Japan Mathematical Olympiad

$$\boxed{\left(\genfrac{}{}{0ex}{}{2022}{1011}\right)}$$ We say that $x$ is a fixed point of $f$ if we have $f(x)=x$. Suppose that a function $f$ belonging to $\mathcal{A}$ has two fixed points $a$ and $b$ with $a<b$. Then, it follows from the first condition that $a=f(a)\ge f(b)=b$, which is a contradiction. Therefore, each function belonging to $\mathcal{A}$ has at most one fixed point.

Suppose that $\mathcal{A}$ is non-empty. Take a function $f$ belonging to $\mathcal{A}$, and define $a=f(f(1))$. Then, $a$ is a fixed point of $f$ since we have $f(f(1))=f(f(f(1)))$ by the second condition. Let $g$ be a function belonging to $\mathcal{A}$. Then, we have $$g(a)=g(f(a))=f(g(f(a)))=f(g(a))$$ by the second condition. It shows that $g(a)$ is also a fixed point of $f$. By the uniqueness of the fixed point of $f$, we conclude that $g(a)=a$. Therefore, we have proved that $a$ is the unique fixed point of any function $g$ belonging to $\mathcal{A}$. Let $f$ and $g$ be functions belonging to $\mathcal{A}$, and $x$ an integer from $1$ to $2023$. Then, the second condition says that $f(g(x))$ is a fixed point of $g$, and hence, we have $f(g(x))=a$. Let $X$ denote the set of integers $x$ from $1$ to $2023$ such that $f(x)=a$ holds for any function $f$ belonging to $\mathcal{A}$. Let $\ell$ be the smallest element of $X$, and $m$ the largest element of $X$. Then, for any function $f$ belonging to $\mathcal{A}$, we have $a=f(\ell )\ge f(\ell +1)\ge \cdots \ge f(m)=a$ by the first condition. Therefore, we conclude that $X$ is the set of integers from $\ell$ to $m$. Note that we already proved that $f(g(x))=a$ for any functions $f$ and $g$ belonging to $\mathcal{A}$ and any integer $x$ from $1$ to $2023$. Therefore, we conclude that $g(x)$ belongs to $X$ for such $g$ and $x$. Summarizing the discussion here, it has been shown that $f$ belonging to $\mathcal{A}$ satisfies both $$(1)m\ge f(1)\ge f(2)\ge \cdots \ge f(2023)\ge \ell ,\text{ and}$$ $$(2)\text{we have }f(x)=a\text{ for any integer }x\text{ from }\ell \text{ to }m.$$ We fix a function $f$ belonging to $\mathcal{A}$. For an integer $x$ from $1$ to $2023$, we define $s_x=f(x)-x$. Then, by (1), the sequence $s_1,s_2,\dots ,s_{2023}$ satisfies $$m-1\ge s_1>s_2>\cdots >s_{2023}\ge \ell -2023.$$ Furthermore, by (2), we have $s_x=a-x$ for any integer $x$ from $\ell$ to $m$. Therefore, for any function that belongs to $\mathcal{A}$, we can associate it with a method of choosing $2023$ integers from the integers from $\ell -2023$ to $m-1$, such that all integers from $a-m$ to $a-\ell$ are chosen. Here, this choice corresponds to choosing $2023-((a-\ell )-(a-m)+1)=2022-m+\ell$ integers from the $2022$ integers that are from $\ell -2023$ to $m-1$ but not between $a-m$ and $a-\ell$. Hence, the number of such choice is $\left(\genfrac{}{}{0ex}{}{2022}{2022-m+\ell }\right)$. Furthermore, since the methods of choosing corresponding to different functions belonging to $\mathcal{A}$ are distinct, the number of elements in $\mathcal{A}$ is at most $\left(\genfrac{}{}{0ex}{}{2022}{2022-m+\ell }\right)$. In addition, for a positive integer $x$ from $0$ to $2021$, we have $$\frac{\left(\genfrac{}{}{0ex}{}{2022}{\ell +1}\right)}{\left(\genfrac{}{}{0ex}{}{2022}{\ell }\right)}=\frac{2022-\ell }{\ell +1},$$ which implies $$\left(\genfrac{}{}{0ex}{}{2022}{0}\right)<\left(\genfrac{}{}{0ex}{}{2022}{1}\right)<\cdots <\left(\genfrac{}{}{0ex}{}{2022}{1010}\right)<\left(\genfrac{}{}{0ex}{}{2022}{1011}\right)>\left(\genfrac{}{}{0ex}{}{2022}{1012}\right)>\cdots >\left(\genfrac{}{}{0ex}{}{2022}{2021}\right)>\left(\genfrac{}{}{0ex}{}{2022}{2022}\right).$$ Therefore, we can conclude that the number of elements in $\mathcal{A}$ is at most $\left(\genfrac{}{}{0ex}{}{2022}{1011}\right)$. On the other hand, consider the set $\mathcal{B}$ of functions $f$ that are defined for integers from $1$ to $2023$ and take integer values from $1$ to $2023$, and satisfy $$1012=f(1)=f(2)=\cdots =f(1012)\ge f(1013)\ge f(1014)\ge \cdots \ge f(2023).$$ This set satisfies the first condition. Furthermore, for any functions $f$ and $g$ in $\mathcal{B}$ and any integer $x$ from $1$ to $2023$, we have $g(x)\le 1012$, so $f(g(x))=1012$ holds. Thus, the second condition is also satisfied. By considering $\ell =1$ and $a=m=1012$ in the argument above, we see that there is a one-to-one correspondence between the functions $f$ in $\mathcal{B}$ and the sequences $s_1,s_2,\dots ,s_{2023}$ of integers satisfying $$s_1=1011,s_2=1010,\dots ,s_{1012}=0,-1\ge s_{1013}>s_{1014}>\cdots >s_{2023}\ge -2022.$$ Therefore, the number of elements in $\mathcal{B}$ is $\left(\genfrac{}{}{0ex}{}{2022}{1011}\right)$. Therefore, we conclude that the maximum possible number of elements in $\mathcal{A}$ is $\left(\genfrac{}{}{0ex}{}{2022}{1011}\right)$.