For which $m$ does a primitive $m^{th}$ root of unity have degree $2$ over $\mathbb{Q}$?

To be honest, I don't really know whether it is okay to post this question here since perhaps the proof is really trivial...

This is the question from Lang's Algebra Chapter VI Q18 part b).

It asks me to find $m$ such that $m^{th}$ root of unity have degree $2$ over $\mathbb{Q}$. The proof is not hard I think, but I am stuck at some point. My proof:

We know that $[\mathbb{Q}(\zeta_{m}):\mathbb{Q}]=\phi(m)$, where $\phi(m)$ is the Euler's Phi Function.

So if we want the degree to be $2$, we need to find all $m$ such that $\phi(m)=2$. I know that only $3,4,6$ satisfy $\phi(m)=2$, but the hard part is to prove that there is no other integer such that $\phi(m)=2$.

So we know that by simple calculation, $\phi(1)=1,\phi(2)=1,\phi(3)=2,\phi(4)=2,\phi(5)=4,\phi(6)=2.$ For all prime $p\geq 5$, $\phi(p)=p-1\geq 5-1=4$. Thus, only one prime number works --- $\phi(3)=2$

However, I have no idea how to prove that $\phi(m)\geq3$ for all $m\geq7$ and are not prime.

It is clear that $\phi(m)\geq2$, but how to improve the lower bound?

Perhaps this question is really stupid... I am sorry for the simple question if the question is too simple.

Any hints and explanations are really appreciated!!

Hint: use the fact that $\phi$ is multiplicative and the fact that $\phi(p^k)=p^k-p^{k-1}$ for any prime $p$ and any positive integer $k$. You will see that you really only need to consider the primes $2$ and $3$.
• Well, if you write $m=p_1^{k_1}\cdots p_n^{k_n}$ for some distinct primes $p_1, \ldots, p_n$ and some positive integers $k_1, \ldots, k_n$, you have $\phi(m)=\phi(p_1^{k_1})\cdots\phi(p_n^{k_n})=(p_1^{k_1}-p_1^{k_1-1})\cdots(p_n^{k_n}-p_n^{k_n-1})$. In particular, if $p_i>3$ for any $i$, then $\phi(m)$ is divisible by $p_i^{k_i}-p_i^{k_i-1}\geq 5^{k_i}-5^{k_i-1}\geq 5-1=4$, so $\phi(m)\geq 4$. So, it suffices to consider $m=2^r3^s$. Does that make sense? – Zilliput Apr 23 '18 at 2:24