$||x||=1$ in $K/\mathbb{Q}$ implies $x$ is a root of unity. Let $K/\mathbb{Q}$ a finite (i.e. algebraic and finitely generated) extension. Let $x \in K$, such that $||x||=1$ for all normalized absolute values of $K$ but at most one. Then $x$ is a root of unity.
By the product formula I get $||x||=1$ for all normalized absolute values of $K$. Hence $x \in S^1$, seen as the embedding $K \subset \mathbb{C}$.
But now I am a bit confused. Let $x = \frac{3}{5} + \frac{4}{5}i$ shows that the conditions $x$ algebraic and $x \in S^1$ are not sufficient for $x$ to be a root of unity, hence I will have to use that the $p$-adic absolutes are $1$ as well. However, I feel a bit clueless  on how to go on. My intuition would tell me that in the example above something will go wrong for $p_i | (5)$, $p_i$ a prime ideal in $\mathbb{Q}(x)$, then taking the $p_i$-valuation. I don't really know if this is true though, as I don't know any statements about absolute values in fields like the given $K$, but the existence, definition and product formula. Also, understanding the example still doesn't give me a solution.
 A: The archimedean places of $K$ correspond to the usual absolute value on $\Bbb C$, after first applying the embeddings $K\hookrightarrow\Bbb C$. Thus if $|x|=1$ for all archimedean places, we know that all of $x$'s conjugates have (the usual) absolute value $1$. Consider the minimal polynomial of $x$. In my Stewart & Tall, the following lemma is used in proving Dirichlet's unit theorem:
Lemma. If $p(t)\in\Bbb Z[t]$ is a monic polynomial all of whose roots have absolute value $1$, then all of its roots are roots of unity. The proof proceeds as follows:


*

*Say $p(t)=(t-a_1)\cdots(t-a_k)$ and define $p_\ell(t)=(t-a_1^\ell)\cdots(t-a_k^\ell)$

*As symmetric polynomials in $a_1,\cdots,a_k$, the coefficients of $p_\ell(t)$ are integers

*$t^j$ coeff. of $p_\ell$ is bounded: $|e_{k-j}(a_1^\ell,\cdots,a_k^\ell)|=|\sum\square|\le\sum|\square|=\sum1=\binom{k}{j}$ (Vieta's)

*Only finitely many $f(t)\in\Bbb Z[t]$ satisfying such bounds, so $p_1,p_2,p_3,\cdots$ has a repeat

*Say $p_\ell(t)=p_m(t)$. So $\exists\pi\in S_k$ such that $a_j^\ell=a_{\pi(j)}^m$ (for each $j=1,\cdots,k$)

*Thus $a_j^{\ell^2}=(a_{\pi(j)}^m)^\ell=(a_{\pi(j)}^\ell)^m=a_{\pi^2(j)}^{m^2}\Rightarrow\cdots\Rightarrow a_j^{\ell^{\large k!}}=a_j^{m^{\large k!}}\Rightarrow a_j^{\ell^{\large k!}-m^{\large k!}}=1$ for each $j$


I am not sure if this is desirable for a homework exercise but it's a fun proof nonetheless.
A: Here's what I finally did (no guarantee for correctness):
First note that as the localizations $R_\mathfrak{p}$ are given by $\{y \in K\mid ||y||_\mathfrak{p} \leq 1\}$, $x$ is contained in all the $R_\mathfrak{p}$ hence also in $R$ (for $R$ being the integral closure of $\mathbb{Z}$ in $K$).
Let $r_1$ be the number of real embeddings of $K$, let $2r_2$ be the number of its complex embeddings into $\mathbb{C}$.
Let $\mathfrak{L}: K^\ast \rightarrow \mathbb{R}^{r_1+r_2}, y \mapsto (\log(|\sigma_1(y)|),...,\log(|\sigma_{r_1+r_2}(y)|)$
Then we had seen as a Lemma for Dirichlet's unit theorem:
Let $B \subset \mathbb{R}^{r_1+r_2}$ compact. Then $\mathfrak{L}^{-1}(B) \cap R$ is finite.
Now in our case: take $B = 0 \in \mathbb{R}^{r_1+r_2}$. 
As for all $n \in \mathbb{N}$:  $x^n \in \mathfrak{L}^{-1}(B) \cap R$, we are done.
