find the smallest $N$ that $\varphi(n)\ge5$ for every $n\ge N$ I know that the solution is with Euler function. I could not understand how to show this. thanks.
 A: Hint: An equivalent problem is

Find the largest $N$ such that $\varphi(N) \le 4$.

Since $\varphi(n) \geq \sqrt{n/2}$ for all $n$ (see here), we only have to consider $N \le 32$.
A: No need to resort to complicated machinery, whatsoever.
Note that, $\forall n\in\mathbb{Z}^+$, $(n-1,n)=(1,n)=1$. Take a prime $q$ such that, $\frac{n}{2}<q<n$ (existence is justified through Bertrand's Postulate). Notice that, $(q,n)=1$ and $(n-q,n)=1$. Hence, we know that, for every $n$, $n-1,q,n-q,1$ are coprime to $n$.
Motivated by this, note that for $n$ sufficiently large (e.g., say $n>32$), look at the intervals, $I=(n/16,n/8),(n/8,n/4),(n/4,n/2),(n/2,n)$, take one prime each, add $1$, we get $\phi(n)\geq 5$, hence, $n<32$ (notice, this is a very naive bound). A few finite case work establishes $n=12$ is the largest such $n$.
A: The smallest $N$ where $\phi(N) \ge 5$ for every $n \ge N$ would be $M + 1$ where $M$ is the largest $M$ where $\phi(M) \le 4$.
Now if $\gcd(n,k) = 1$ then $\phi(nk) = \phi(n)\phi(k)$ and for any $p^m$ for a prime $p$, then $\phi (p^m) = (p-1)p^{m-1} \ge p-1$.  So if $\phi(M) \le 4$ then $p \le 5$ so the only possible prime factors of $M$ are $2,3$ or $5$.
So $M = 2^a3^b5^c$ and $\phi(M) = \phi (2^a)\phi(3^b)\phi(5^c)$.  
Case 1: Now if $c \ge 1$ then $\phi(5^c) = 4*5^{c-1}$ so $c=1$ and $\phi(M) = 4$.  And $\phi(3^b) = \phi(2^a) =1$.
If $b\ge 1$ then $\phi(3^b) = (3-1)3^{b-1} = 2*3^{b-1}$ that's impossible so $b = 0$.
If $a \ge 1$ then $\phi(2^a) = (2-1)2^{a-1} = 2^{a-1}$.  That is possible only if $a =1$ so if $M=10$, $\phi(M) =4$ and that's the largest such number that is a multiple of $5$.
Case 2: $c=0$ and If $b\ge 1$: we can have $\phi(3^b) = 2*3^{b-1}$.  If this is to be less than or equal to $4$ we must have $b=1$ and $\phi(3) = 2$ so $M= 2^a*3$ and $\phi(M) = \phi (2^a)\phi 3 = 2\phi(2^a)$.  
SO $\phi(2^a) \le 2$ and as $\phi(2^a)= 2^{a-1}$ we may  have $a = 2$ and $M = 12$ and $\phi(12) = 4$
And that's the largest such number that is a multiple of $3$.
Case 3: $c=0; b=0$ and $M = 2^k$ so $\phi(M) = \phi(2^{k})=2^{k-1}\le 4$.  This means $k \le 3$ and $M \le 8$.  If $M = 8$ then $\phi 8 = 4$.
And there we have it.
$M =12$ is the largest number so that $\phi(M) \le 4$ and $N = 13$ is the smallest number so that for all $n\ge N$, $\phi(n) \ge 5$.
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Worth noting the only time $\phi(k)$ is odd is if $k=1,2$ so $\phi(k) =5$ is impossible.  (If $p$ is an odd factor of $k$ then $p-1|\phi(k)$.  If $k = 2^k$ then $\phi(2^k) = 2^{k-1}$.)
$\phi(k) \ge 5; k \le 12$ occur with $\phi(7) = 6; \phi(9)=6; \phi(11)=10$
$\phi(k) = 4$  occur with $\phi(5)=\phi(8)= \phi(10)=\phi(12)= 4$
And the rest $\phi(3)=\phi(4)=\phi(6) = 2$.
