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Difference between revisions of "2023 USAMO Problems/Problem 2"

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Let <math>\mathbb{R}^{+}</math> be the set of positive real numbers. Find all functions <math>f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}</math> such that, for all <math>x, y \in \mathbb{R}^{+}</math>,<cmath>f(xy + f(x)) = xf(y) + 2</cmath>
 
Let <math>\mathbb{R}^{+}</math> be the set of positive real numbers. Find all functions <math>f:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}</math> such that, for all <math>x, y \in \mathbb{R}^{+}</math>,<cmath>f(xy + f(x)) = xf(y) + 2</cmath>
  
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== Solution ==
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== Solution 1 ==
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First, let us plug in some special points; specifically, plugging in <math>x=0</math> and <math>x=1</math>, respectively:
 
  
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<cmath>
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Make the following substitutions to the equation:
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\begin{align}
 
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    f(f(0)) &= 2 \\
 
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    f(y + f(1)) &= f(y) + 2
 
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\end{align}
 
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</cmath>
 
  
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Next, let us find the first and second derivatives of this function. First, with (2), we isolate <math>f(y)</math> one one side
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1. <math>(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2</math>
  
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<cmath>
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2. <math>(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4</math>
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\begin{align*}
 
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  f(y) = f(y + f(1)) - 2
 
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\end{align*}
 
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</cmath>
 
  
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and then take the derivative:
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3. <math>(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2</math>
  
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<cmath>
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It then follows from (2) and (3) that <math>f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}</math>, so we know that this function is linear for <math>x > 1</math>. Substitute <math>f(x) = ax+b</math> and solve for <math>a</math> and <math>b</math> in the functional equation; we find that <math>f(x) = x + 1 \forall x > 1</math>.
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\begin{align*}
 
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    \dfrac{\mathrm{d}f}{\mathrm{d}y}
 
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    &=\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1)) - 2\right] \\
 
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    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f(y + f(1))\right] - \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[2\right] \\
 
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    &= f'(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
 
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    &= f'(y + f(1))\cdot(1)\\
 
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    f'(y) &= f'(y + f(1))\\
 
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\end{align*}
 
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</cmath>
 
  
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The second derivative is as follows:
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Now, we can let <math>x > 1</math> and <math>y \le 1</math>. Since <math>f(x) = x + 1</math>, <math>xy + f(x) > x > 1</math>, so <math>f(xy + f(x)) = xy + x + 2 = xf(y) + 2</math>. It becomes clear then that <math>f(y) = y + 1</math> as well, so <math>f(x) = x + 1</math> is the only solution to the functional equation.
  
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<cmath>
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~jkmmm3
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\begin{align*}
 
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    \dfrac{\mathrm{d}^2f}{\mathrm{d}y^2}
 
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    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[\dfrac{\mathrm{d}f}{\mathrm{d}y}\right] \\
 
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    &= \dfrac{\mathrm{d}f}{\mathrm{d}y}\left[f'(y + f(1))\right] \\
 
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    &= f''(y + f(1))\cdot\dfrac{\mathrm{d}f}{\mathrm{d}y}\left[y + f(1)\right] \\
 
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    f''(y) &= f''(y + f(1))\\
 
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\end{align*}
 
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</cmath>
 
  
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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 <math>f'</math> and <math>f''</math> must be constants, and <math>f</math> must be a linear equation or a constant. We know it is not a constant because if it was, the problem could be reduced to a linear equation with two unknowns, <math>f</math> and <math>x</math>, making <math>f</math> depend on <math>x</math>, which is not a constant function. That means we can model <math>f(x)</math> like so:
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==See Also==
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{{USAMO newbox|year=2023|num-b=1|num-a=3}}
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<cmath>
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{{MAA Notice}}
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\begin{align*}
 
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    f(x) = ax + b
 
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\end{align*}
 
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</cmath>
 
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Via (1), we get the following:
 
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<cmath>
 
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\begin{align*}
 
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    f(f(0)) &= 2 \\
 
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    a(a(0) + b) + b &= 2 \\
 
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    ab + b &= 2
 
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\end{align*}
 
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</cmath>
 
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And via (2),
 
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<cmath>
 
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\begin{align*}
 
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    f(y + f(1)) &= f(y) + 2 \\
 
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    a(y + a(1) + b) + b &= ay + b + 2 \\
 
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    ay + a^2 + ab + b &= ay + b + 2 \\
 
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    a^2 + ab &= 2 \\
 
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\end{align*}
 
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</cmath>
 
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Setting these equations equal to each other,
 
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<cmath>
 
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\begin{align*}
 
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    ab + b &= a^2 + ab \\
 
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    b &= a^2 \\
 
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\end{align*}
 
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</cmath>
 
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Therefore,
 
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<cmath>
 
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\begin{align*}
 
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    ab + b &= 2 \\
 
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    a^3 + a^2 &= 2 \\
 
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\end{align*}
 
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</cmath>
 
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There are three solutions to this equation: <math>a = 1</math>, <math>a = -1 + i</math>, and <math>a = -1 - i</math>. Knowing that <math>b = a^2</math>, the respective <math>b</math> values are <math>b = 1</math>, <math>b = -2i</math>, and <math>b = 2i</math>. Thus, <math>f(x)</math> could be the following:
 
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<cmath>
 
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\begin{align*}
 
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    f(x) &= x + 1 \\
 
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    f(x) &= x(-1 + i) - 2i \\
 
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    f(x) &= x(-1 - i) + 2i \\
 
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\end{align*}
 
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</cmath>
 
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Because only the first function maps strictly to positive real numbers, the only solution that works is <math>f(x) = x + 1</math>. <math>\square</math>
 
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~cogsandsquigs
 

Latest revision as of 12:25, 1 January 2024

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 $x, y \in \mathbb{R}^{+}$,\[f(xy + f(x)) = xf(y) + 2\]

Solution 1

Make the following substitutions to the equation:

1. $(x, 1) \rightarrow f(x + f(x)) = xf(1) + 2$

2. $(1, x + f(x)) \rightarrow f(x + f(x) + f(1)) = f(x + f(x)) + 2 = xf(1) + 4$

3. $(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf\biggl(1 + \frac{f(1)}{x}\biggr) + 2$

It then follows from (2) and (3) that $f(1 + \frac{f(1)}{x}) = f(1) + \frac{2}{x}$, so we know that this function is linear for $x > 1$. Substitute $f(x) = ax+b$ and solve for $a$ and $b$ in the functional equation; we find that $f(x) = x + 1 \forall x > 1$.

Now, we can let $x > 1$ and $y \le 1$. Since $f(x) = x + 1$, $xy + f(x) > x > 1$, so $f(xy + f(x)) = xy + x + 2 = xf(y) + 2$. It becomes clear then that $f(y) = y + 1$ as well, so $f(x) = x + 1$ is the only solution to the functional equation.

~jkmmm3

See Also

2023 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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