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Source: Nepal TST 2025, Problem 3

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#1 h Apr 11, 2025, 1:27 PM • 1 Y

Y by khan.academy

Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $f(f(x)) + xf(xy) = x + f(y)$ for all positive real numbers $x$ and $y$ .

(Andrew Brahms, USA)

This post has been edited 1 time. Last edited by Mathdreams, Apr 11, 2025, 1:28 PM

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#2 h Apr 11, 2025, 1:46 PM • 4 Y

Y by khan.academy, KevinYang2.71, abrahms, Alex-131

The only solutions are $f(x) = \frac cx$ for some constant $c$ and $f\equiv 1$ , which clearly work.

Let $P(x,y)$ be the given assertion. Clearly $1$ is the only constant solution, so assume $f$ is not constant.

Claim: $f$ is injective.
Proof: Suppose $f(a) = f(b)$ for some positive reals $a,b$ .

$P(x,a)$ with $P(x,b)$ gives that $f(xa) = f(xb)$ for all $x \in \mathbb R^{+}$ .

Now, $P(a,x)$ compared with $P(b,x)$ gives $af(ax) - a = b f(bx) - b$ , so $a(f(ax) - 1) = b (f(bx) - 1)$ . But, since $f(ax) = f(bx)$ , we have $a(f(ax) - 1) = b(f(ax) - 1)$ Since $f$ isn't constant, we can choose $x$ where $f(ax) \ne 1$ , so $a = b$ . $\square$

$P(x, f(x)): f(f(x)) + xf(xf(x)) = x + f(f(x))$ , so $xf(xf(x)) = x \implies f(xf(x)) = 1$ .

$P(1, y): f(f(1)) = 1$ .

Injectivity implies $xf(x) = f(1)$ for all $x$ , so $f(x) = \frac{f(1)}{x}$ , and setting $c = f(1)$ gives the desired result.

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#3 h Apr 11, 2025, 3:08 PM • 4 Y

Y by khan.academy, Maksat_B, Sedro, abrahms

Mathdreams wrote:

Find all functions $f : \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $f(f(x)) + xf(xy) = x + f(y)$ for all positive real numbers $x$ and $y$ .

Let $P(x,y)$ be the assertion $f(f(x))+xf(xy)=x+f(y)$
Let $c=f(1)$ and $d=f(2)$

Subtracting $P(x,1)$ from $P(x,2)$ , we get $f(2x)=f(x)+\frac{d-c}x$
Subtracting $P(2,1)$ from $P(2,x)$ , we get $f(2x)=d+\frac{f(x)-c}2$

Subtracting : $f(x)=2\frac{c-d}x+2d-c$

Plugging $f(x)=\frac ax+b$ in original equation, we get $(a,b)=(\text{anything},0)$ or $(0,1)$ and solutions :
$\boxed{\text{S1 : }f(x)=1\quad\forall x>0}$ , which indeed fits

$\boxed{\text{S2 : }f(x)=\frac ax\quad\forall x>0}$ , which indeed fits, whatever is $a>0$

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#4 h Apr 11, 2025, 5:23 PM • 1 Y

Y by AlexCenteno2007

Let $P(x,y)$ be the assertion $f(f(x))+xf(xy)=x+f(y)$ .
$P(x,f(x))\Rightarrow f(xf(x))=1$
$P(f(x),x)\Rightarrow f(f(f(x)))=f(x)$
$P(f(x),y)\Rightarrow f(x)f(yf(x))=f(y)$ , in particular we have $f(f(x))=\frac{f(1)}{f(x)}$
$P(x,f(y))\Rightarrow xf(x)^2-(xf(y)+f(1))f(x)+f(y)f(1)=0\Rightarrow f(x)\in\left\{f(y),\frac{f(1)}x\right\}$ for each $x,y\in\mathbb R^+$ (we solved this as a quadratic in $f(x)$ )
If $f(x)\ne\frac{f(1)}x$ for some $x$ then by varying $y$ over $\mathbb R^+$ we get that $f$ is constant, and testing, the only constant solution is $\boxed{f(x)=1}$ for all $x$ .
Otherwise, $\boxed{f(x)=\frac cx}$ which works for any $c\in\mathbb R^+$ .

This post has been edited 1 time. Last edited by jasperE3, Apr 11, 2025, 5:24 PM

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#6 h Apr 12, 2025, 12:35 AM

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if $f \equiv c$ then $c$ must be $1$ further assume $f$ is not constant:
it's easy to get $f(f(1))=1$
and observe that $f$ is injective:
now $P(x,f(x)): f(f(x))+xf(xf(x))=x+f(f(x)) \implies f(xf(x))=1$
from injectivity $xf(x)=f(1) \implies f(x)=\frac {f(1)}{x}$
hence $f(x)=1 \forall x \in R$ and $f(x)=\frac {f(1)}{x} \forall x \in R$ are the only solutions

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#7 h Apr 12, 2025, 1:17 AM

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Tony_stark0094 wrote:

it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

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#8 h Apr 12, 2025, 1:59 AM

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jasperE3 wrote:

Tony_stark0094 wrote:

it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

$P(1,1): f(f(1))+f(1)=1+f(1) \implies f(f(1))=1$
for injectivity assume $f(a)=f(b)$
then subtract $P(a,1)$ from $P(b,1)$

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#9 h Apr 12, 2025, 2:29 AM

Y by

Tony_stark0094 wrote:

jasperE3 wrote:

Tony_stark0094 wrote:

it's easy to get $f(f(1))=1$
and observe that $f$ is injective:

how?

$P(1,1): f(f(1))+f(1)=1+f(1) \implies f(f(1))=1$
for injectivity assume $f(a)=f(b)$
then subtract $P(a,1)$ from $P(b,1)$

and what if $f(a)=f(b)=1$ (which is indeed the particular case $f(xf(x))=f(f(1))=1$ that you use injectivity for)

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#10 h Apr 12, 2025, 2:15 PM

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Claim : the only solutions are $f(x) = 1$ and $f(x) = \frac{c}{x}$ .
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 $P(x, \frac{f(1)}{x}) : f(f(x)) = f(\frac{f(1)}{x}) \implies f(x) = \frac{f(1)}{x}$

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#11 h Apr 15, 2025, 7:20 AM

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The answers are $f(x) = 1$ for all $x\in \mathbb{R}^+$ and $f(x)= \frac{c}{x}$ for all $x\in \mathbb{R}^+$ for some fixed constant $c \in \mathbb{R}^+$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(f(x))+xf(xy)=x+f(y)$ for positive real numbers $x$ and $y$ .

In what follows we assume that $f$ is not constant one. Say there does now exist some $x_0 \ne 1$ such that $f(x_0) \ne 1$ (i.e for all $x_0\ne 1$ we have $f(x_0)=1$ ). As $f$ is not constant one this indicates that $f(1) \ne 1$ . Then, $P\left(x,\frac{1}{x}\right)$ yields,
$\begin{align*} f(f(x)) + xf(1) &= x+f\left(\frac{1}{x}\right)\\ f(1) + xf(1) &= x+1 \\ (x+1)f(1) &= x+1 \end{align*}$ which is a clear contradiction since $x+1>0$ and $f(1) \ne 1$ . Thus, there indeed exists some $x_0 \ne 1$ such that $f(x_0) \ne 1$ . Now, $P(1,1)$ implies that
$f(f(1))+f(1)=1+f(1)$ from which we have $f(f(1))=1$ . We now make the following observation.

Claim : The function $f$ is injective.

Proof : We first show that it is injective at all points except 1. For this, note that if there exists $t_1 \ne t_2$ such that $f(t_1) =f(t_2) \ne 1$ . Then, $P(t_1,1)$ and $P(t_2,1)$ yeild,
$f(f(t_1))+t_1f(t_1)=t_1+f(1)$ $f(f(t_2))+t_2f(t_2)=t_2+f(1)$ whose difference implies
$(t_1-t_2)f(t_1)=t_1-t_2$ which since $f(t_1) \ne 1$ implies that $t_1=t_2$ which is a contradiction. Hence, $f$ is indeed injective at all points except 1.

With this observation in hand, consider $x_0 \ne 1$ such that $f(x_0) \ne 1$ and $\alpha$ such that $f(\alpha)=1$ . Then, from $P\left(x_0 , \frac{\alpha}{x_0}\right)$ we have
$\begin{align*} f(f(x_0)) + x_0f(\alpha) &= x_0 + f\left(\frac{\alpha}{x_0}\right)\\ f(f(x_0)) &= f\left(\frac{\alpha}{x_0}\right) \end{align*}$ Now,
$f(f(f(x_0))) = f\left(f\left(\frac{\alpha}{x_0}\right)\right)=f(x_0)$ which since $f(x_0) \ne 1$ implies $f(f(x_0))=x_0 \ne 1$ . Thus,
$\begin{align*} f(f(x_0)) &= f\left(\frac{\alpha}{x_0}\right)\\ f(x_0) &= \frac{\alpha}{x_0} \end{align*}$ In particular, if there exists $\alpha_1,\alpha_2 \in \mathbb{R}^+$ such that $f(\alpha_1)=f(\alpha_2)=1$ we have
$\frac{\alpha_1}{x_0} = f(x_0) = \frac{\alpha_2}{x_0}$ which implies $\alpha_1=\alpha_2$ proving the claim.

Thus, $f$ is injective and
$f(x) = \frac{f(1)}{x}$ for all $x \ne f(1)$ and $f(f(1))=1$ . This implies that indeed $f(x) = \frac{c}{x}$ for some fixed constant $c \in \mathbb{R}^+$ as desired.

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#12 h Apr 15, 2025, 8:11 AM

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cutie patootie problem

We claim either $f$ is one or $\frac{c}{x}$ . These clearly work. Let $P(x,y)$ denote the assertion.
Firstly, $P(1,1)$ gives $f(f(1))=1$ .
$P(x,1)$ yields $f(f(x))+xf(x)=x+f(1)$ . Hence $f(f(x))=xf(1)-xf(x)$ . Thus, subbing back, $xf(xy)+f(1)-xf(x)=f(y)$ . Now take $x\to f(1)$ so we get
$f(1)f(f(1)y)=f(y)$ Now, we will take $y\to f(1)y$ in $xf(xy)+f(1)-xf(x)=f(y)$ to get $\frac{xf(xy)}{f(1)}+f(1)-xf(x)=\frac{f(y)}{f(1)}$ . Comparing this with $xf(xy)+f(1)-xf(x)=f(y)$ , we have
$xf(xy)\left(\frac{1}{f(1)}-1\right)=f(y)\left(\frac{1}{f(1)}-1\right)$ This gives us two cases. Either $f(1)=1$ or $xf(xy)=f(y)$ for all $x,y$ . In the latter case, we would have $xf(x)=f(1)$ , and $f(x)=\frac{c}{x}$ which is a solution.

Now, suppose $f(1)=1$ . Then, taking $P\left(x,\frac{1}{x}\right)$ , we have $f(f(x))+x=x+f\left(\frac{1}{x}\right)$ hence $f(f(x))=f\left(\frac{1}{x}\right)$ . Assume that $f$ is not always constant, as if it was constant then we would have $cx=x$ hence $c=1$ as desired. We shall prove that $f$ is injective, which would finish as then cancelling one $f$ we get $f(x)=\frac{1}{x}$ .

Let $f(a)=f(b)$ such that WLOG $\frac{b}{a}>1$ . Take $y=a,b$ , so we have $f(f(x))+xf(ax)=x+f(a)$ and $f(f(x))+xf(bx)=x+f(b)$ . Subtracting we have
$x(f(ax)-f(bx))=f(a)-f(b)=0$ Hence, as $x\neq 0$ , we have
$f(ax)=f(bx)\quad f(x)=f(cx)$ where $c=\frac{b}{a}>1$ by scaling down $x$ .

Now, consider $P(cx,y)$ so we get $f(f(x))+cxf(xy)=cx+f(y)$ . Comparing with $P(x,y)$ we have
$xf(xy)(c-1)=x(c-1)$ Yet $c-1>0$ . Thus, we have $xf(xy)=x$ or $f(xy)=1$ . Taking $y=1$ gives $f(x)=1$ for all $x$ , a contradiction as we have dealt with constant $f$ .

Thus, $f$ must be injective, and we are done.

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