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<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 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

<?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:

This article is contributed by Shubham Agrawal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

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

C++

Java

Python3

C#

PHP

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