How to find the square root of an integer using binary search

import java.math.BigInteger;
class SquareRoot {
   public static final BigInteger NOT_A_PERFECT_SQUARE = new BigInteger("-1");
   public static BigInteger squareRoot(long n){
       /*Base condition when n=0 because for this case, left would
       be greater than the right and sqaure root wouldn't be computed
       for 0 and -1 will b returned.*/
       if (n==0)
       {
           return BigInteger.valueOf(0);
       }
       // For n > 0
       BigInteger left = new BigInteger("1");
       BigInteger right = BigInteger.valueOf(n);
       BigInteger middle = new BigInteger("1");
       while((left.compareTo(right)) == 0 || (left.compareTo(right)) == -1) {
          middle = (left.add(right)).divide(BigInteger.valueOf(2));
          BigInteger square = middle.multiply(middle);
          int check1 = square.compareTo(BigInteger.valueOf(n));
          if (check1==0) {
              return middle;
          }
          if (check1 == -1) {
              left = middle.add(BigInteger.valueOf(1));
          } else {
              right = middle.subtract(BigInteger.valueOf(1));
          }
       }
       return NOT_A_PERFECT_SQUARE;       
   }
 
   public static void main( String args[] ) {
       System.out.println( "Square root of 24 : " + squareRoot(24));
       System.out.println( "Square root of 1: " + squareRoot(1));
       System.out.println( "Square root of 4: " + squareRoot(4));
       System.out.println( "Square root of 9: " + squareRoot(9));
       System.out.println( "Square root of 16: " + squareRoot(16));
       System.out.println( "Square root of 25: " + squareRoot(25));
       System.out.println( "Square root of 36: " + squareRoot(36));
       System.out.println( "Square root of 49: " + squareRoot(49));
       System.out.println( "Square root of 64: " + squareRoot(64));
       System.out.println( "Square root of 81: " + squareRoot(81));
       System.out.println( "Square root of 100: " + squareRoot(100));
   }
}