Prove, that the product of $3$, and $4$ following natural numbers can never be a number with the form of $x^k$ , where $x$ and $k$ are natural numbers, and $k>1$ (for example $9$ has this form, because it's $3^2$ , but $10$ doesn't have this form).

I tried to write them down as $n-1$ , $n$ , $n+1$ , so we get $(n^2-n)(n+1)$, which is $(n^3-n^2+n^2-n)=(n^3-n)$ , but can't really go further with this, same, if we have $4$ numbers.

  • 1
    $\begingroup$ You should at least require that $k>1,$ otherwise your statement is trivially false. $\endgroup$ Mar 4, 2015 at 11:52
  • $\begingroup$ Sorry, that is true. :) Yeah, k>1 $\endgroup$
    – Atvin
    Mar 4, 2015 at 11:53
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    $\begingroup$ see this question and a comment by a user marty cohen. $\endgroup$
    – mathlove
    Mar 4, 2015 at 11:54
  • $\begingroup$ I remember seeing quite some time one (not easy) proof that no product of consecutive integers can be a perfect power. $\endgroup$
    – Piquito
    Jun 25, 2016 at 2:27


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