TITLE: Typo in Marcus' $\textit{Number Fields}$?
QUESTION [6 upvotes]: I am doing Problem 5.10 of Marcus where it is given that $m$ is a square-free negative integer and that $\mathcal{O}_K$ is a PID where $K = \Bbb{Q}(\sqrt{m})$.  Now in part (b) of this problem he asks: 


5.10 (b): Suppose $p$ is an odd prime such that $m < 4p$. Show that $m$ is a non-square mod $p$.


Is there a typo in this question? Firstly I am confused because $m$ in question 5.10 is supposed to be a negative integer. Then sure any negative integer will be less than $4p$ for any given odd prime $p$. Further if we take $p = 3$ and the negative integer $m  = -2$ clearly $-2 = 1\mod{3}$ that is a square. It is also clear that $\mathcal{O}_K$ where $K= \Bbb{Q}(\sqrt{-2})$ is a PID.
Are there supposed to be some further hypothesis on $m$ and $p$ or is the problem just plain wrong? Or is there a typo in the inequality $m < 4p$?  
Thanks.
Edit: Brett Frankel has requested that I add part (a) of the problem (which I have done) :


5.10(a): Show that $m \equiv 5\mod{8}$ except when $m \equiv -1,-2,$ or $-7$. 
    (Consider a prime lying over $2$).

REPLY [2 votes]: I think what Marcus meant was: Suppose $\mathcal{O}_K$ is a PID. Then prove under the assumption that $p < \operatorname{disc}\mathcal{O}_K/4$, we have $m$ a non-square mod $p$. 
This is proved easily as follows. If $p < \operatorname{disc}\mathcal{O}_K/4$ then there is no element of norm $p$ in $\mathcal{O}_K$, thus no non-trivial factor of $p$ is principal. This forces $m$ to be a quadratic non-residue mod $p$. Otherwise if $m \equiv n^2 \mod{p}$ then
$$p\mathcal{O}_K = \begin{cases} (p,\sqrt{m}), &\text{if $p |m$} \\
(p,n +\sqrt{m})(p,n-\sqrt{m}),& \text{if $p\nmid m$} \end{cases}$$
which contradicts $\mathcal{O}_K$  being a PID.