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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.

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1  
I think you should write $\prod_{k\in S} k=2m^2$. –  Luiz Cordeiro Jan 1 at 20:07
    
Where did you find this? (Given the proof in Gerry's link, this problem seems to be really hard) –  chubakueno Jan 1 at 23:13
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I'm still trying to puzzle out that P.S. –  Byron Schmuland Jan 1 at 23:32

2 Answers 2

up vote 13 down vote accepted

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

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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 $n \le 200$.

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This is nice. I gave a bad reason for why there should be fewer than $2n$ prime factors amongst the integers in the interval. The requirement that $[1,0,\ldots,0]$ be in the span unfortunately makes it so that knowing (or estimating) the dimension is not enough. –  Hugh Denoncourt Jan 2 at 7:11

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