Count number of solutions of x^2 = 1 (mod p) in given range

Given two integers n and p, find the number of integral solutions to x2 = 1 (mod p) in the closed interval [1, n].

Examples:

Input : n = 10, p = 5
Output : 4
There are four integers that satisfy the equation
x2 = 1. The numbers are 1, 4, 6 and 9.

Input : n = 15, p = 7
Output : 5
There are five integers that satisfy the equation
x2 = 1. The numbers are 1, 8, 15, 6 and 13.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

One simple solution is to go through all numbers from 1 to n. For every number, check if it satisfies the equation. We can avoid going through the whole range. The idea is based on the fact that if a number x satisfies the equation, then all numbers of the form x + i*p also satisfy the equation. We traverse for all numbers from 1 to p and for every number x that satisfies the equation, we find the count of numbers of the form x + i*p. To find the count, we first find the largest number for given x and then add (largest-number – x)/p to the result.

Below is the implementation of the idea.

C++

 // C++ program to count number of values // that satisfy x^2  = 1 mod p where x lies // in range [1, n] #include using namespace std; typedef long long ll;    int findCountOfSolutions(int n, int p) {     // Initialize result     ll ans = 0;        // Traverse all numbers smaller than     // given number p. Note that we don't     // traverse from 1 to n, but 1 to p     for (ll x=1; x n)                 last -= p;                // Add count of numbers of the form              // x + p*i. 1 is added for x itself.             ans += ((last-x)/p + 1);         }     }     return ans; }    // Driver code int main() {     ll n = 10, p = 5;     printf("%lld\n", findCountOfSolutions(n, p));     return 0; }

Java

 // Java program to count  // number of values that  // satisfy x^2 = 1 mod p  // where x lies in range [1, n] import java.io.*;    class GFG { static int findCountOfSolutions(int n,                                  int p) {     // Initialize result     int ans = 0;        // Traverse all numbers      // smaller than given      // number p. Note that      // we don't traverse from      // 1 to n, but 1 to p     for (int x = 1; x < p; x++)     {         // If x is a solution,          // then count all numbers         // of the form x + i*p          // such that x + i*p is          // in range [1,n]         if ((x * x) % p == 1)         {             // The largest number              // in the form of x +              // p*i in range [1, n]             int last = x + p * (n / p);             if (last > n)                 last -= p;                // Add count of numbers              // of the form x + p*i.              // 1 is added for x itself.             ans += ((last - x) / p + 1);         }     }     return ans; }    // Driver code public static void main (String[] args)  {     int n = 10;     int p = 5;     System.out.println(                findCountOfSolutions(n, p)); } }    // This code is contributed by ajit

Python3

 # Program to count number of  # values that satisfy x^2 = 1  # mod p where x lies in range [1, n]    def findCountOfSolutions(n, p):            # Initialize result     ans = 0;        # Traverse all numbers smaller      # than given number p. Note      # that we don't traverse from      # 1 to n, but 1 to p     for x in range(1, p):                    # If x is a solution, then          # count all numbers of the          # form x + i*p such that          # x + i*p is in range [1,n]         if ((x * x) % p == 1):                            # The largest number in the             # form of x + p*i in range             # [1, n]             last = x + p * (n / p);             if (last > n):                 last -= p;                # Add count of numbers of              # the form x + p*i. 1 is              # added for x itself.             ans += ((last - x) / p + 1);     return int(ans);    # Driver code n = 10; p = 5; print(findCountOfSolutions(n, p));        # This code is contributed by mits

C#

 // C# program to count  // number of values that  // satisfy x^2 = 1 mod p  // where x lies in range [1, n] using System;    class GFG { static int findCountOfSolutions(int n,                                  int p) {     // Initialize result     int ans = 0;        // Traverse all numbers      // smaller than given      // number p. Note that      // we don't traverse from      // 1 to n, but 1 to p     for (int x = 1; x < p; x++)     {         // If x is a solution,          // then count all numbers         // of the form x + i*p          // such that x + i*p is          // in range [1,n]         if ((x * x) % p == 1)         {             // The largest number              // in the form of x +              // p*i in range [1, n]             int last = x + p * (n / p);             if (last > n)                 last -= p;                // Add count of numbers              // of the form x + p*i.              // 1 is added for x itself.             ans += ((last - x) / p + 1);         }     }     return ans; }    // Driver code static public void Main () {     int n = 10;     int p = 5;     Console.WriteLine(             findCountOfSolutions(n, p)); } }    // This code is contributed by ajit

PHP

 \$n)                 \$last -= \$p;                // Add count of numbers of              // the form x + p*i. 1 is              // added for x itself.             \$ans += ((\$last - \$x) / \$p + 1);         }     }     return \$ans; }    // Driver code \$n = 10; \$p = 5; echo findCountOfSolutions(\$n, \$p);        // This code is contributed by ajit ?>

Output:

4

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