square numbers Java

Question

"An array is used to store ten integer numbers. Write a Java program that determines and print the square numbers which are also odd numbers from the given array."

The problem I have is how to figure out if the number in the array is a square number. I tried this way but it's not correct!

import java.math.*;
public class JavaApplication43 {

    public static void main(String[] args) {

        int[] no = {22, 44, 25, 89, 81, 55, 23, 25, 55};

        for (int i = 0; i < no.length; i++) {

            int x = no[i];
            double y;
            if (x % 2 != 0) {
                y = Math.sqrt(x);
                if (x == (Math.pow(y, 2))) 
                    System.out.println(no[i]);
            }
       }
   }
}

This is the output it gives me

run:
25
81
55
25
55

55 is there too that means this method I used is not successful!

Answer 1

You could just do:

for (int i = 0; i < no.length; i++) {
    int x = no[i];
    if (x % 2 == 0) continue;
    int y = (int) Math.sqrt(x);
    if (x == y * y) { 
        System.out.println(x);
    }
}

Answer 2

You can determine if a number is a square by checking if its square root is an integer.

double sqrt = Math.sqrt(x);
long sqrt2 = Math.round(sqrt);
if (Math.abs(sqrt - sqrt2) / sqrt < 1e-15)
   // we have a square.

If you know x is an int you shouldn't get a rounding error and you can do

int x = ...
double sqrt = Math.sqrt(x);
if ((int) sqrt == sqrt)
    // we have a square.