Algebraic number with bounded coefficients How many algebraic numbers $z$ are there satisfying $P(z)=0$ where $P(z)$ is some polynomial with integer coefficients of degree less than or equal to $n$ such that the absolute value of every coefficient is less than or equal to $x$?
In other words, define a function $N(x,n)$ as the number of such algebraic numbers as described above. What conclusions, particularly asymptotic, can we draw about $N(x,n)$? Can we extend $N(x,n)$ to real or complex values of $x,n$?
Clearly $N(x,0)=1$ for all $x$. Can we find a closed form for $N(x,1)$, the number of rational number with numerator and denominator less than or equal to $x$? An asymptotic formula? What if we were to hold $x$ constant with $n$ variable? Clearly $N(x,n)$ grows quickly in both variables.
I can't see how to attack this function. Anyone have any ideas?
(Sorry about tagging, there don't seem to be many appropriate ones.)

Just for the note, pick $m \ge 1$. $N(1,m)$ is the number of zeroes of polynomials of the form $aX+b$ with $a,b \in I_m=\{-m,-m+1,...0,m-1,m\}$, i.e. the number of rationals $p/q$ with $p,q \in I_m$. 
You can focus on the $(p,q)$ where $p,q \ge 0$ (you just need to multiply the result by two to take into account the negatives) and $\text{gcd}(p,q)=1$ (the other ones would be double-counting). 
As hinted below, $J_k = \text{Card}\{(p,q) \;\; \text{s.t.} \;\; p,q \in \{1,...,k\}, \text{gcd}(p,q)=1\} \sim \frac{6}{\pi^2}k^2 $ as $k \to \infty$. That should imply: $$N(1,m) \sim \frac {12} {\pi^2} m^2 \simeq 1.22\, m^2$$
To get the result above, you need Dirichlet convolution, Dirichlet series and the Moebius function. 
Write 


*

*$J_k = 2 \text{ Card }\{(p,q) \; ; \; 1 \le p \le q \le k,\; \text{gcd}(p,q)=1\} = 2\sum_{q=1}^k \varphi(q)$

*use $\mu \star \text{id} = \varphi$ to get to $\displaystyle \sum_1^N \varphi(n) = \sum_d^N \sum_p^{N/d} \mu(d)p$

*with some work you get to $\displaystyle\frac{1}{N^2}\sum_1^N \varphi(n) \to  \sum^{\infty} \frac{\mu(d)}{d^2} = \frac{1}{\zeta(2)}$

