使用给定的四个整数求素数P

给定四个整数X，Y，X ² ％P，Y ² ％P，其中P是质数。任务是找到素数P。

注意：答案总是存在的。

例子：

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

方法 ：
从以上数字我们可以得出，

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

现在，从这两个方程中找到所有公有质因数，并检查它是否满足原始方程，如果满足(其中一个将始终存在答案)，那就是答案。

下面是上述方法的实现：

C++

// CPP program to possible prime number
#include 
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;
}

Java

// 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

Python3

# Python3 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

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..

输出：

3