Team selection tests

Because the polynomial $P(x)$ is non-constant and has the highest coefficient equal to $1$, there exists a constant $c$ such that $P(x)$ can take on all values above $[c,+\infty )$. From equation (1), we get $$f(f(x)+y+2023f(y))=x+2024f(y),\forall x\ge c,y\in \mathbb{R}.(2)$$ We will prove ${\lim }_{x\to +\infty }|f(x)|=+\infty$. Indeed, assuming the existence of real numbers $M,m$ (we can assume $m\ge c$) such that $|f(x)|\le M$ for all $x\ge m$. Fix a real number $y_0$. For $x\ge m$, we have $$-M+y_0+2023f(y_0)\le f(x)+y_0+2023f(y_0)\le M+y_0+2023f(y_0).$$ In addition, since $f$ is continuous on $\mathbb{R}$, there exists a real number $N$ such that $f(x)\le N$ for all real numbers $x$ belonging to the interval $[-M+y_0+2023f(y_0),M+y_0+2023f(y_0)]$. From here and from (2), we deduce $$N\ge f(f(x)+y_0+2023f(y_0))=x+2024f(y_0),$$ or $N\ge x+2024f(y_0)$ for all $x\ge m$, contradiction. Therefore ${\lim }_{x\to +\infty }|f(x)|=+\infty$. From (2), it is easy to see that $f$ is injective on the domain $[c,+\infty )$. Since $f$ is a continuous function on $\mathbb{R}$, $f$ is a truly monotone function on the domain $[c,+\infty )$. Thus, there are two possible cases.

Case 1: ${\lim }_{x\to +\infty }f(x)=-\infty$. We will prove ${\lim }_{x\to -\infty }f(x)=+\infty$. Indeed, suppose there exist real numbers $L,\ell$ such that $f(x)\le L$ for all $x\le \ell$. Fix the real number $y_0$. Since ${\lim }_{x\to +\infty }f(x)=-\infty$, there exists a real number $x_0\ge c$ such that $x_0>L-2024f(y_0)$ and $f(x_0)+y_0+2023\le \ell$. Then, from (2), we have $$L<x_0+2024f(y_0)=f(f(x_0)+y_0+2023f(y_0))\le L.$$ From the received contradiction, we deduce ${\lim }_{x\to -\infty }f(x)=+\infty$. Because ${\lim }_{x\to +\infty }f(x)=-\infty$, ${\lim }_{x\to -\infty }f(x)=+\infty$ and $f$ is continuous on $\mathbb{R}$ so $f$ is an all-reflection on $\mathbb{R}$. Now, fixing the real number $x_1\ge c$. We see that there exists a real number $y_1$ such that $f(y_1)=-\frac{f(x_1)}{2023}$. Then, from equation (2), we deduce $$x_1-\frac{2024}{2023}f(x_1)=x_1+2024f(y_1)=f(f(x_1)+y_1+2023f(y_1))=f(y_1)=-\frac{f(x_1)}{2023},$$ therefore $f(x_1)=x_1$. Thus, $f(x)=x$ for all $x\ge c$, which is a contradiction because ${\lim }_{x\to +\infty }f(x)=-\infty$.

Case 2: ${\lim }_{x\to +\infty }f(x)=+\infty$. In this case, it is easy to see that $f$ is a strictly increasing function on the domain $[c,+\infty )$. Now, suppose there are two real numbers $a,b$ such that $f(a)=f(b)$. Since ${\lim }_{x\to +\infty }f(x)=+\infty$ so there exists a real number $d\ge c$ such that $f(d)+a+2023f(a)\ge c$ and $f(d)+b+2023f(b)\ge c$. Then, from equation (2), we have $$f(f(d)+a+2023f(a))=d+2024f(a)=d+2024f(b)=f(f(d)+b+2023f(b)).$$ Because $f$ increases strictly over the domain $[c,+\infty )$, $f(d)+a+2023f(a)=f(d)+b+2023f(b)$, implies $a=b$. Thus, the function $f$ is one-to-one on $\mathbb{R}$. Because $f$ is continuous on $\mathbb{R}$ and $f$ is strictly increasing on the domain $[c,+\infty )$ so from here, we deduce that $f$ increases strictly on $\mathbb{R}$. Next, we will prove ${\lim }_{x\to -\infty }f(x)=-\infty$. Indeed, suppose there exist real numbers $V,\nu$ such that $f(x)\ge V$ for all $x\le \nu$. Substituting $y=-f(x)$ into equation (2), we get $$f(2023f(-f(x)))=x+2024f(-f(x)),\forall x\ge c.(3)$$ Since ${\lim }_{x\to +\infty }f(x)=+\infty$, there exists a real number $x_2\ge c$ such that $x_2>f(2023f(\nu ))-2024V$ and $f(x_2)\ge -\nu$. We have $-f(x_2)\le \nu$ so $f(-f(x_2))\ge V$, follows output $$x_2+2024f(-f(x_2))\ge x_2+2024V.$$ There is again $-f(x_2)\le \nu$ so $f(-f(x_2))\le f(\nu )$, infers $$f(2023f(-f(x_2)))\le f(2023f(\nu )).$$ Combined with (3), we get $$\begin{array}{rl}f(2023f(\nu ))& \ge f(2023f(-f(x_2)))=x_2+2024f(-f(x_2))\\ & \ge x_2+2024L>f(2023f(\nu )).\end{array}$$ From the received contradiction, we deduce ${\lim }_{x\to -\infty }f(x)=-\infty$. We have ${\lim }_{x\to +\infty }f(x)=+\infty$, ${\lim }_{x\to -\infty }f(x)=-\infty$ and $f$ is continuous on $\mathbb{R}$ so $f$ is an all-reflection on $\mathbb{R}$. At here, doing the same substitution as case 1 above, we have $f(x)=x$ for all $x\ge c$. Now, in equation (2), fix the real number $x$ and choose $x\ge c$ enough so that $f(x)+y+2023f(y)\ge c$. We have $$\begin{array}{rl}x+y+2023f(y)& =f(x)+y+2023f(y)\\ & =f(f(x)+y+2023f(y))=x+2024f(y).\end{array}$$ It follows that $f(y)=y$. Thus, we have $f(x)=x$ for all real numbers $x$. It is easy to verify that $f(x)=x$ indeed satisfies the condition of the problem.