Nepal TST 2025 Day 1 Problem 3

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Nepal TST 2025 Day 1 Problem 3

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Source: Nepal TST 2025, Problem 3

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#1 h Apr 11, 2025, 1:29 PM • 1 Y

Y by khan.academy

Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that
$f(f(x)) + x f(xy) = x + f(y)$ for all positive real numbers $x$ and $y$ .(Andrew Brahms, USA)

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#2 h Apr 11, 2025, 3:26 PM

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Bata325 wrote:

Find all functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that
$f(f(x)) + x f(xy) = x + f(y)$ for all positive real numbers $x$ and $y$ .(Andrew Brahms, USA)

https://artofproblemsolving.com/community/c6h3546497p34517439

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#3 h Apr 12, 2025, 2:08 PM

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Claim : the only solutions are $f(x) = 1$ and $f(x) = \frac{1}{x}$ .
Pf : Its easy to check these work, we now show these are only solutions.

Claim 1: If $f$ is not injective, its identically $1$ .
Let $P(x, y)$ be the assertion. $P(1,1) \implies f(f(1)) = 1$ Assuming $f(a) = f(b)$ for $a \neq b$ . Let $r = \frac{b}{a} \neq 1$ . $P(a, y) - P(b, y) \implies a(f(ay) - 1) = b(f(by)-1)$ Putting $y = \frac{f(1)}{a}$ we can conclude: $f(rf(1)) = 1$ $P(\frac{x}{f(1)}, f(1)) - P(\frac{x}{f(1)}, rf(1)) \implies f(x) = f(rx)$ $P(x, \frac{t}{x}) - P(rx, \frac{t}{x}) \implies f(t) = 1$ Since $r \neq 1$ .

Now, assuming $f$ is injective consider $P(x, \frac{f(1)}{x}) : f(f(x)) = f(\frac{f(1)}{x}) \implies f(x) = \frac{f(1)}{x}$ Substituting this back in the original equation we have $(f(1) - 1)((f(1) + 1)x + \frac{1}{y}) = 0$ so $f(1)=1$ so we are done.

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#4 h Apr 12, 2025, 2:13 PM

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Oh skull, i messed up $\frac{c}{x}$ works. Just change the beginning to have the correct solution set and remove the text concluding $f(1) = 1$ at end, this is why you don't try to manipulate equations in head.

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