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

  • Difficulty Level : Medium
  • Last Updated : 30 Mar, 2021

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 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
?>

Javascript




<script>
 
// Javascript 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
    let 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(let 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]
            let 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
let n = 10;
let p = 5;
 
document.write(findCountOfSolutions(n, p));
            
// This code is contributed by susmitakundugoaldanga
     
</script>

Output: 

4

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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