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  "body": "\n## Statement\n\nLet $p$ be a [[prime number|0SN0MAEKYX]], and let $f$ be a polynomial with integer coefficients, with derivative $f'$.\nLet $r$ be an integer such that $f(r) \\equiv 0 \\pmod p$ and $f'(r) \\not\\equiv 0 \\pmod p$.\nThen for each positive integer $k$ there is an integer $r_k$ with $r_k \\equiv r \\pmod p$ and $f(r_k) \\equiv 0 \\pmod{p^k}$.\nThis $r_k$ is unique modulo $p^k$.\n\n## Remarks\n\nThe proof gives the lifts step by step. If $f(r_k) \\equiv 0 \\pmod{p^k}$ and $t$ is an inverse of $f'(r)$ modulo $p$, then $r_{k+1} = r_k - t f(r_k)$ is a root modulo $p^{k+1}$.\nFor example, $3^2 \\equiv 2 \\pmod 7$ and $f'(3) = 6$ for $f = x^2 - 2$. The step gives $3 - 6 \\cdot 7 = -39 \\equiv 10 \\pmod{49}$, and $10^2 = 100 \\equiv 2 \\pmod{49}$.\nIn the language of $p$-adic numbers, a simple root modulo $p$ lifts to a unique root in the $p$-adic integers.\nThe lemma is the analogue of Newton's method for congruences.\n\n## History\n\nThe lemma is named after Kurt Hensel.\n\n## Explanations\n\n### L1 human\n\nA solution of a polynomial congruence modulo $p$ often gives solutions modulo $p^2$, $p^3$, and so on.\nThe condition $f'(r) \\not\\equiv 0 \\pmod p$ says that $r$ is a simple root modulo $p$.\nUnder this condition, each step of the lifting has exactly one answer.\n",
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