuVRit2010-11-13T14:40:47Z2012-07-28T08:37:50Z
<p>While skimming the book <em>Concrete Mathematics</em>, (edit: first edition) I came across the following problem, which is listed there as a Research Problem: (Chapter 5, Exercise 96)</p>
<blockquote>
<p>Is ${2n \choose n}$ divisible by the square of a prime for all $n > 4$.</p>
</blockquote>
<p>This problem looked to me much simpler than a divisibility problem that I found on MO <a href="http://mathoverflow.net/questions/11335/a-binomial-sum-is-divisible-by-p2" rel="nofollow">(look here)</a>, but then again, I guess in number theory, the simpler the problems looks, the harder it usually is!</p>
<p>The nice form of this problem has made me very curious to find out more about it. But because I do not have more than a fleeting acquaintance with number theory, I don't know what search keywords would be useful to gain more information about this problem. </p>
<p>Thus, could somebody please tell me more about this problem and its current status?</p>
http://mathoverflow.net/questions/45923/divisibility-of-a-binomial-coefficient-by-p2-current-status/45929#45929Answer by Mark Grant for Divisibility of a binomial coefficient by $p^2$ -- current statusMark Grant2010-11-13T15:19:25Z2012-07-28T08:37:50Z<p>This is/was known as the Erdős square-free conjecture, and seems to now be solved. See the bottom of <a href="http://en.wikipedia.org/wiki/Square-free_integer#Erd.C5.91s_Squarefree_Conjecture" rel="nofollow">this page</a>.</p>