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

(→‎Solution 1: Removing this solution because it is flawed and therefore is not good to use, so people shouldn't see it. Oh well, can't get them all the time :$)
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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 2 ==
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== Solution 1 ==
  
 
Make the following substitutions to the equation:
 
Make the following substitutions to the equation:
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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>
 
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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3. <math>(x, 1 + \frac{f(1)}{x}) \rightarrow f(x + f(x) + f(1)) = xf(1 + \frac{f(1)}{x}) + 2</math>
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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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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>. Solving for the coefficients (in the same way as solution 1), we find that <math>f(x) = x + 1 \forall x > 1</math>.
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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>.
  
 
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.
 
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.
  
 
~jkmmm3
 
~jkmmm3
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==See Also==
 +
{{USAMO newbox|year=2023|num-b=1|num-a=3}}
 +
{{MAA Notice}}

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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