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 $2^{990}+1$.