Collections such as HashMap and HashSet use a hashcode value of an object to determine how it should be stored inside a collection, and the hashcode is used again in order to locate the object in its collection.

Hashing retrieval is a two-step process:

1. Find the right bucket (using hashCode())
2. Search the bucket for the right element (using equals() )

Here is a small example on why we should overrride equals() and hashcode().

Consider an Employee class which has two fields: age and name.

public class Employee {

    String name;
    int age;

    public Employee(String name, int age) {
        this.name = name;
        this.age = age;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public boolean equals(Object obj) {
        if (obj == this)
            return true;
        if (!(obj instanceof Employee))
            return false;
        Employee employee = (Employee) obj;
        return employee.getAge() == this.getAge()
                && employee.getName() == this.getName();
    }

    // commented    
    /*  @Override
        public int hashCode() {
            int result=17;
            result=31*result+age;
            result=31*result+(name!=null ? name.hashCode():0);
            return result;
        }
     */
}

Now create a class, insert Employee object into a HashSet and test whether that object is present or not.

public class ClientTest {
    public static void main(String[] args) {
        Employee employee = new Employee("rajeev", 24);
        Employee employee1 = new Employee("rajeev", 25);
        Employee employee2 = new Employee("rajeev", 24);

        HashSet<Employee> employees = new HashSet<Employee>();
        employees.add(employee);
        System.out.println(employees.contains(employee2));
        System.out.println("employee.hashCode():  " + employee.hashCode()
        + "  employee2.hashCode():" + employee2.hashCode());
    }
}

It will print the following:

false
employee.hashCode():  321755204  employee2.hashCode():375890482

Now uncomment hashcode() method , execute the same and the output would be:

true
employee.hashCode():  -938387308  employee2.hashCode():-938387308

Now can you see why if two objects are considered equal, their hashcodes must also be equal? Otherwise, you'd never be able to find the object since the default hashcode method in class Object virtually always comes up with a unique number for each object, even if the equals() method is overridden in such a way that two or more objects are considered equal. It doesn't matter how equal the objects are if their hashcodes don't reflect that. So one more time: If two objects are equal, their hashcodes must be equal as well.

According to the doc, the default implementation of hashCode will return some integer that differ for every object

As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation
technique is not required by the JavaTM programming language.)

However some time you want the hash code to be the same for different object that have the same meaning. For example

Student s1 = new Student("John", 18);
Student s2 = new Student("John", 18);
s1.hashCode() != s2.hashCode(); // With the default implementation of hashCode

This kind of problem will be occur if you use a hash data structure in the collection framework such as HashTable, HashSet. Especially with collection such as HashSet you will end up having duplicate element and violate the Set contract.