December 29th, 2018, 12:58 AM  #11 
Senior Member Joined: Oct 2009 Posts: 696 Thanks: 235 
The first thing you'd need to prove is that $\varphi(n)$ is not $<5$ infinitely many times. So you'd need to prove there actually IS an N such that for $n\geq N$, $\varphi(n)\geq 5$. Use the prime decompositions for that, and the formula of $\varphi$ on such decompositions. When proving this, you'll be able to constructively find a value of $N$. Then it is just a matter of checking all values from $1$ to $N$. 
December 29th, 2018, 10:38 PM  #12 
Newbie Joined: Dec 2018 From: Euclidean Plane Posts: 7 Thanks: 3 
Here's a start: For any positive integer $n$, if $n$ is divisible by a prime factor $p \ge 7$, then $\phi(n)$ contains $p1 \ge 6$ as a factor. So any number $n$ with $\phi(n) < 5$ can only have 2, 3, and 5 as prime factors.


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