Japan Mathematical Olympiad

Denote the equation $f(m + n)^2 = f(m|f(n)|) + f(n^2)$ by $P(m, n)$. $P(0, 0)$ implies $f(0)^2 = 2f(0)$, thus $f(0) \in \{0, 2\}$.

When $f(0) = 0$, $P(m, 0)$ implies $f(m) = 0$ for any $m \in \mathbb{N}_0$. Since this $f$ satisfies the condition, there is just one possible tuple $(f(0), f(1), \dots, f(2024))$ in this case.

We consider the case $f(0) = 2$. $P(0, 1)$ implies $f(1)^2 = f(1)+2$, thus we have $f(1) \in \{-1, 2\}$. If $f(1) = -1$, $P(1, 1)$ implies $f(2)^2 = -2$ which contradicts $f(2) \in \mathbb{Z}$. Therefore we have $f(1) = 2$. Then $P(m, 1)$ and $P(m, 0)$ imply that $f(m+1)^2 = f(2m)+2 = f(m)^2$. By induction we have $|f(m)| = 2$ for any $m \in \mathbb{N}_0$. Moreover, $f(2m) = f(m)^2 - 2 = 2$ holds. On the other hand, $P(0, n)$ implies $f(n^2) = f(n)^2 - f(0) = 2$ for any $n \in \mathbb{N}_0$.

Conversely, if we choose $f: \mathbb{N}_0 \to \{-2, 2\}$ so that $f(n) = 2$ whenever $n$ is even or square, $f$ satisfies the condition because $f(m+n)^2 = 4$, $f(m|f(n)|) = f(2m) = 2$ and $f(n^2) = 2$ for any $m, n \in \mathbb{N}_0$. Since there exist 990 integers between 0 to 2024 which are neither even nor square, the number of possible tuples $(f(0), f(1), \dots, f(2024))$ in this case is $2^{990}$.

Therefore the answer is $2^{990} + 1$.