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++

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// 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;
}

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Java

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// 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 

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Python3

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# Pyhton 3 program to possible prime number
from math import sqrt
  
# Function to check if a 
# number is prime or not
def Prime(n):
    for j in range(2, int(sqrt(n)) + 1, 1):
        if (n % j == 0):
            return False
  
    return True
  
# Function to find possible prime number
def find_prime(x, xsqmodp, y, ysqmodp):
    n = x * x - xsqmodp
    n1 = y * y - ysqmodp
      
    # Find a possible prime number
    for j in range(2, max(int(sqrt(n)), 
                          int(sqrt(n1))), 1):
        if (n % j == 0 and (x * x) % j == xsqmodp and 
            n1 % j == 0 and (y * y) % j == ysqmodp):
            if (Prime(j)):
                return j
                  
        j1 = n // j
        if (n % j1 == 0 and (x * x) % j1 == xsqmodp and
            n1 % j1 == 0 and (y * y) % j1 == ysqmodp):
            if (Prime(j1)):
                return j1
          
        j1 = n1 // j
        if (n % j1 == 0 and (x * x) % j1 == xsqmodp and 
            n1 % j1 == 0 and (y * y) % j1 == ysqmodp):
            if (Prime(j1)):
                return j1
      
    # Last condition
    if (n == n1):
        return n
  
# Driver code
if __name__ == '__main__':
    x = 3
    xsqmodp = 0
    y = 5
    ysqmodp = 1
      
    # Function call
    print(find_prime(x, xsqmodp, y, ysqmodp))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// 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..

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Output:

3


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