Solution: Let the sets be $$\mathbb{A} = \{0, 1, 4, 5, 16, 17, 20, 21, ...\}$$ $$\mathbb{B} = \{0, 2, 8, 10, 32, 34, 40, 42, ...\}.$$ Note that \(\mathbb{A}\) is the sum of any finite amount of powers of 4, and \(\mathbb{B}\) is twice every element \(a \in\mathbb{A}\). We'll now prove these two subsets satisfy the problem statement. Let \(k\) be a non-negative integer, and let \((d_m)\) be its digits in base 4. Make a table by writing the digits in columns like so: $$ \begin{array}{r|r|} & d_m & d_{m-1} & \cdots & d_2 & d_1 \\ \hline A & & & & & \\ \hline B & & & & & \\ \hline \end{array} $$ For each column, write a 0 or a 1 in row \(A\) and a 0 or a 2 in row \(B\) following these cases: $$ \begin{array}{r|r|} & 0 \\ \hline A & 0 \\ \hline B & 0 \\ \hline \end{array} \hspace{1cm} \begin{array}{r|r|} & 1 \\ \hline A & 1 \\ \hline B & 0 \\ \hline \end{array} \hspace{1cm} \begin{array}{r|r|} & 2 \\ \hline A & 0 \\ \hline B & 2 \\ \hline \end{array} \hspace{1cm} \begin{array}{r|r|} & 3 \\ \hline A & 1 \\ \hline B & 2 \\ \hline \end{array} $$ Note that for each possible digit there's a distinct pair of numbers in rows \(A\) and \(B\) that add to it, so by uniqueness of digit representations each possible \(k\) gives a unique table. By doing this procedure (ignoring leading zeroes), we'll be writing in base 4 two numbers in rows \(A\) and \(B\) that add to \(k\), necessarily belonging to sets \(\mathbb{A}\) and \(\mathbb{B}\) respectively. This procedure is also reversible by first putting an \(a\in\mathbb{A}\) in row \(A\) and a \(b\in\mathbb{B}\) in row \(B\), so any pair of members \(\mathbb{A}\) and \(\mathbb{B}\) sum to a unique non-negative integer. □