Davi has a very original calculator that performs only two operations, the usual addition $(+)$ and another operation, denoted by $*$, which satisfies

(i) $a * a=a$,

(ii) $a * 0=2a$ and

(iii) $(a * b) + (c * d) = (a + c) * (b + d)$,

for any integers $a$ and $b$. What are the results of the operations $(2 * 3) + (0 * 3)$ and $1024 * 48$?
Solution 1: To calculate $(2 * 3)+(0 * 3)$, we use properties (i), (ii), and (iii), obtaining

$$
\begin{array}{rll}
(2 * 3)+(0 * 3) & \stackrel{(\mathrm{iii})}{=} & (2+0) *(3+3) \\
= & (6+(-4)) *(6+0) \\
& \stackrel{(\mathrm{iii})}{=} & (6 * 6)+((-4) * 0) \\
& \stackrel{(\mathrm{i})}{=}(\mathrm{ii}) & 6+(-4) \times 2 \\
& =6-8=-2
\end{array}
$$

To calculate $1024 * 48$, observe that $1024=976+48$. Thus,

$$
\begin{aligned}
1024 * 48 & =(976+48) *(0+48) \\
& =(976 * 0)+(48 * 48) \\
& =976 \times 2+48 \\
& =1952+48=2000
\end{aligned}
$$

Solution 2: By properties (i), (ii), and (iii),

$$
\begin{aligned}
a * b & =((a-b)+b) *(0+b) \\
& =((a-b) * 0)+(b * b) \\
& =(a-b) \times 2+b \\
& =2 a-2 b-b \\
& =2 a-b
\end{aligned}
$$

for any integers $a$ and $b$. Thus,

$$
(2 * 3)+(0 * 3)=(2 \times 2-3)+(2 \times 0-3)=1-3=-2
$$

$$
1024 * 48=2 \times 1024-48=2048-48=2000
$$

Observation: There is a unique operation $*$ on the integers with properties (i), (ii), and (iii) of the statement, namely, $a * b=2 a-b$, as we showed in the second solution. However, even restricting the domain of $*$ to non-negative integers, it can be shown that an operation with the properties (i), (ii), and (iii) of the statement exists and is unique, given by $a * b=2 a-b$, but now we need to restrict ourselves to non-negative integers $a, b$ such that $2 a \geq b$, so that the result of the operation is still a non-negative integer. Let us denote the set of non-negative integers by $\mathbb{N}^{*}$.

It is clear that the deduction made in the second solution applies only if $a \geq b$ because, in this case, $a-b \in \mathbb{N}^{*}$. To prove the existence and uniqueness of the operation $a * b$ on non-negative integers such that $2 a \geq b$, we need a more subtle argument, as follows.

Assuming that the operation $a * b$ is defined on $\mathbb{N}^{*}$ whenever $2 a \geq b$, giving a result in $\mathbb{N}^{*}$ and satisfying properties (i), (ii), and (iii) of the statement, we claim that $a * b=2 a-b$ holds always. Indeed, given $c \in \mathbb{N}^{*}$, we have $c *(2 c)=0$, since we can cancel $2 c$ from both sides of the equality

$$
2 c=(2 c) *(2 c)=(c+c) *(2 c+0)=(c *(2 c))+(c * 0)=(c *(2 c))+2 c
$$

Now, given $a, b \in \mathbb{N}^{*}$, with $2 a \geq b$ and $b>a$, we have $2 a-b, b-a \in \mathbb{N}^{*}$ and

$$
\begin{aligned}
a * b & =((2 a-b)+(b-a)) *((2 a-b)+(2 b-2 a)) \\
& =(2 a-b) *(2 a-b)+(b-a) *(2(b-a)) \\
& =(2 a-b)+0=2 a-b
\end{aligned}
$$

Finally, for $a, b \in \mathbb{N}^{*}$, with $2 a \geq b$ and $a \geq b$, this has already been shown in the second solution. Thus, $a * b=2 a-b$ holds for any $a, b \in \mathbb{N}^{*}$ such that $2 a \geq b$.