Java(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 the Java code to solve this problem: ``` import java.util.*; public class PerfectSquare { public static void main(String[] args) { Scanner input = new Scanner(System.in); System.out.print("Enter an integer m: "); int m = input.nextInt(); List<Integer> factors = getSmallestFactors(m); int n = 1; for (int factor : factors) { if (getFrequency(factors, factor) % 2 != 0) { n *= factor; } } System.out.println("The smallest number n for m * n to be a perfect square is " + n); System.out.println("m * n is " + (m * n * n)); } public static List<Integer> getSmallestFactors(int n) { List<Integer> factors = new ArrayList<>(); int factor = 2; while (n > 1) { if (n % factor == 0) { factors.add(factor); n /= factor; } else { factor++; } } return factors; } public static int getFrequency(List<Integer> list, int value) { int count = 0; for (int element : list) { if (element == value) { count++; } } return count; } } ``` Explanation: The `getSmallestFactors` method takes an integer `n` as input and returns a list of its smallest factors. It does this by repeatedly dividing `n` by its smallest factor until `n` becomes 1. The `getFrequency` method takes a list of integers and a value as input and returns the number of times the value appears in the list. In the `main` method, we first prompt the user to enter an integer `m`. Then we call `getSmallestFactors` to get the list of smallest factors of `m`. We then iterate through the list of factors and multiply the ones that appear an odd number of times. This gives us the smallest integer `n` such that `m * n` is a perfect square. Finally, we print out the value of `n` and `m * n * n`.