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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]#includeusing 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 codeint 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 codepublic 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 coden = 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 codestatic 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?>

## Javascript



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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