Consider the integral binary quadratic form $$f(x,y) = 2Axy+Bx^2$$ with $A,B \in \mathbb{Z}$ different from $0$. In Cassel's book "Rational quadratic forms" page 237 he claims that for $p \neq 2$ every small enough $a_p \in \mathbb{Z}_p$ is primitively representable by $f$. Further, he says it can be done with $x=1$, namely the quadratic equation $$By^2 + 2Ay - a_p = 0$$ has a solution in $\mathbb{Z}_p$ for all small enough $a_p$. The discriminant of the polynomial and it is $\Delta = 4A^2-4Ba_p$ and so for small $a_p$ we have $\left| \Delta \right|_p = \left| 4A^2 \right|_p$. $4A^2$ is of course a square, but I don't think this implies anything about $\Delta$.
Of course, "small enough" means $p$-adically small throughout the question. Can someone tell me why is this true?
If $A \ne 0$ then you can take $a_p$ small enough so that $|a_p|_p < |2A|^2_p$. Then you apply the (general form of) Hensel's Lemma as follows:
$$f(y) = B y^2 + 2 A y - a_p,$$
Let $a = 0$, then
$$|f(a)|_p = |f(0)|_p = |a_p|_p < |2 A|^2_p = |f'(0)|^2_p = |f'(a)|^2_p.$$
See: (https://en.wikipedia.org/wiki/Hensel%27s_lemma#Hensel%27s_lemma_for_p-adic_numbers)
Another way of thinking about it is that any element $t$ in $\mathbf{Q}_p$ sufficiently close to $a^2 \ne 0$ is also a square, as can be proved by applying Hensel's Lemma to $f(x) = x^2 - t$ with the seed $x = a$, and then making sure $f(a) = a^2 - t$ is sufficiently small compared to $f'(a) = 2a$. Thus for $a_p$ small enough, the discriminant $\Delta$ will be a square.