# how to prove that $2^n = (n+1)(n+2)(n+3)$ has no solution when $n>0$?

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?

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

$2^n$ is not a multiple of $3$ and $(n+1)(n+2)(n+3)$ is a multiple of $3$. Thus given equation has no solution.
• 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.
• @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. Apr 1, 2018 at 7:35
• Make that $60$ :) Apr 1, 2018 at 7:39