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Check if a number is Euler Pseudoprime
  • Last Updated : 19 May, 2021

Given an integer N and a base number A, the task is to check whether N is a Euler Pseudoprime to the given base A
An integer N is called Euler Pseudoprime to the base A, if 
 

  1. A > 0 and N is an odd composite number.
  2. A and N are co-prime i.e. GCD(A, N) = 1.
  3. A(N – 1) / 2 % N is either 1 or N – 1.

Examples: 
 

Input: N = 121, A = 3 
Output: Yes
Input: N = 343, A = 2 
Output: No 
 

 

Approach: Check all the given conditions for Euler Pseudoprime. If any one of the conditions is not true then print No else print Yes.
Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if n is composite
bool isComposite(int n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0)
            return true;
    }
    return false;
}
 
// Iterative Function to calculate
// (x^y) % p in O(log y)
int Power(int x, int y, int p)
{
 
    // Initialize result
    int res = 1;
 
    // Update x if it is greater
    // than or equal to p
    x = x % p;
 
    while (y > 0) {
 
        // If y is odd, multiply x with result
        if (y & 1) {
            res = (res * x) % p;
        }
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
// Function that returns true if N is
// Euler Pseudoprime to the base A
bool isEulerPseudoprime(int N, int A)
{
 
    // Invalid base
    if (A <= 0)
        return false;
 
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
 
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
 
    int mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
 
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
 
// Driver code
int main()
{
 
    int N = 121, A = 3;
 
    if (isEulerPseudoprime(N, A))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
// Function that returns true if n is composite
static boolean isComposite(int n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (int i = 2; i <= Math.sqrt(n); i++)
    {
        if (n % i == 0)
            return true;
    }
    return false;
}
 
// Iterative Function to calculate
// (x^y) % p in O(log y)
static int Power(int x, int y, int p)
{
 
    // Initialize result
    int res = 1;
 
    // Update x if it is greater
    // than or equal to p
    x = x % p;
 
    while (y > 0)
    {
 
        // If y is odd, multiply x with result
        if (y % 2 == 1)
        {
            res = (res * x) % p;
        }
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
// Function that returns true if N is
// Euler Pseudoprime to the base A
static boolean isEulerPseudoprime(int N, int A)
{
 
    // Invalid base
    if (A <= 0)
        return false;
 
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
 
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
 
    int mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
 
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Driver code
public static void main(String []args)
{
    int N = 121, A = 3;
 
    if (isEulerPseudoprime(N, A))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program for nth Fuss–Catalan Number
from math import gcd, sqrt
 
# Function that returns true if n is composite
def isComposite(n) :
 
    # Check if there is any divisor of n.
    # we only need check divisor till sqrt(n)
    # because if there is divisor which is greater
    # than sqrt(n) then there must be a divisor
    # which is less than sqrt(n)
    for i in range(2, int(sqrt(n)) + 1) :
        if (n % i == 0) :
            return True;
 
    return False;
 
# Iterative Function to calculate
# (x^y) % p in O(log y)
def Power(x, y, p) :
 
    # Initialize result
    res = 1;
 
    # Update x if it is greater
    # than or equal to p
    x = x % p;
 
    while (y > 0) :
 
        # If y is odd, multiply x with result
        if (y & 1) :
            res = (res * x) % p;
 
        # y must be even now
        y = y >> 1; # y = y/2
        x = (x * x) % p;
 
    return res;
 
# Function that returns true if N is
# Euler Pseudoprime to the base A
def isEulerPseudoprime(N, A) :
 
    # Invalid base
    if (A <= 0) :
        return False;
 
    # N is not a composite odd number
    if (N % 2 == 0 or not isComposite(N)) :
        return False;
 
    # If A and N are not coprime
    if (gcd(A, N) != 1) :
        return false;
 
    mod = Power(A, (N - 1) // 2, N);
    if (mod != 1 and mod != N - 1) :
        return False;
 
    # All the conditions for Euler
    # Pseudoprime are satisfied
    return True;
 
# Driver code
if __name__ == "__main__" :
 
    N = 121; A = 3;
 
    if (isEulerPseudoprime(N, A)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that returns true if n is composite
static bool isComposite(int n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (int i = 2; i <= Math.Sqrt(n); i++)
    {
        if (n % i == 0)
            return true;
    }
    return false;
}
 
// Iterative Function to calculate
// (x^y) % p in O(log y)
static int Power(int x, int y, int p)
{
 
    // Initialize result
    int res = 1;
 
    // Update x if it is greater
    // than or equal to p
    x = x % p;
 
    while (y > 0)
    {
 
        // If y is odd, multiply x with result
        if (y % 2 == 1)
        {
            res = (res * x) % p;
        }
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
// Function that returns true if N is
// Euler Pseudoprime to the base A
static bool isEulerPseudoprime(int N, int A)
{
 
    // Invalid base
    if (A <= 0)
        return false;
 
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
 
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
 
    int mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
 
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
 
static int __gcd(int a, int b)
{
    if (b == 0)
        return a;
    return __gcd(b, a % b);
     
}
 
// Driver code
public static void Main(String []args)
{
    int N = 121, A = 3;
 
    if (isEulerPseudoprime(N, A))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function that returns true if n is composite
function isComposite(n)
{
    // Check if there is any divisor of n.
    // we only need check divisor till sqrt(n)
    // because if there is divisor which is greater
    // than sqrt(n) then there must be a divisor
    // which is less than sqrt(n)
    for (let i = 2; i <= Math.sqrt(n); i++) {
        if (n % i == 0)
            return true;
    }
    return false;
}
 
// Iterative Function to calculate
// (x^y) % p in O(log y)
function Power(x, y, p) {
 
    // Initialize result
    let res = 1;
 
    // Update x if it is greater
    // than or equal to p
    x = x % p;
 
    while (y > 0) {
 
        // If y is odd, multiply x with result
        if (y & 1) {
            res = (res * x) % p;
        }
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
function __gcd(a, b) {
    if (b == 0)
        return a;
    return __gcd(b, a % b);
}
 
// Function that returns true if N is
// Euler Pseudoprime to the base A
function isEulerPseudoprime(N, A) {
 
    // Invalid base
    if (A <= 0)
        return false;
 
    // N is not a composite odd number
    if (N % 2 == 0 || !isComposite(N))
        return false;
 
    // If A and N are not coprime
    if (__gcd(A, N) != 1)
        return false;
 
    let mod = Power(A, (N - 1) / 2, N);
    if (mod != 1 && mod != N - 1)
        return false;
 
    // All the conditions for Euler
    // Pseudoprime are satisfied
    return true;
}
 
// Driver code
 
let N = 121, A = 3;
 
if (isEulerPseudoprime(N, A))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by _saurabh_jaiswal
 
</script>
Output: 
Yes

 

Time Complexity: O(sqrt(N))
 

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