Let $V = \Pi_{1 \le i < j \le n} (a_j  a_i)$ be the determinant of the Vandermonde matrix where $1 = a_1 < \cdots < a_n = d$ (with $d >> n$). What is the smallest prime $p$ (or the lower bound) such that $p \nmid V$? Preferably $p < n$.

closed as not a real question by Gjergji Zaimi, Andres Caicedo, Mark Sapir, Felipe Voloch, Qiaochu Yuan Jan 6 '11 at 12:33
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Not really clear about what is being asked. If the $a_i$ are all divisible by the same p (choose one) then this p does divide V. Suppose the $a_i$ are 1 ... n, then if $p < n$ then then with $a_i=1$ there is an $a_j$ with $a_ja_i=p$. If $p \ge n$ then $p \nmid V$. If $p < n$ then in any n numbers there are two with the same residue mod p (pigeonhole) so $p \mid V$. Is a more general context intended? 


To extend Mark Bennet's answer, one could have a_2 = a_1 + P_m, the mth primorial, giving that V is a multiple of P_m. So without parameters, there is no bound. If you want something in terms of V or the a_i, you might start with the idea that such a prime need be not much larger than the largest of (a_i  a_j), and is likely to be smaller. Gerhard "Ask Me About System Design" Paseman, 2011.01.05 


If $p<n$ then it must be that $p \mid V$. However if $p \ge n$ then it can be arranged that $p \nmid V$. If you set $a_i=2^m(i1)+1$ then no prime greater than $n1$ divides $V$. You could replace $2^m$ by $(n1)!$ or anything else with all divisors less than $n$. 


Considered as a polynomial in the $a_i$'s, $V$ is never divisible by p, since the monomial $a_1^{n1}a_2^{n2}\cdots a_{n1}$ always appears with coefficient 1. However, by the magic of Fermat's little theorem, it can be that all of its values are divisible by p, even if the polynomial itself isn't. As Mark points out, this happens if and only if $p< n$. 

