Let $p$ be a prime, let $\mathbb{Q}_p$ be the field of $p$-adic numbers. I want to show that it contains a square root of $4p+1$, but does not contain a square root of $4p$.
I could only manage the case for $4p+1$. First of all, assume that $p\neq 2$, we know that a necessary and sufficient condition for $4p+1$ to have a square root is the existence of $a\in \mathbb{Q}_p$, such that $$ |4p+1-a^2|_p<1. $$ Taking $a=1$, we get $$ |4p|_p=\frac{1}{p}. $$ Now, if $p=2$, we want $a$ to satisfy $$ |4p+1-a^2|_p<\frac{1}{4}, $$ but here the same $a=1$ will do, since $$ |8|_2=\frac{1}{8}<\frac{1}{4}. $$ For $4p$, again assume $p \neq 2$ first, we need to show that for any $a\in \mathbb{Q}_p$ we have $$ |4p-a^2|_p\geq1. $$ Using the reverse triangle inequality, $$ |4p-a^2|_p\geq |4p|_p-|a^2|_p=\frac{1}{p}-|a^2|_p, $$ and so, it follows that $|4p-a^2|_p\geq \frac{1}{p}$, so it does not help that much. How do I prove this?
Hint: Because $\Bbb{Q}_p$ always contains $\sqrt4$ you really only need to worry about the (non)-existence of $\sqrt p$. But if $a^2=p$, then $|a|_p^2=|p|_p=1/p$.
You're doing fine with $\sqrt{4p+1}$.