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I came across this post which says that multiplication of $4$ consecutive integers is $1$ less than a perfect square. And I played around with numbers and polynomials to see if this is only when we exactly take $4$ integers, to no avail.

I think it is then valid to ask if this is uniquely a property of $4$. I don't know how to approach this funny question. I appreciate any help and insights.

Math-fun
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    If it works for any $n$ consecutive numbers, it must work for both $n!$ and $(n+1)! ; ; , ; ;$ and this is rare. https://en.wikipedia.org/wiki/Brocard%27s_problem – Will Jagy Mar 31 '21 at 15:25
  • @WillJagy Thank you very much for the nice and useful information! – Math-fun Mar 31 '21 at 15:44
  • @WillJagy The Brocard's problem seems to be a bit more restricted compared to my question. In my question for example $10\times11\times12\times13+1=131^2$ and any other "$4$" consecutive numbers can be chosen. – Math-fun Apr 01 '21 at 12:55

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