wardtnt1234 wrote: Find all function $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that: $f(\frac{f(x)}{x}+y)=1+f(y), \forall x,y \in \mathbb R^+$ Let $P(x,y)$ be the assertion $f(\frac{f(x)}x+y)=1+f(y)$ If $f(x)$ is injective : $P(x,y)$ and $P(1,y)$ imply $f(\frac{f(x)}x+y)=f(f(1)+y)$ and so, since injective, $f(x)=xf(1)$ Plugging this back in original equation, we get $f(1)=1$ and so : $\boxed{\text{The only injective solution is }f(x)=x\quad\forall x>0}$, which indeed fits. If $f(x)$ is not injective : $\exists a>b$ such that that $f(a)=f(b)=c$ Comparaison of $P(a,x)$ with $P(b,x)$ implies $f(x+\frac ca)=f(x+\frac cb)$ $\forall x>0$ And so $f(x)=f(x+T)$ $\forall x>\frac ca$ and $T=\frac cb-\frac ca$ Note then that easy induction from $P(1,x)$ implies $f(x+nf(1))=f(x)+n$ $\forall n\in\mathbb Z_>0$ And so $f(x)>n$ $\forall x>nf(1)$ Let then $x>\frac ca$ and $k_n\in\mathbb Z_{>0}$ such that $x+k_nT>nf(1)$ : We have $f(x)=f(x+k_nT)>n$ $\forall x>\frac ca$ and $n\in\mathbb Z_{>0}$, which is clearly impossible. And so $\boxed{\text{No non injective solution}}$

pco wrote: wardtnt1234 wrote: Find all function $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that: $f(\frac{f(x)}{x}+y)=1+f(y), \forall x,y \in \mathbb R^+$ Let $P(x,y)$ be the assertion $f(\frac{f(x)}x+y)=1+f(y)$ If $f(x)$ is injective : $P(x,y)$ and $P(1,y)$ imply $f(\frac{f(x)}x+y)=f(f(1)+y)$ and so, since injective, $f(x)=xf(1)$ Plugging this back in original equation, we get $f(1)=1$ and so : $\boxed{\text{The only injective solution is }f(x)=x\quad\forall x>0}$, which indeed fits. If $f(x)$ is not injective : $\exists a>b$ such that that $f(a)=f(b)=c$ Comparaison of $P(a,x)$ with $P(b,x)$ implies $f(x+\frac ca)=f(x+\frac cb)$ $\forall x>0$ And so $f(x)=f(x+T)$ $\forall x>\frac ca$ and $T=\frac cb-\frac ca$ Note then that easy induction from $P(1,x)$ implies $f(x+nf(1))=f(x)+n$ $\forall n\in\mathbb Z_>0$ And so $f(x)>n$ $\forall x>nf(1)$ Let then $x>\frac ca$ and $k_n\in\mathbb Z_{>0}$ such that $x+k_nT>nf(1)$ : We have $f(x)=f(x+k_nT)>n$ $\forall x>\frac ca$ and $n\in\mathbb Z_{>0}$, which is clearly impossible. And so $\boxed{\text{No non injective solution}}$ wow elaborate solution!

I wonder if there is a solution using comparison of $\frac{f(a)}{a}$ and $\frac{f(b)}{b}$ (also my idea)

Assume that there exist $a$ and $b$ such that $\frac{f(a)}{a} \neq \frac{f(b)}{b}$. We can prove that $f$ is periodic. Then I think we can prove that $f$ can be arbitrary large, which is contradiction.

tanlam wrote: Assume that there exist $a$ and $b$ such that $\frac{f(a)}{a} \neq \frac{f(b)}{b}$. We can prove that $f$ is periodic. Then I think we can prove that $f$ can be arbitrary large, which is contradiction. Your very short "explanation" does not show exactly what you have in mind. The fact that periodicity is in contradiction with unbound is indeed the solution (see my post). But this contradiction is not trivial (see my post). A lot of periodic functions are unbounded. I'm afraid you claimed that periodicity and unboundness is a general contradiction, which is wrong (this is a contradiction just in case of the very small subset of functions which are continuous).

The following nice lemma also kills the problem! https://artofproblemsolving.com/community/c6h2399644p19686093

Pco . Why have Plugging this back in original equation, we get $f(1)=1$

Lukariman wrote: Pco . Why have Plugging this back in original equation, we get $f(1)=1$ Uhh ? Just plug $f(x)=ax$ in original equation and you get $a(\frac{ax}x+y)=1+ay$ $\forall x,y>0$ And so $a^2=1$ and since codomain is $\mathbb R^+$, it remains $a=1$

We prove that f(x)/x is an constant. Assume WLOG z= f(x)/x > f(y)/y = g for some x and y. Lemma 1. f(x) > (least positive integer less than or equal to) x/g. Which is pretty obvious, since you can take away g from a function while losing only 1 of its value. Let’s f(1) = a. Then for all integers x -> f(1+zx) = a+x > (1+zx)/g which doesn’t work for really big x. Contradiction

$P(1,y) : f(f(1) + y) = 1 + f(y) = f(\frac{f(x)}{x}+y)$. Case $1 : f$ is injective. we have $f(f(1) + y) = f(\frac{f(x)}{x}+y)$ so $f(1) = \frac{f(x)}{x}$ so $f(x) = xa$ now by putting $a$ in main equation we have $a = 1$ so $f(x) = x$. Case $2 : f$ isn't injective. Let $a,b$ be such that $f(a) = f(b) = k$ and WLOG $a > b$. $P(a,y) , P(b,y) : f(\frac{k}{a} + y) = 1 + f(y) = f(\frac{k}{b} + y)$ so $f(x) = f(x+t)$ and $t = \frac{k}{a} - \frac{k}{b}$. $P(1,y) : f(f(1) + y) = 1 + f(y)$ now by putting $y = f(1) + y$ for times we have $f(nf(1) + y) = f(y) + n$ so $n = 0$ which isn't true so no answers.

Here is my solution: https://calimath.org/pdf/VNO2022-2.pdf And I uploaded the solution with motivation to: https://youtu.be/CqURVIqt3FY

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Plugging in $1$’s shows that $f$ has a fixed point, plugging this fixed point for $x$ gives $f(y+1)=f(y)+1$, and easy induction to $f(y+n)=f(y)+n$. Hence, for $y>1$ we have \[ f\left(y + \left( \frac{f(x)}{x}-1 \right)\right) = f(y). \]Assume for some $t$ we have $f(t)\neq t$ and let $\left|\frac{f(t)}{t}-1\right|=T$. Then $f$ is eventually periodic with period $T$. Letting $x:= t+n$ in the above as $n\to\infty$ we get that $f$ is eventually continuous. Then plugging in $x:=x+nT$ into the original equation as $n\to\infty$ yields a contradiction for large enough $y$. Edit: "we get that $f$ is eventually continuous" incorrect