Past exam of the short options course of the University of Oxford / 2023 / 1 / c

Let $a$ be a root modulo $p^j$. Every lift to a residue modulo $p^{j+1}$ is uniquely $a+t{p}^j$ with $t(\mathrm{mod}p)$. The polynomial expansion givesBecause $a(\mathrm{mod}p)$ is a root and there are no common roots of $f$ and $f^{\prime }$, $f^{\prime }(a)$ is nonzero modulo $p$. Thus precisely one $t$ solvesEvery root at one level therefore has exactly one lift, and every root at the next level reduces to a root at the previous level. Induction proves equality of the counts from $p$ to $p^l$. This is simple-root Hensel lifting by successive residues at finite precision; completeness is unnecessary for this finite-level assertion.

For $f(x)=x^2-n$, a root modulo the odd prime $p$ is nonzero because $p\nmid n$. Its derivative $2x$ is therefore nonzero modulo $p$. Part (i) gives exactly one lift of each root. A nonzero quadratic residue modulo an odd prime has two distinct roots, $x$ and $-x$; a nonresidue has none. Hence