为什么java重写equals的时候要重写hashCode？

Student类

package test.equalshashcode;

public class Student {
    private String id;
    private String name;

    public Student(String id, String name) {
        this.id = id;
        this.name = name;
    }

    @Override
    public boolean equals(Object anObject) {
        if (this == anObject) {
            return true;
        }
        if (anObject instanceof Student) {
            Student anotherStudent = (Student) anObject;
            return id.equals(anotherStudent.id);
        }
        return false;
    }
}

测试类

package test.equalshashcode;

import java.util.HashMap;
import java.util.Map;

public class Test {
    public static void main(String[] args) {
        Student student1 = new Student("STU001", "小明");
        Student student2 = new Student("STU001", "小明");
        // id相等，打印true
        System.out.println(student1.equals(student2));

        Map<Student, Integer> studentAgeMap = new HashMap<>();
        studentAgeMap.put(student1, 20);
        studentAgeMap.put(student2, 20);

        // put方法会检查hashCode(key)是否相同，相同则覆盖value
        // 此处用Object#hashCode方法，判定student1,student2不等
        // 打印2
        System.out.println(studentAgeMap.size());

        // 根据与student1和student2相等的new Student("STU001", "小明")未获取到数据
        // 打印null
        System.out.println(studentAgeMap.get(new Student("STU001", "小明")));

        // 总结
        // 当我们new两个学生对象时，根据id是否相等重写了equals方法,但是两个学生对象的hashcode不等
        // Map是根据hash(key)来判断key是否相等的
        // 上面例子出现了悖论：
        // equals相等的学生竟然能在Map的key中出现两次；
        // 根据equals相等的new学生对象，未能从Map中获取数据

        // 反证法得出结论：重写equals()必要时要重写hashCode()
    }
}