Find the prime P using given four integers

Given four integers X, Y, X2%P, Y2%P, where P is a prime number. The task is to find prime P.

Note: Answer always exsits.

Examples:

Input : X = 3, XsqmodP = 0, Y = 5, YsqmodP = 1

Output : 3
When x = 3, x2 = 9, and 9 modulo P is 0. So possible value of p is 3
When x = 5, x2 = 25, and 25 modulo P is 1. So possible value of p is 3

Input : X = 4, XsqmodP = 1, Y = 5, YsqmodP = 0
Output : 5

Approach :
From above numbers we get,

X2 – XsqmodP = 0 mod P
Y2 – YsqmodP = 0 mod P

Now find all the common prime factors from both the equation, and check if it satisfies the original equation, If it does (one of them will since answer always exists) then that’s the answer.

Below is the implementation of the above approach :

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to possible prime number
#include <bits/stdc++.h>
using namespace std;
  
// Function to check if a 
// number is prime or not
bool Prime(int n)
{
    for (int j = 2; j <= sqrt(n); j++)
        if (n % j == 0)
            return false;
  
    return true
}
  
  
// Function to find possible prime number
int find_prime(int x, int xsqmodp , int y, int ysqmodp)
{
      
    int n = x*x - xsqmodp;
    int n1 = y*y - ysqmodp;
      
    // Find a possible prime number
    for (int j = 2; j <= max(sqrt(n), sqrt(n1)); j++) 
    {
        if (n % j == 0 && (x * x) % j == xsqmodp && 
                  n1 % j == 0 && (y * y) % j == ysqmodp)
            if (Prime(j))
                return j;
                  
        int j1 = n / j;
        if (n % j1 == 0 && (x * x) % j1 == xsqmodp 
             && n1 % j1 == 0 && (y * y) % j1 == ysqmodp)
            if (Prime(j1))
                return j1;
          
        j1 = n1 / j;
        if (n % j1 == 0 && (x * x) % j1 == xsqmodp && 
                n1 % j1 == 0 && (y * y) % j1 == ysqmodp)
            if (Prime(j1))
                return j1;
    }
      
    // Last condition
    if (n == n1)
        return n;
}
  
// Driver code
int main()
{
    int x = 3, xsqmodp = 0, y = 5, ysqmodp = 1;
      
    // Function call
    cout << find_prime(x, xsqmodp, y, ysqmodp);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to possible prime number
import java.util.*;
class GFG
{
  
// Function to check if a 
// number is prime or not
static boolean Prime(int n)
{
    for (int j = 2
             j <= Math.sqrt(n); j++)
        if (n % j == 0)
            return false;
  
    return true
}
  
// Function to find possible prime number
static int find_prime(int x, int xsqmodp , 
                      int y, int ysqmodp)
{
    int n = x * x - xsqmodp;
    int n1 = y * y - ysqmodp;
      
    // Find a possible prime number
    for (int j = 2
             j <= Math.max(Math.sqrt(n), 
                           Math.sqrt(n1)); j++) 
    {
        if (n % j == 0 && (x * x) % j == xsqmodp && 
            n1 % j == 0 && (y * y) % j == ysqmodp)
            if (Prime(j))
                return j;
                  
        int j1 = n / j;
        if (n % j1 == 0 && (x * x) % j1 == xsqmodp && 
            n1 % j1 == 0 && (y * y) % j1 == ysqmodp)
            if (Prime(j1))
                return j1;
          
        j1 = n1 / j;
        if (n % j1 == 0 && (x * x) % j1 == xsqmodp && 
            n1 % j1 == 0 && (y * y) % j1 == ysqmodp)
            if (Prime(j1))
                return j1;
    }
      
    // Last condition
    if (n == n1)
        return n;
    return Integer.MIN_VALUE;
}
  
// Driver code
public static void main(String[] args) 
{
    int x = 3, xsqmodp = 0
        y = 5, ysqmodp = 1;
      
    // Function call
    System.out.println(find_prime(x, xsqmodp, 
                                  y, ysqmodp));
}
}
  
// This code is contributed by PrinciRaj1992 

chevron_right


C#

// C# program to possible prime number
using System;

class GFG
{

// Function to check if a
// number is prime or not
static bool Prime(int n)
{
for (int j = 2;
j <= Math.Sqrt(n); j++) if (n % j == 0) return false; return true; } // Function to find possible prime number static int find_prime(int x, int xsqmodp , int y, int ysqmodp) { int n = x * x - xsqmodp; int n1 = y * y - ysqmodp; // Find a possible prime number for (int j = 2; j <= Math.Max(Math.Sqrt(n), Math.Sqrt(n1)); j++) { if (n % j == 0 && (x * x) % j == xsqmodp && n1 % j == 0 && (y * y) % j == ysqmodp) if (Prime(j)) return j; int j1 = n / j; if (n % j1 == 0 && (x * x) % j1 == xsqmodp && n1 % j1 == 0 && (y * y) % j1 == ysqmodp) if (Prime(j1)) return j1; j1 = n1 / j; if (n % j1 == 0 && (x * x) % j1 == xsqmodp && n1 % j1 == 0 && (y * y) % j1 == ysqmodp) if (Prime(j1)) return j1; } // Last condition if (n == n1) return n; return int.MinValue; } // Driver code public static void Main() { int x = 3, xsqmodp = 0, y = 5, ysqmodp = 1; // Function call Console.WriteLine(find_prime(x, xsqmodp, y, ysqmodp)); } } // This code is contributed by anuj_67.. [tabbyending]

Output:

3


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : princiraj1992, vt_m