The ring is $\mathbb Z[\sqrt D]$, where $D$ is a square free integer. Define the norm $N: \mathbb Z[\sqrt D] \rightarrow \mathbb N$, $N(a+b\sqrt D)= |a^2-Db^2|$.
To show that there is no element of a certain norm, my Abstract Algebra book uses clever tricks involving modulo.
It takes some effort to show it, but in $\mathbb Z[\sqrt 10]$, there are no elements of norm 2 or norm 5. At first I thought that this was because $10=2*5$. But the next example showed me that this guess was probably wrong: In $\mathbb Z[\sqrt -5]$, there are no elements of norm 3.
Can we generalize a statements about these rings $\mathbb Z[\sqrt D]$? That is, I want to know if there is a better method than guessing and checking just by looking at the integer $D$.
Note:
$D$ is a squarefree integer if one of the following holds:
(i) $D$ is a prime number $p$ or a finite product of distinct prime numbers.
(ii) $D$ is $-p$ or negative of finite product of distinct prime numbers.
My Assumption: $D$ belongs to (i)
Now, Let $a^2-Db^2=m$ where $m\in \mathbb{Z}$.
Now taking $mod$ $D$ on both sides we get
$a^2\equiv m$ $mod$ $D$
$\iff m\equiv a^2$ $mod$ $D$
Now, Use Chinese Remainder Theorem for simultaneous linear congruences to get the solutions for $m$.
Remarks:
We will always get a square integer as the norm( it may not be possibly indicated by the solutions of Chinese remainder theorem for simultaneous congruences)