Suppose that $p^k$ is a prime power $\ne 2 , 4$. Suppose $\exists x, y\in \Bbb Z$ not both divisible by p, for which $ax^2 + by^2 ≡ c\ (mod\ p^k )$.
Prove $\exists u, v \in \Bbb Z_p$ the $p$-adics not both divisible by $p$, such that $au^2 + bv^2 = c$ in the p-adics.
What I did:
We look for $r,s\in \Bbb Z$ such that $u_k=x+rp^k, v_k=x+sp^k$ verifies: $au_k^2+bv_k^2\equiv c (mod\ p^{k+1})$
So $ax^2+by^2 +2p^k(axr+bys)\equiv c (mod\ p^{k+1})$,
So $\frac{ax^2+by^2-c}{p^k}+2(axr+bys)\equiv 0(\ mod\ p)$
Denote $\alpha := \frac{ax^2+by^2-c}{p^k}$
If we choose $p\nmid y$ we can take $r=0$, then $s=2^{-1}\alpha (by)^{-1}$ which is well defined since $b$ and $p$ are not invertible $mod\ p$ (I suppose it is the case for b even there is no such assumption in the question).
Thank you for your feedback.