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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.  



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 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<bits/stdc++.h>
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<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]
            ll last = x + p * (n/p);
            if (last > n)
                last -= p;
  
            // Add count of numbers of 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 ", 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

/div>

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

<?php
// Program to count number of 
// values that satisfy x^2 = 1 
// mod p where x lies in range [1, n]
  
function 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 = 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]
            $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
$n = 10;
$p = 5;
echo findCountOfSolutions($n, $p);
      
// This code is contributed by ajit
?>


Output:

4

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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