(Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.) Here are some sample runs: Enter an integer m: 1500 The smallest number n for m * n to be a perfect square is 15 m * n is 22500

Here's a possible solution in Python:

import math

# Function that returns all the factors of a given number
def factors(n):
    result = []
    i = 2
    while i <= n:
        if n % i == 0:
            result.append(i)
            n = n // i
        else:
            i += 1
    return result

# Prompt the user to enter an integer m
m = int(input("Enter an integer m: "))

# Compute the factors of m
m_factors = factors(m)

# Compute the product of the factors that appear an odd number of times
n = 1
for factor in set(m_factors):
    if m_factors.count(factor) % 2 == 1:
        n *= factor

# Compute the smallest integer n such that m * n is a perfect square
mn = m * n
sqrt_mn = int(math.sqrt(mn))
while sqrt_mn * sqrt_mn != mn:
    mn += m
    sqrt_mn = int(math.sqrt(mn))

# Print the result
print("The smallest number n for m * n to be a perfect square is", n)
print("m * n is", mn)

Sample output:

Enter an integer m: 1500
The smallest number n for m * n to be a perfect square is 15
m * n is 22500