Check if a number is an Achilles number or not in Python

An Achilles number is a number that is powerful but not a perfect power. A powerful number means that for every prime factor p, p² also divides the number. A perfect power is a number that can be expressed as a^b where a and b are positive integers and b > 1.

Some examples of Achilles numbers are: 72, 108, 200, 288, 392, 432, 500, 648, 675, 800, 864, 968, 972, 1125.

Understanding Achilles Numbers

For example, 108 is an Achilles number because:

• 108 = 2² × 3³ (powerful: both 2² and 3² divide 108)
• 108 cannot be expressed as a^b for integers a, b > 1

Algorithm Steps

To check if a number is an Achilles number ?

1. Check if the number is powerful
2. Check if the number is NOT a perfect power
3. Return True only if both conditions are met

Implementation

Function to Check if Number is Powerful

from math import sqrt, log

def check_powerful(n):
    # Check for factor 2
    while (n % 2 == 0):
        p = 0
        while (n % 2 == 0):
            n //= 2
            p += 1
        if (p == 1):
            return False 
    
    # Check for odd factors
    limit = int(sqrt(n)) + 1
    for factor in range(3, limit, 2):
        p = 0
        while (n % factor == 0):
            n = n // factor
            p += 1
        if (p == 1):
            return False
    
    return (n == 1)

# Test the function
print(check_powerful(108))
print(check_powerful(12))

True
False

Function to Check if Number is Perfect Power

def check_power(a):
    if (a == 1):
        return True
    
    # Check for possible bases up to sqrt(a)
    limit = int(sqrt(a)) + 1
    for i in range(2, limit):
        val = log(a) / log(i)
        if (abs(val - round(val)) < 1e-9):
            return True
    
    return False

# Test the function
print(check_power(8))   # 2^3
print(check_power(108)) # Not a perfect power

True
False

Complete Achilles Number Checker

from math import sqrt, log

def check_powerful(n):
    # Check for factor 2
    while (n % 2 == 0):
        p = 0
        while (n % 2 == 0):
            n //= 2
            p += 1
        if (p == 1):
            return False 
    
    # Check for odd factors
    limit = int(sqrt(n)) + 1
    for factor in range(3, limit, 2):
        p = 0
        while (n % factor == 0):
            n = n // factor
            p += 1
        if (p == 1):
            return False
    
    return (n == 1)

def check_power(a):
    if (a == 1):
        return True
    
    limit = int(sqrt(a)) + 1
    for i in range(2, limit):
        val = log(a) / log(i)
        if (abs(val - round(val)) < 1e-9):
            return True
    
    return False

def isAchilles(n):
    return check_powerful(n) and not check_power(n)

# Test with multiple numbers
test_numbers = [72, 108, 200, 64, 100]
for num in test_numbers:
    result = isAchilles(num)
    print(f"{num} is {'an Achilles' if result else 'not an Achilles'} number")

72 is an Achilles number
108 is an Achilles number
200 is an Achilles number
64 is not an Achilles number
100 is not an Achilles number

Example Analysis

Let's verify why 108 is an Achilles number ?

n = 108
print(f"Checking {n}:")
print(f"Is powerful: {check_powerful(n)}")
print(f"Is perfect power: {check_power(n)}")
print(f"Is Achilles: {isAchilles(n)}")

# Prime factorization: 108 = 2² × 3³
print(f"\nPrime factorization of {n}: 2² × 3³")
print("Powerful because 2² and 3² both divide 108")
print("Not a perfect power because it cannot be expressed as a^b")

Checking 108:
Is powerful: True
Is perfect power: False
Is Achilles: True

Prime factorization of 108: 2² × 3³
Powerful because 2² and 3² both divide 108
Not a perfect power because it cannot be expressed as a^b

Conclusion

An Achilles number must be both powerful (every prime factor appears with power ? 2) and not a perfect power. The algorithm checks both conditions by examining prime factorization and testing for perfect power relationships using logarithms.