Let $n>30$. Then prove there exists a natural number $1
$(m,n)$ denotes the greatest common divisor of $m$ and $n$. Thanks
Let $n>30$. Then prove there exists a natural number $1
$(m,n)$ denotes the greatest common divisor of $m$ and $n$. Thanks
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
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 $m=p^2$.
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
Hint: do you know a proof that there are an infinite number of primes? If you find two of them not factors of $n, \dots$