TITLE: Prove for all $m \in \mathbb{N}$ exists $n \in \mathbb{N}$ such that $\varphi(n)-\varphi(n+1)>m$ and $\varphi(n)-\varphi(n-1)>m$
QUESTION [7 upvotes]: Number Theory :
Prove for all $m \in \mathbb{N}$ exists $n \in \mathbb{N}$ such that :
$\varphi(n)-\varphi(n+1)>m$ and $\varphi(n)-\varphi(n-1)>m$
Attempt:
For $m\in \mathbb{N}$ let $q_m$ be prime number form of $4k+3$ for some $k\in \mathbb{N}$ such that $2m+3<q_m$

REPLY [2 votes]: Hint: Suppose $n$ is an odd prime. Then $\varphi(n) = n-1$. Also, $n-1$ and $n+1$ are both even, and thus, $\varphi(n-1) \le \dfrac{n-1}{2}$ and $\varphi(n+1) \le \dfrac{n+1}{2}$. Do you see why this is true?
Using these results, how large does $n$ need to be to guarantee that $\varphi(n)-\varphi(n-1) > m$ and $\varphi(n)-\varphi(n+1) > m$?