Context
Let $n ≥ 3$ be an odd positive integer that is not a perfect square, and let $R =\mathbb{Z}[\sqrt{−n}]$, with multiplicative norm $N(r + s\sqrt{−n} ) = r^2 + ns^2$ for every element $r + s\sqrt{−n} ∈ R$. Let $p ∈ \mathbb{Z}, p ≥ 2$, be a prime in $\mathbb{Z}$ such that $p$ divides $n + 1$. $p$ is not prime in $R$.
Question
Show that there is no element $r + s\sqrt{−n} ∈ R$, for $r, s ∈ \mathbb{Z}$, with norm equal to $p$.
Hint: determine first which of the two numbers $p$, $n$ is smaller than the other.
My answer
Let $$r^2+ns^2=p$$ $$ns^2=p-r^2$$ $p$ divides $n+1$ therefore $p\leq n+1$ $$(p-1)s^2\leq ns^2=p-r^2$$ $$(p-1)s^2\leq p-r^2$$ $$ps^2-s^2\leq p-r^2$$ $$p(s^2-1)\leq s^2-r^2$$ $$2\leq p \leq \frac{s^2-r^2}{s^2-1}$$ $$2(s^2-1)\leq s^2-r^2$$ $$s^2-2\leq -r^2$$ $$s^2+r^2\leq 2$$ Only integer square numbers less than $2$ are $0$ and $1$. When you sub the 4 combinations of $r^2=0$ or $1$ and $s^2=0$ or $1$ into $r^2+ns^2$ none of them can be true. Hence there is no element in $R$, that satisfies $N(r+s\sqrt{-n})=p$.
I believe this to be a satisfactory answer to the question, but it felt very long winded. Is there a shorter way? Thanks
If $p=2,$ we have $n \geq 3$ so $p < n.$
Otherwise, $n+1$ is even but $p$ is odd. So $p \neq n+1.$ in fact, as $p$ is a divisor, $p \leq (n+1)/ 2$ and $p < n.$
detail: we are told $p$ divides $n+1.$ In both cases, $p \neq n+1.$ There is some integer $q$ with $pq=n+1.$ As $p \neq n+1,$ we get $q \neq 1$ and $q \geq 2.$ So $n+1 = pq \geq 2p.$
If we assume we have $r^2 + n s^2 = p,$ this shows $s = 0.$ But then $r^2 = p$ in the ordinary integers, while $p$ is prime. Contradiction.