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A 1939 paper of Erdos (Note on Products of Consecutive Integers, J. London Math. Soc. 14 (1939), 194–198) shows that a product of consecutive positive integers cannot be a perfect square. He cites a 1917 paper by Narumi which proves that a product of at most 202 consecutive positive integers cannot be a perfect square. I cannot seem to easily find Narumi's paper.

Although this result is known, I am curious about self-contained elementary proofs of special cases.

It's not too difficult to come up with fairly quick proofs for two, three, four, five, or seven consecutive integers.

Is there a short self-contained elementary proof that the product of six consecutive positive integers cannot be a perfect square? Or is it perhaps fair to say that this is the first "tricky" case?

  • 0
    I do not have so much experience to prove things, but I would start by the following steps: $k(k + 1) = u$, so the square root of $u$ would be the square of $k$ multiplied by $k + 1$, and this makes us think of two consecutive numbers that satisfy that condition. Extend this for 6 numbers, and this seems to be hard to think in numbers that can satisfy that.2011-12-12
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    The square root of $k$ multiplied by the square root of $k + 1$ - sorry.2011-12-12
  • 0
    You're referring to "Narumi, S.: An extension of a theorem of Liouville’s. Tohoku Math. J. 11, 128–142 (1917)". I doubt if you'll find it online ; it should be available in your library.2011-12-12
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    @P23 The paper by Narumi is [available here](http://www.journalarchive.jst.go.jp/english/jnlabstract_en.php?cdjournal=tmj1911&cdvol=11&noissue=0&startpage=128).2011-12-12
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    Please unaccept my wrong answer, so that I can delete it.2015-12-01

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