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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and $f = \\sum_{n=0}^{D} c_n x^n$ with integers $c_n$. Its derivative is the [[formal derivative|0AD4VS02H3]] $f' = \\sum_{n=1}^{D} n c_n x^{n-1}$. Let $r$ satisfy $f(r) \\equiv 0$ and $f'(r) \\not\\equiv 0 \\pmod p$.\n2. Claim T: for integers $a$ and $h$, $f(a + h) = f(a) + f'(a) h + h^2 G$ for an integer $G$. By the [[binomial theorem|0EHWPWGHB2]], $(a + h)^n = a^n + n a^{n-1} h + h^2 \\sum_{i=2}^{n} \\binom{n}{i} a^{n-i} h^{i-2}$. Multiply by $c_n$ and sum over $n$.\n3. Claim C: if $a \\equiv b \\pmod p$ and $g$ has integer coefficients, then $g(a) \\equiv g(b) \\pmod p$. This follows from the compatibility of congruence with [[sums|00FWBN7F0Y]], [[products|072D5G4SK6]] and [[powers|0J5VGJYYC3]].\n4. Since $p$ is prime and $p \\nmid f'(r)$, we have $\\gcd(f'(r), p) = 1$. By [[Bézout's identity|0DQVHMYAGA]] there is an integer $t$ with $t f'(r) \\equiv 1 \\pmod p$, a [[modular inverse|00GXTQMY8Z]].\n\nExistence. We argue by induction on $k$.\n\n5. For $k = 1$, take $r_1 = r$.\n6. Assume $r_k \\equiv r \\pmod p$ and $p^k \\mid f(r_k)$. Let $h = -t f(r_k)$ and $r_{k+1} = r_k + h$. Then $p^k \\mid h$.\n7. By claim T, $f(r_{k+1}) = f(r_k) \\left(1 - t f'(r_k)\\right) + h^2 G$ for an integer $G$.\n8. By claim C, $f'(r_k) \\equiv f'(r) \\pmod p$, so $1 - t f'(r_k) \\equiv 1 - t f'(r) \\equiv 0 \\pmod p$. With $p^k \\mid f(r_k)$, the first term is divisible by $p^{k+1}$.\n9. Also $p^{2k} \\mid h^2$, and $2k \\ge k + 1$. So $p^{k+1} \\mid f(r_{k+1})$.\n10. Since $p \\mid h$, we have $r_{k+1} \\equiv r_k \\equiv r \\pmod p$. This completes the induction.\n\nUniqueness.\n\n11. Let $s$ and $s'$ satisfy $s \\equiv s' \\equiv r \\pmod p$ and $f(s) \\equiv f(s') \\equiv 0 \\pmod{p^k}$. We show $s \\equiv s' \\pmod{p^j}$ for $j = 1, \\dots, k$ by induction on $j$. The case $j = 1$ holds by assumption.\n12. Assume $p^j \\mid h$ with $h = s' - s$ and $j \u003c k$. By claim T, $f(s') = f(s) + f'(s) h + h^2 G$ for an integer $G$.\n13. Here $p^{j+1}$ divides $f(s)$ and $f(s')$, as $j + 1 \\le k$. It divides $h^2$, as $2j \\ge j + 1$. So $p^{j+1} \\mid f'(s) h$.\n14. Write $h = p^j h'$. Then $p \\mid f'(s) h'$. By claim C, $f'(s) \\equiv f'(r) \\not\\equiv 0 \\pmod p$. By [[Euclid's lemma|051VYKA1EX]], $p \\mid h'$, so $p^{j+1} \\mid h$.\n15. With $j = k$, this gives $s \\equiv s' \\pmod{p^k}$. So $r_k$ is unique modulo $p^k$.\n\n## Idea\n\nThis is Newton's method for congruences. Near a root, $f(a + h) \\approx f(a) + f'(a) h$, and the error $h^2 G$ is invisible to the next power of $p$. Since $f'(r)$ is invertible modulo $p$, one can always solve for the correction $h$, and the same fact shows the correction is forced.\n",
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