let $n\geq 3$ be odd positive integer that is not a perfect square and let $R= \Bbb{Z}[\sqrt {-n}]$ witn multiplicative norm $N(r+s\sqrt{ -n})=r^2+ns^2$ for all elements $r+s\sqrt{-n}$ of $R$
let P be a prime be an integer , $p\geq2$ such that p divides n+1
a) show that p is not prime in $R$ by finding distinct factorisations of $n+1$
b) show there is no element in $R$ with norm equal to $p$
c) with the help of b) show $p$ is irreducible element of $R$
for a) i have said $p$ divides $n+1=(1+\sqrt{-n})(1-\sqrt{-n})$ but $p$ does not divide either of these factors so $p$ not prime. is this ok?if so why do i need an alternative factorisation as alluded to in question?
as for part (b) im at a loss.
i have to show $r^2+ns^2=p$ has no solution. I have $pk=(n+1)$ for some integer $k$ and $p$ is prime but how to proceed? i put $pk-1=n$ into the norm equation to get
$r^2+(pk)s^2-s^2=p$
so $p$ would divide $r^2-s^2=(r+s)(r-s)$
but $p$ prime means
$p$ divides $(r+s)$ or $(r-s)$
but is this leading anywhere?
thanks
Suppose $\;z=a+b\sqrt{-n}\;,\;\;a,\,b\in\Bbb Z\;$ , has norm $\;p\;$ , then $\;a^2+b^2n=p\;$ . But we're given that $\;p\,\mid\,(n+1)\implies n=kp-1\;$ , for some positive integer (why?) $\;k\;$ , then:
$$p=a^2+b^2n=a^2+b^2(kp-1)=a^2-b^2+b^2kp\implies (1-b^2k)p=(a-b)(a+b)\implies$$
either $\;p\,\mid(a-b)\;$ or $\;p\,\mid\,(a+b)\;$ . Anyway, we'd get $\;a=\pm b+mp\;$ , and thus:
$$p=a^2+b^2n=b^2\pm2bmp+m^2p^2+b^2(kp-1)=(m^2+b^2k\pm2bm)p\implies$$
$$m^2\pm2bm+b^2k=\pm1\iff m^2\pm2b\,m+b^2k-1=0\;\color{red}{(*)}$$
The last one is a quadratic in $\;m\;$ whose discriminant is
$$\Delta=4b^2-4(b^2k-1)=4\left[b^2(1-k)+1\right]\implies$$
For $\;\color{red}{(*)}\;$ to be true it must be that
$$b^2(1-k)+1\ge0\iff b^2(k-1)\le1$$
But the last inequality is possible iff $\; k=1,\,2,\,b=0,\,\pm1\;$ , so the only possibilities for $\;z\;$ are $\;z=a,\,z=a\pm\sqrt{-n}\;$ , and $\;n=p-1\,\text{or}\;\,2p-1\;$ , but this would imply
$$\begin{cases}p=a^2\;,\;\;\text{or}\;\;p=a^2+n=a^2+p-1\;\;\text{or}\;\;p=a^2+2p-1\end{cases}$$
The first possibility is absurd as $\;\sqrt p\;$ cannot be integer, and the second possibilites are absurd since they would imply either
$$\begin{cases}p=a^2+p-1\implies a=\pm1\\{}\\ p=a^2+2p-1\implies a^2+p-1=0\end{cases}$$
and both possibilites are impossible (why?)