Let n>30. Then prove there exists a natural number $ 1< m\leq n$ such that $(n,m)=1$ and $m$ is not prime. Let $n>30$. Then prove there exists a natural number $1<m\leq n$ such that $(n,m)=1$ and $m$ is not prime.
$(m,n)$ denotes the  greatest common divisor of $m$ and $n$.
Thanks
 A: Since $30=2\cdot3\cdot5$ and $2^2,3^2,5^2<30$ it's enough to consider the case $30\mid n$ (if not take $m=2^2$ or $m=3^2$ or $m=5^2$). Assume that to be the case.
Let $p$ be the smallest prime with $p\nmid n$. Then $(n,p^2)=1$ and $p^2<n$. This is because $31$ is the largest primorial prime less than the next prime(of the primes in the primorial) squared (see this and for a proof Hagen von Eitzen's answer). Take $m=p^2$.
A: Let $p_k$ be the largest prime for which $n$ is divisible by $p_k\#=2\cdot 3\cdot 5\cdot\ldots\cdot p_{k-1}\cdot p_k$.
Clearly $(n,m)=1$ if we let $m=p_{k+1}^2$. We need to show that $m<n$.
By the Bertrand postulate, $\frac12p_k<p_{k-1}<p_k<p_{k+1}<2p_k$. 
But this follows from $m<4p_k^2$ and


*

*$n\ge p_k\#\ge 30p_{k-1}p_k> 15p_k^2$ if $p_k\ge 11$

*$n\ge p_k\# = 210>11^2=m$ if $p_k=7$

*$30|n$ and $n>30$, hence $n\ge 60$, and $m=49$ if $p_k=5$

*$m\le 25<n$ if $p_k\le 3$

A: Hint:  do you know a proof that there are an infinite number of primes?  If you find two of them not factors of $n, \dots$
