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

We are given with integers x and p. The goal is to find the number of solutions of the equation −x²=1 ( mod p ) such that x lies in range [1,N].

We will do this by traversing from 1 to N and take each number as x check if (x*x)%p==1. If yes then increment the count.

Let’s understand with examples.

Input − n=5, p=2

Output − Number of Solutions − 3

Explanation − Between the range 1 to 5.

1<sup>2</sup>=1%2=1, count=1
2<sup>2</sup>=4%2=0, count=1
3<sup>2</sup>=9%2=1, count=2
4<sup>2</sup>=16%2=0, count=2
5<sup>2</sup>=25%2=1, count=3
Total number of solutions=3.

Input − n=3, p=4

Output − Number of Solutions − 2

Explanation − Between the range 1 to 3.

1<sup>2</sup>=1%4=1, count=1
2<sup>2</sup>=4%4=0, count=1
3<sup>2</sup>=9%4=1, count=2
Total number of solutions=2

Approach used in the below program is as follows

• We take two variables n and p.

• Function solutionsCount(int n,int p) takes both parameters n and p and returns the number of solutions for equation: x²%p==1 (or x²=1 ( mod p ) ).

• Starting from x=1 to x=n, check if x*x==1, if yes increment count.

• At the end of the loop, count will have the number of solutions.

• Return count as result.

Example

#include<bits/stdc++.h>
using namespace std;
int solutionsCount(int n, int p){
   int count = 0;
   for (int x=1; x<=n; x++){
      if ((x*x)%p == 1)
         { ++count; }
   }
   return count;
}
int main(){
   int n = 8, p = 3;
   cout<<"Number of solutions :"<<solutionsCount(n, p);
   return 0;
}

Output

If we run the above code it will generate the following output −

Number of solutions : 6