5. Each pair of numbers $A$ and $B$ is assigned a number $A * B$. Find $2021 * 1999$, if it is known that for any three numbers $A, B, C$ the identities are satisfied: $A * A=0$ and $A *(B * C)=(A * B)+C$.
Solution.

$A *(A * A)=A * 0=A *(B * B)=A * B+B$,

$A *(A * A)=(A * A)+A=0+A=A$, then $A * B+B=A$.

Therefore, $A * B=A-B$. Hence, $2021 * 1999=2021-1999=22$.

Answer: 22.