product is twice a square For every positive integer $n$, there exists a set $S\subset \{n^2+1,n^2+2,\dotsc,(n+1)^2-1\}$, such that 
$$\prod_{k\in S}k=2m^2$$
for some positive integer $m$
I have no clue about it. Could anyone help me? Thanks a lot.
p.s. Whether or not the problem was a conclusion of an paper is unknown.
 A: Consider (as Hugh Denoncourt did in his deleted answer) the vector space $V$ over ${\mathbb F}_2$ with basis corresponding to the prime factors $p_1, \ldots, p_N$ 
 of $n^2 + 1, \ldots, (n+1)^2 - 1$.  The number $N$ of such prime factors
 is OEIS sequence A143346.  It appears that $N$ is approximately $1.52 n$ for large $n$, and is less than $2n$ for $6 < n \le 10000$ (the maximum $n$ in the data file).  Presumably this is true for all $n > 6$.  Let $v(k) = [\alpha_1, \ldots, \alpha_N] \mod 2$ where $k = p_1^{\alpha_1} \ldots p_N^{\alpha_N}$.   Now the question is whether  $[1,0,\ldots,0]$ is in the linear span of $v(n^2+1), \ldots, v((n+1)^2-1)$.  In fact, it appears that this linear span is the whole vector 
space $V$.  That is true for at least all $n \le 200$.
A: Granville and Selfridge, Product of integers in an interval, modulo squares: "We prove a conjecture of Irving Kaplansky which asserts that between any pair of consecutive positive squares there is a set of distinct integers whose product is twice a square." 
The details are Electronic Journal of Combinatorics, Volume 8(1), 2001. 
Paper is available at http://www.dms.umontreal.ca/~andrew/PDF/selfridge.pdf
