ACM-light, more light problem

According to the "Mathematical Induction", we found out that when the serial are 1,4,9,16 ...the final bulb is on ,which means that number has a square root . On the other hand, besides those numbers, the bulb is off. Therefore , following is my Java solution:

ACM-Greatest Common Denominator

It is a Java based solution. The operation is dividing the two numbers mutually until one of them equals zero. The green part is the main mechanism .

package acm;

import java.io.*;
import java.util.*;

public class GCD {
 public static void main(String[] args) throws IOException {
  ArrayList<String> userInput = new ArrayList<String>();
  while (true) {
   int intCases = 0;
   System.out.println("How many test cases to run?(from 1 to 1000)?");
   BufferedReader bs = new BufferedReader(new InputStreamReader(
     System.in));
   String cases = bs.readLine();
   intCases = Integer.parseInt(cases);
   if (intCases < 1 || intCases > 1000) {
    System.out.println("Please enter a valid number!!");
   } else {
    System.out.println("Please enter 2 numbers m and n,separate them by dot.");
    System.out.println("ex: 12,60");
    while (intCases != userInput.size()) {
     userInput.add(bs.readLine());
    }
    break;
   }
  }
  Iterator it=userInput.iterator();
  while(it.hasNext()){
   String numbers=(String)(it.next());
   int[] ns=getInts(numbers);
   System.out.println(gcd(ns[0],ns[1]));
  }

 }

 public static int gcd(int a, int b) {
  while (a != 0 && b != 0) {
   if (a > b)
    a %= b;
   else
    b %= a;
  }
  if (a == 0)
   return b;
  else
   return a;
 }

 
 public static int[] getInts(String str){
  int totalNumbers=str.split(",").length;
  int[] ints=new int[totalNumbers];
  for(int i=0;i<totalNumbers;i++){
   ints[i]=Integer.parseInt(str.split(",")[i]);
  }
  return ints;
 }

}