I need help proving that for every positive integer $n$, there exist $n$ consecutive positive integers, each of which is composite. The hint that came with the problem is: Consider the numbers $$(n+1)!+2,(n+1)!+3,…,(n+1)!+n,(n+1)!+n+1$$
I honestly have no idea how to begin here. I am specifically confused about what it means when it says "there exist $n$ consecutive positive integers". Does this mean that if $n = 5$, for example, then somewhere in the positive integers there are 5 consecutive composite integers? And that we want to prove that?
I get that composite means that they are not prime, hence they have more factors than 1 and themselves.
Any help would be greatly appreciated.
Thank you!
Best Answer
To expand on the other solution already given,
Proof. Assume $m$ and $n$ exist in the positive integers and that $m$ is less than $n$. If $m$ is less than $n$, then $$ n!=1\cdot2\cdot3\cdot\dotsb\cdot m\cdot\dotsb\cdot n, $$
which is to say that $m$ is a factor of $n!$. So, \begin{align} m+n! &= m+(1\cdot2\cdot\dotsb\cdot m\cdot\dotsb\cdot n) \\ &= m\left(\frac{1\cdot 2\cdot \dotsb\cdot n}{m + 1}\right) \end{align} Remember that $m$ is a factor of $n!$ and so $n!/m$ is still an integer. So since $m$ is an integer that is bounded between $1$ and $n$, it stands that whatever number you pick up to $n$ can divide $m+n!$ making it composite till the $n$th integer, but $n!$ has that $n$th integer in it so the $n$th integer is also composite which means that you can pick any integer between $1$ and $n$ inclusively and it will be composite.