Blum Integer

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Blum Integer is a semi-prime number, suppose p and q are the two factors (i.e. n = p * q), they(p and q) are of the form 4t + 3, where t is some integer.
First few Blum Integers are 21, 33, 57, 69, 77, 93, 129, 133, 141, 161, 177, …

Note: Because of the condition that both the factors should be semi-primes, even numbers can not be Blum integers neither can be the numbers below 20,
So we have to check only for an odd integer greater than 20 if it is a Blum Integer or not.

Examples :

Input: 33
Output: Yes
Explanation: 33 = 3 * 11, 3 and 11 are both 
semi-primes as well as of the form 4t + 3 
for t = 0, 2

Input: 77
Output:  Yes
Explanation: 77 = 7 * 11, 7 and 11 both are 
semi-prime as well as of the form 4t + 3 
for t = 1, 2

Input: 25
Output: No
Explanation: 25 = 5*5, 5 and 5 are both 
semi-prime  but are not of the form 4t 
+ 3, Hence 25 is not a Blum integer.

Approach: For a given odd integer greater than 20, we calculate the prime numbers from 1 to that odd integer. If we find any prime number that divides that odd integer and its quotient both are prime and follow the form 4t + 3 for some integer, then the given odd integer is Blum Integer.

Below is the implementation of above approach :

C++

// CPP program to check if a number is a Blum
// integer
#include <bits/stdc++.h>
using namespace std;
 
// Function to cheek if number is Blum Integer
bool isBlumInteger(int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));
 
    // to store prime numbers from 2 to n
    for (int i = 2; i * i <= n; i++) {
 
        // If prime[i] is not
        // changed, then it is a prime
        if (prime[i] == true) {
 
            // Update all multiples of p
            for (int j = i * 2; j <= n; j += i)
                prime[j] = false;
        }
    }
 
    // to check if the given odd integer
    // is Blum Integer or not
    for (int i = 2; i <= n; i++) {
        if (prime[i]) {
 
            // checking the factors
            // are of 4t+3 form or not
            if ((n % i == 0) && ((i - 3) % 4) == 0) {
                int q = n / i;
                return (q != i && prime[q] && 
                           (q - 3) % 4 == 0);
            }
        }
    }
 
    return false;
}
 
// driver code
int main()
{
    // give odd integer greater than 20
    int n = 249;
 
    if (isBlumInteger(n))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java implementation to check If
// a number is a Blum integer
import java.util.*;
class GFG {
    public static boolean isBlumInteger(int n)
    {
        boolean prime[] = new boolean[n + 1];
        for (int i = 0; i < n; i++)
            prime[i] = true;
 
        // to store prime numbers from 2 to n
        for (int i = 2; i * i <= n; i++) {
 
            // If prime[i] is not changed,
            // then it is a prime
            if (prime[i] == true) {
 
                // Update all multiples of p
                for (int j = i * 2; j <= n; j += i)
                    prime[j] = false;
            }
        }
 
        // to check if the given odd integer
        // is Blum Integer or not
        for (int i = 2; i <= n; i++) {
            if (prime[i]) {
 
                // checking the factors are
                // of 4t + 3 form or not
                if ((n % i == 0) && ((i - 3) % 4) == 0) {
                    int q = n / i;
                    return (q != i && prime[q] && 
                            (q - 3) % 4 == 0);
                }
            }
        }
        return false;
    }
 
    // driver code
    public static void main(String[] args)
    {
        // give odd integer greater than 20
        int n = 249;
 
        if (isBlumInteger(n) == true)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# python 3 program to check if a
# number is a Blum integer
 
# Function to cheek if number 
# is Blum Integer
def isBlumInteger(n):
 
    prime = [True]*(n + 1)
 
    # to store prime numbers from 2 to n
    i = 2
    while (i * i <= n):
 
        # If prime[i] is not
        # changed, then it is a prime
        if (prime[i] == True) :
 
            # Update all multiples of p
            for j in range(i * 2, n + 1, i):
                prime[j] = False
        i = i + 1
     
    # to check if the given odd integer
    # is Blum Integer or not
    for i in range(2, n + 1) :
        if (prime[i]) : 
 
            # checking the factors
            # are of 4t+3 form or not
            if ((n % i == 0) and
                        ((i - 3) % 4) == 0):
                q = int(n / i)
                return (q != i and prime[q] 
                       and (q - 3) % 4 == 0)
             
    return False
 
# driver code
# give odd integer greater than 20
n = 249
if (isBlumInteger(n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Smitha.

C#

// C# implementation to check If
// a number is a Blum integer
using System;
 
class GFG {
     
    public static bool isBlumInteger(int n)
    {
        bool[] prime = new bool[n + 1];
        for (int i = 0; i < n; i++)
            prime[i] = true;
 
        // to store prime numbers from 2 to n
        for (int i = 2; i * i <= n; i++) {
 
            // If prime[i] is not changed,
            // then it is a prime
            if (prime[i] == true) {
 
                // Update all multiples of p
                for (int j = i * 2; j <= n; j += i)
                    prime[j] = false;
            }
        }
 
        // to check if the given odd integer
        // is Blum Integer or not
        for (int i = 2; i <= n; i++) {
            if (prime[i]) {
 
                // checking the factors are
                // of 4t + 3 form or not
                if ((n % i == 0) && ((i - 3) % 4) == 0) 
                {
                    int q = n / i;
                    return (q != i && prime[q] && 
                           (q - 3) % 4 == 0);
                }
            }
        }
        return false;
    }
     
    // Driver code
    static public void Main ()
    {
        // give odd integer greater than 20
        int n = 249;
 
        if (isBlumInteger(n) == true)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by Ajit.

PHP

<?php
// PHP program to check if a
// number is a Blum integer
 
// Function to cheek if 
// number is Blum Integer
function isBlumInteger($n)
{
    $prime = array_fill(0, $n + 1, true);
 
    // to store prime 
    // numbers from 2 to n
    for ($i = 2; $i * $i <= $n; $i++) 
    {
 
        // If prime[i] is not
        // changed, then it is a prime
        if ($prime[$i] == true) 
        {
 
            // Update all multiples of p
            for ($j = $i * 2; $j <= $n; $j += $i)
                $prime[$j] = false;
        }
    }
 
    // to check if the given
    // odd integer is Blum 
    // Integer or not
    for ($i = 2; $i <= $n; $i++) 
    {
        if ($prime[$i]) 
        {
 
            // checking the factors
            // are of 4t+3 form or not
            if (($n % $i == 0) && 
               (($i - 3) % 4) == 0) 
            {
                $q = (int)$n / $i;
                return ($q != $i && $prime[$q] && 
                             ($q - 3) % 4 == 0);
            }
        }
    }
 
    return false;
}
 
// Driver code
 
// give odd integer
// greater than 20
$n = 249;
 
if (isBlumInteger($n))
    echo "Yes";
else
    echo "No";
 
// This code is contributed by mits.
?>

Javascript

<script>
 
// Javascript implementation to check If
// a number is a Blum integer
function isBlumInteger(n)
{
    let prime = new Array(n + 1);
    for(let i = 0; i < n; i++)
        prime[i] = true;
 
    // To store prime numbers from 2 to n
    for(let i = 2; i * i <= n; i++) 
    {
         
        // If prime[i] is not changed,
        // then it is a prime
        if (prime[i] == true)
        {
             
            // Update all multiples of p
            for(let j = i * 2; j <= n; j += i)
                prime[j] = false;
        }
    }
 
    // To check if the given odd integer
    // is Blum Integer or not
    for(let i = 2; i <= n; i++)
    {
        if (prime[i])
        {
             
            // Checking the factors are
            // of 4t + 3 form or not
            if ((n % i == 0) && ((i - 3) % 4) == 0)
            {
                let q = parseInt(n / i, 10);
                return (q != i && prime[q] &&
                       (q - 3) % 4 == 0);
            }
        }
    }
    return false;
}
 
// Driver code
 
// Give odd integer greater than 20
let n = 249;
 
if (isBlumInteger(n) == true)
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by decode2207
 
</script>

Output:

Yes

Time Complexity: O(nsqrtn)
Auxiliary Space: O(n)

Please suggest if someone has a better solution which is more efficient in terms of space and time.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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