2023 USAMO Problems/Problem 2

1. Problem 2

Let $\mathbb{R}^{+}$ be the set of positive real numbers. Find all functions $f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ such that, for all $x, y \in \mathbb{R}^{+}$ , $f(xy + f(x)) = xf(y) + 2$

1. Solution

First, let us plug in some special points; specifically, plugging in $x=0$ and $x=1$ , respectively:

\begin{align}

   f(f(0)) &= 2 \\
   f(y + f(1)) &= f(y) + 2

\end{align}

Next, let us find the first and second derivatives of this function. First, with (2), we isolate $f(y)$ one one side

\begin{align*}

  f(y) = f(y + f(1)) - 2

\end{align*}

and then take the derivative:

\begin{align*}

   \dv{f}{y} 
   &= \dv{f}{y}\left[f(y + f(1)) - 2\right] \\
   &= \dv{f}{y}\left[f(y + f(1))\right] - \dv{f}{y}\left[2\right] \\
   &= f'(y + f(1))\cdot\dv{f}{y}\left[y + f(1)\right] \\
   &= f'(y + f(1))\cdot(1)\\
   f'(y) &= f'(y + f(1))\\

\end{align*}

The second derivative is as follows:

\begin{align*}

   \dv[2]{f}{y} 
   &= \dv{f}{y}\left[\dv{f}{y}\right] \\
   &= \dv{f}{y}\left[f'(y + f(1))\right] \\
   &= f(y + f(1))\cdot\dv{f}{y}\left[y + f(1)\right] \\
   f(y) &= f(y + f(1))\\

\end{align*}

For both of these derivatives, we see that the input to the function does not matter: it will return the same result regardless of input. Therefore, the functions $f'$ and $f''$ must be constants, and $f$ must be a linear equation. That means we can model $f(x)$ like so:

\begin{align}

   f(x) = ax + b

\end{align}

Via (1), we get the following:

\begin{align*}

   f(f(0)) &= 2 \\
   a(a(0) + b) + b &= 2 \\
   ab + b &= 2

\end{align*}

And via (2),

\begin{align*}

   f(y + f(1)) &= f(y) + 2 \\
   a(y + a(1) + b) + b &= ay + b + 2 \\
   ay + a^2 + ab + b &= ay + b + 2 \\
   a^2 + ab &= 2 \\

\end{align*}

Setting these equations equal to each other,

\begin{align*}

   ab + b &= a^2 + ab \\
   b &= a^2 \\

\end{align*}

Therefore,

\begin{align*}

   ab + b &= 2 \\
   a^3 + a^2 &= 2 \\

\end{align*}

There are three solutions to this equation: $a = 1$ , $a = -1 + i$ , and $a = -1 - i$ . Knowing that $b = a^2$ , the respective $b$ values are $b = 1$ , $b = -2i$ , and $b = 2i$ . Thus, $f(x)$ could be the following:

\begin{align*}

   f(x) &= x + 1 \\
   f(x) &= x(-1 + i) - 2i \\
   f(x) &= x(-1 - i) + 2i \\

\end{align*}

$\square$

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2023 USAMO Problems/Problem 2