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Hensel’s Lemma
• Last Updated : 13 May, 2021

Hensel’s Lemma is a result that stipulates conditions for roots of polynomials modulo powers of primes to be “lifted” to roots modulo higher powers. The lifting method outlined in the proof is reminiscent of Newton’s Method for solving equations. Say the equations of the below type is to be solved:

The idea is to use Hensel’s Lemma to solve this type of congruence. It is also known as Hensel’s “lifting” lemma, which is a result of modular arithmetic. If f is a polynomial function and p is a Prime Number, then if f(a1) = 0 (mod p) and [f'(a1)] mod p != 0, then it’s possible to “lift” this solution to the solution for f(x) = 0 (mod pk) by using the below recursion. Note that the [f'(a1)]-1 refers to the modular inverse of f ‘(a1) modulo p.

Example 1:

First, solve for (x = 2 is the solution for mod 5)

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Example 2:

This has no solution as (x3 – 3) % mod 7 = 0 has no solution

Below is the implementation of Hensel’s Lemma

## C++

 // C++ program to illustrate the// Hensel's Lemma#include using namespace std; // Consider f(x) = x ^ 3 - k,// where k is a constant // Function to find the modular// inverse of a modulo mint inv(int a, int m){    int m0 = m, t, q;    int x0 = 0, x1 = 1;    if (m == 1)        return 0;     // Apply the Euclidean algorithm,    // to find the modular inverse    while (a > 1) {        q = a / m;        t = m;        m = a % m;        a = t;        t = x0;        x0 = x1 - q * x0;        x1 = t;    }    if (x1 < 0)        x1 += m0;    return x1;} // Function to find the derivative of// f(x) and f'(x) = 3 * (x ^ 2)int derivative(int x){    return 3 * x * x;} // Function to find the image of// x in f(x) = x ^ 3 - k.int Image(int x, int k){    return x * x * x - k;} // Function to find the next power// of the numberint next_power(int a_t, int t, int a1,               int prime, int k){    // Next power of prime for which    // solution is to be found    int power_p = (int)pow(prime, t + 1);     // Using Hensel's recursion to    // find the solution (next_a) for    // next power of prime    int next_a = (a_t                  - Image(a_t, k)                        * inv(derivative(a1),                              prime))                 % power_p;     // If next_a < 0 return equivalent    // positive remainder modulo p    if (next_a < 0)        return next_a += power_p;     // Return the next power of a    return next_a;} // Function to find the solution of// the required exponent of primeint powerOfPrime(int prime, int power,                 int k, int a1){    // The lemma does not work for    // derivative of f(x) at a1    if (derivative(a1) != 0) {        int a_t = a1;         // Looping from 1 to power        // of prime whose solution        // is to be found        for (int p = 1; p < power; p++) {            a_t = next_power(a_t, p,                             a1, prime, k);        }         // Final answer after evaluating        // all the exponents up till        // the required exponent        return a_t;    }     return -1;} // Driver Codeint main(){    int prime = 7, a1 = 3;    int power = 2, k = 3;     // Function Call    cout << powerOfPrime(prime, power,                         k, a1);     return 0;}

## Java

 // Java program to illustrate the// Hensel's Lemmaimport java.util.*; class GFG{ // Consider f(x) = x ^ 3 - k,// where k is a constant // Function to find the modular// inverse of a modulo mstatic int inv(int a, int m){    int m0 = m, t, q;    int x0 = 0, x1 = 1;         if (m == 1)        return 0;     // Apply the Euclidean algorithm,    // to find the modular inverse    while (a > 1)    {        q = a / m;        t = m;        m = a % m;        a = t;        t = x0;        x0 = x1 - q * x0;        x1 = t;    }    if (x1 < 0)        x1 += m0;             return x1;} // Function to find the derivative of// f(x) and f'(x) = 3 * (x ^ 2)static int derivative(int x){    return 3 * x * x;} // Function to find the image of// x in f(x) = x ^ 3 - k.static int Image(int x, int k){    return x * x * x - k;} // Function to find the next power// of the numberstatic int next_power(int a_t, int t, int a1,                      int prime, int k){         // Next power of prime for which    // solution is to be found    int power_p = (int)Math.pow(prime, t + 1);     // Using Hensel's recursion to    // find the solution (next_a) for    // next power of prime    int next_a = (a_t - Image(a_t, k) *                  inv(derivative(a1), prime)) %                  power_p;     // If next_a < 0 return equivalent    // positive remainder modulo p    if (next_a < 0)        return next_a += power_p;     // Return the next power of a    return next_a;} // Function to find the solution of// the required exponent of primestatic int powerOfPrime(int prime, int power,                        int k, int a1){         // The lemma does not work for    // derivative of f(x) at a1    if (derivative(a1) != 0)    {        int a_t = a1;         // Looping from 1 to power        // of prime whose solution        // is to be found        for(int p = 1; p < power; p++)        {            a_t = next_power(a_t, p,                             a1, prime, k);        }         // Final answer after evaluating        // all the exponents up till        // the required exponent        return a_t;    }    return -1;} // Driver Codepublic static void main(String []args){    int prime = 7, a1 = 3;    int power = 2, k = 3;         // Function Call    System.out.print(powerOfPrime(prime, power,                                  k, a1));}} // This code is contributed by rutvik_56

## C#

 // C# program to illustrate the// Hensel's Lemmausing System;class GFG{ // Consider f(x) = x ^ 3 - k,// where k is a constant // Function to find the modular// inverse of a modulo mstatic int inv(int a, int m){    int m0 = m, t, q;    int x0 = 0, x1 = 1;      if (m == 1)        return 0;     // Apply the Euclidean algorithm,    // to find the modular inverse    while (a > 1)    {        q = a / m;        t = m;        m = a % m;        a = t;        t = x0;        x0 = x1 - q * x0;        x1 = t;    }    if (x1 < 0)        x1 += m0;          return x1;} // Function to find the derivative of// f(x) and f'(x) = 3 * (x ^ 2)static int derivative(int x){    return 3 * x * x;} // Function to find the image of// x in f(x) = x ^ 3 - k.static int Image(int x, int k){    return x * x * x - k;} // Function to find the next power// of the numberstatic int next_power(int a_t, int t, int a1,                      int prime, int k){         // Next power of prime for which    // solution is to be found    int power_p = (int)Math.Pow(prime, t + 1);     // Using Hensel's recursion to    // find the solution (next_a) for    // next power of prime    int next_a = (a_t - Image(a_t, k) *                  inv(derivative(a1), prime)) %                  power_p;     // If next_a < 0 return equivalent    // positive remainder modulo p    if (next_a < 0)        return next_a += power_p;     // Return the next power of a    return next_a;} // Function to find the solution of// the required exponent of primestatic int powerOfPrime(int prime, int power,                        int k, int a1){         // The lemma does not work for    // derivative of f(x) at a1    if (derivative(a1) != 0)    {        int a_t = a1;         // Looping from 1 to power        // of prime whose solution        // is to be found        for(int p = 1; p < power; p++)        {            a_t = next_power(a_t, p,                             a1, prime, k);        }         // Final answer after evaluating        // all the exponents up till        // the required exponent        return a_t;    }    return -1;} // Driver Codepublic static void Main(string []args){    int prime = 7, a1 = 3;    int power = 2, k = 3;         // Function Call    Console.Write(powerOfPrime(prime, power,                                  k, a1));}} // This code is contributed by pratham76.

## Javascript

 
Output:
6

Time Complexity: O(log N)
Auxiliary Space: O(1)

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