Why must you override both equals() and hashCode() in Java?

The contract: equal objects must have the same hash code. If you override equals() but not hashCode(), two equal objects get different hash codes, and a HashSet keeps both:

import java.util.HashSet;
import java.util.Objects;
import java.util.Set;

public class Main {
    static class Bad {
        final String name;
        Bad(String name) { this.name = name; }
        @Override public boolean equals(Object o) { return o instanceof Bad b && b.name.equals(name); }
    }

    static class Good {
        final String name;
        Good(String name) { this.name = name; }
        @Override public boolean equals(Object o) { return o instanceof Good g && g.name.equals(name); }
        @Override public int hashCode() { return Objects.hash(name); }
    }

    public static void main(String[] args) {
        Set<Bad> bad = new HashSet<>();
        bad.add(new Bad("Riya"));
        bad.add(new Bad("Riya"));
        System.out.println("Without hashCode: " + bad.size());     // wrong: 2

        Set<Good> good = new HashSet<>();
        good.add(new Good("Riya"));
        good.add(new Good("Riya"));
        System.out.println("With hashCode: " + good.size());       // right: 1
    }
}

The easy way: use a record, which writes both methods for you (Java Records).

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