A question referring the proof of class number 
In this proof, second paragraph, why do we just need to consider the primes for $2\le x\le \sqrt{|d|}$, how about the primes larger than $\sqrt{|d|}$ ? And why the the primes for $2\le x\le \sqrt{|d|}$ inert implies the class number 1? (the reason he gives doesn't make sense to me, what does he mean that no primes inert? At least we don't know the case larger than $\sqrt{|d|}$, and even all the primes are inert, why is the class number 1?)
 A: There are theorems to the effect that there is an effectively computable bound $A_K$ depending on the number field $K$ such that each ideal class in its ring of integers has a representative ideal of norm at most $A_K$. The most famous of these is Minkowski's bound. For an imaginary quadratic field it is $(2/\pi)\sqrt{|\Delta_K|}$ where $\Delta_K$ is the discriminant. I suspect the book here (Borevich and Shafarevich?) is using a slightly weaker bound than that.
A: It follows from the observation he made in page 149

Which in turn follows from Minkowski's theorem and Corolary I (in pag 135)

This is because if $A$ is an ideal class and $\mathfrak{a}$ is a representative  such that $N(\mathfrak{a})< \sqrt{\mid d \mid}$. Then for a prime ideal $\mathfrak{p}$ dividing $\mathfrak{a}$ with $ p \,\mathbb{Z}= \mathfrak{p} \cap \mathbb{Z} $, we have $p\leq N(\mathfrak{p})\mid N(\mathfrak{a})< \sqrt{\mid d \mid }$ so  $\mathfrak{a}$ is a product of prime ideals $\mathfrak{p} $ each factor of some prime $p< \sqrt{\mid d \mid}$ and so $A$ is the product of the corresponding classes.
In this case all this primes $p$ are inertial (thus their only factor is principal) and therefore every class is the identity.
