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Find values of P and Q satisfying the equation N = P^2.Q

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  • Last Updated : 16 Mar, 2023
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Given a number N (1 ≤ N ≤ 9×1018), the task is to find P and Q satisfying the equation N = P 2. Q where P and Q will be prime numbers.

Examples:

Input: N = 175
Output: P = 5, Q = 7

Input: N = 2023
Output: P = 17, Q = 7

Method 1:

Approach: The problem can be solved based on the following idea:

As N = P2. Q then min(P, Q) ≤ 3√N .So we can find at least one of P and Q by doing iterations up to  3√N.

Follow the below steps to implement the idea:

  • Let M be the maximum integer i among i = 1, 2, …, [ 3√N] that divides N (actually, M=P or Q).
  • Check N/M divided by M . If it is divisible, P = M and Q = N/M2; otherwise, P = √(N/M) and Q = M2.

Below is the implementation of the above approach: 

C++




// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find P and Q
void solvePQ(int n)
{
 
    // Taking long long for big integer
    long long int p = 0, q = 0;
 
    for (long long int i = 2; (i * i * i) <= n; i++) {
 
        // If N does not divisible by i
        if (n % i != 0)
 
            continue;
 
        // if N/i divisible by i
        if ((n / i) % i == 0) {
 
            p = i;
 
            // Value of p
            q = n / (i * i);
 
            // Value of q
        }
 
        else {
            q = i;
 
            // Value of q
            p = (long long int)round(sqrt(n / i));
 
            // Value of p taking round
            // figure of square root value
        }
    }
    cout << "P is " << p << "\n"
         << "Q is " << q << endl;
}
 
// Driver code
int main()
{
 
    long long int N = 2023;
 
    // Function call
    solvePQ(N);
    return 0;
}


Java




import java.math.BigDecimal;
import java.math.RoundingMode;
 
public class GFG
{
   
  // Function to find P and Q
  public static void solvePQ(int n)
  {
 
    // Taking long long for big integer
    long p = 0, q = 0;
 
    for (long i = 2; (i * i * i) <= n; i++) {
 
      // If N does not divisible by i
      if (n % i != 0)
        continue;
 
      // if N/i divisible by i
      if ((n / i) % i == 0) {
 
        p = i;
 
        // Value of p
        q = n / (i * i);
 
        // Value of q
      }
      else {
        q = i;
 
        // Value of q
        p = (long)Math.round(Math.sqrt(n / i));
 
        // Value of p taking round
        // figure of square root value
      }
    }
    System.out.println("P is " + p);
    System.out.println("Q is " + q);
  }
 
  // Driver code
  public static void main(String[] args)
  {
 
    int N = 2023;
 
    // Function call
    solvePQ(N);
  }
}
 
// This code is contributed by hkdass001.


Python3




import math
 
def solvePQ(n):
    p = 0
    q = 0
    for i in range(2, int(math.pow(n, 1 / 3)) + 1):
        if n % i != 0:
            continue
        if (n / i) % i == 0:
            p = i
            q = n / (i * i)
        else:
            q = i
            p = int(math.sqrt(n / i))
    print(p, q)
 
N = 2023
solvePQ(N)


C#




// C# program for above approach
 
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
    // Function to find P and Q
    static void solvePQ(long n)
    {
     
        // Taking long long for big integer
        long p = 0, q = 0;
     
        for (long i = 2; (i * i * i) <= n; i++) {
     
            // If N does not divisible by i
            if (n % i != 0)
     
                continue;
     
            // if N/i divisible by i
            if ((n / i) % i == 0) {
     
                p = i;
     
                // Value of p
                q = n / (i * i);
     
                // Value of q
            }
     
            else {
                q = i;
     
                // Value of q
                p = (long)(Math.Sqrt(n / i));
     
                // Value of p taking round
                // figure of square root value
            }
        }
        Console.Write("P is " + p + "\n"
             + "Q is " + q);
    }
     
    // Driver code
    static public void Main()
    {
     
        long N = 2023;
     
        // Function call
        solvePQ(N);
    }
}


Javascript




// JavaScript code for the above approach
 
function solvePQ(n)
{
 
// Declare variables for P and Q
let p = 0;
let q = 0;
 
// Iterate from 2 to the cube root of n
for (let i = 2; i * i * i <= n; i++)
{
 
// If n is not divisible by i, continue
if (n % i !== 0) {
continue;
}
 
 
// If n/i is divisible by i
if ((n / i) % i === 0) {
  p = i;
  q = n / (i * i);
}
// Otherwise
else {
  q = i;
  p = Math.round(Math.sqrt(n / i));
}
}
console.log("P is " + p);
console.log("Q is " + q);
}
 
// Test with n = 2023
solvePQ(2023);
 
//This code is contributed by ik_9


Output

P is 17
Q is 7

Time Complexity: O(3√n)
Auxiliary space: O(1)

Method 2:

Approach:

  1. Define a function isPrime that takes an integer n and returns a boolean indicating whether n is prime or not.
    • If n is less than 2, return false.
    • For each integer i from 2 up to the square root of n, do:
      • If n is divisible by i, return false.
      • Return true.
  2. Define a function findPQ that takes an integer N and two integer references P and Q.
    • Create an empty vector primes.
    • For each integer i from 2 up to the square root of N, do:
      • If i is prime (i.e., isPrime(i) returns true), add i to the primes vector.
    • For each integer p in the primes vector, do:
      • If N is divisible by p * p and (N / (p * p)) is prime, do:
        • Set P = p and Q = N / (p * p).
        • Return from the function.
    • If no such P and Q are found, set P and Q to 0 (or any other value to indicate failure).
  3. Call the findPQ function with the input value of N and references to P and Q.
  4. Output the resulting values of P and Q.

Below is the implementation of the above approach:

C++




// CPP code of the above approach
#include <iostream>
#include <vector>
using namespace std;
 
bool isPrime(int n)
{
    if (n < 2)
        return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
void findPQ(int N, int& P, int& Q)
{
    vector<int> primes;
    for (int i = 2; i * i <= N; i++) {
        if (isPrime(i))
            primes.push_back(i);
    }
    for (int p : primes) {
        if (N % (p * p) == 0 && isPrime(N / (p * p))) {
            P = p;
            Q = N / (p * p);
            return;
        }
    }
}
 
int main()
{
    int N = 175;
    int P, Q;
    findPQ(N, P, Q);
    cout << "P = " << P << ", Q = " << Q << endl;
    return 0;
}
 
// This code is contributed by Susobhan Akhuli


Java




// Java code of the above approach
import java.util.*;
 
public class Main {
    static boolean isPrime(int n)
    {
        if (n < 2)
            return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
    static void findPQ(int N, int[] pq)
    {
        ArrayList<Integer> primes = new ArrayList<>();
        for (int i = 2; i * i <= N; i++) {
            if (isPrime(i))
                primes.add(i);
        }
        for (int p : primes) {
            if (N % (p * p) == 0 && isPrime(N / (p * p))) {
                pq[0] = p;
                pq[1] = N / (p * p);
                return;
            }
        }
    }
 
    public static void main(String[] args)
    {
        int N = 175;
        int[] pq = new int[2];
        findPQ(N, pq);
        System.out.println("P = " + pq[0] + ", Q = " + pq[1]);
    }
}
 
// This code is contributed by Susobhan Akhuli


Python3




# Python3 code of the above approach
import math
 
def isPrime(n):
    if (n < 2):
        return False
    for i in range(2, int(math.sqrt(n))+1):
        if (n % i == 0):
            return False
    return True
 
def findPQ(N):
    P, Q = None, None
    primes = []
    for i in range(2, int(math.sqrt(N))+1):
        if (isPrime(i)):
            primes.append(i)
    for p in primes:
        if (N % (p * p) == 0 and isPrime(N // (p * p))):
            P = p
            Q = N // (p * p)
            break
    return P, Q
 
N = 175
P, Q = findPQ(N)
print(f"P = {P}, Q = {Q}")


C#




//C# code of the above approach
using System;
using System.Collections.Generic;
 
public class MainClass {
    static bool IsPrime(int n)
    {
        if (n < 2)
            return false;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
 
    static void FindPQ(int N, int[] pq)
    {
        List<int> primes = new List<int>();
        for (int i = 2; i * i <= N; i++) {
            if (IsPrime(i))
                primes.Add(i);
        }
        foreach(int p in primes)
        {
            if (N % (p * p) == 0 && IsPrime(N / (p * p))) {
                pq[0] = p;
                pq[1] = N / (p * p);
                return;
            }
        }
    }
 
    public static void Main(string[] args)
    {
        int N = 175;
        int[] pq = new int[2];
        FindPQ(N, pq);
        Console.WriteLine("P = " + pq[0]
                          + ", Q = " + pq[1]);
    }
}


Javascript




// Javascript code of the above approach
function isPrime(n) {
  if (n < 2) return false;
  for (let i = 2; i * i <= n; i++) {
    if (n % i === 0) return false;
  }
  return true;
}
 
function findPQ(N) {
  const primes = [];
  for (let i = 2; i * i <= N; i++) {
    if (isPrime(i)) primes.push(i);
  }
  for (const p of primes) {
    if (N % (p * p) === 0 && isPrime(N / (p * p))) {
      return [p, N / (p * p)];
    }
  }
  return null;
}
 
const N = 175;
const [P, Q] = findPQ(N);
if (P !== undefined && Q !== undefined) {
  console.log(`P = ${P}, Q = ${Q}`);
} else {
  console.log("No valid P and Q found.");
}
 
// This code is contributed by Prajwal Kandekar


Output

P = 5, Q = 7

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