I'm having trouble proving that $2^n = (n+1)(n+2)(n+3)$ has no solution when $n>0$. I tried showing there is only one critical point for $(n+1)(n+2)(n+3) - 2^n$. But, I couldn't do it. Can anyone help?

  • 4
    $\begingroup$ What do you mean by 'solution'? Do you mean a solution in the integers, or in the real numbers? $\endgroup$
    – Toby Mak
    Apr 1, 2018 at 7:28
  • $\begingroup$ $2^n$ will outstrip any polynomial eventually. Try to find a value of $n$ with $2^n>(n+1)(n+2)(n+3)$ $\endgroup$
    – Empy2
    Apr 1, 2018 at 7:34
  • $\begingroup$ It does have a solution somewhere in $(11,12)$. Not an integer solution, but then you never mentioned that. $\endgroup$
    – dxiv
    Apr 1, 2018 at 7:42
  • $\begingroup$ By 'critical point', do you really mean positive real number solution? Please state clearly in your question. $\endgroup$
    – Tony Ma
    Apr 1, 2018 at 8:16

1 Answer 1


$2^n$ is not a multiple of $3$ and $(n+1)(n+2)(n+3)$ is a multiple of $3$. Thus given equation has no solution.

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    $\begingroup$ You seem to be assuming that $n$ is an integer, which the question doesn't state. Otherwise it's enough to note that $2^n=k(n+1)(n+2)$ has no integer solutions, since $2^n$ has no odd factors. $\endgroup$
    – dxiv
    Apr 1, 2018 at 7:32
  • 2
    $\begingroup$ @dxiv You're right, because OP used the alphabet $n$ and the category "number-theory". So I will delete the answer if $n$ is not necessarily an integer, though losing 50 points is so painful. $\endgroup$ Apr 1, 2018 at 7:35
  • $\begingroup$ Make that $60$ :) $\endgroup$
    – Mr Pie
    Apr 1, 2018 at 7:39

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