Find all functions $f:\mathbb{R^+} \to \mathbb{R^+}$ such that for all $x,y,z\in \mathbb{R^+}$: $$\biggl\{\frac{f(x)}{f(y)}\biggl\}+\biggl\{\frac{f(y)}{f(z)}\biggl\}+ \biggl\{\frac{f(z)}{f(x)}\biggl\}= \biggl\{\frac{x}{y}\biggl\} +\biggl\{\frac{y}{z}\biggl\}+ \biggl\{\frac{z}{x}\biggl\}$$Note: For any real number $x$, let $\{x\}$ denote the fractional part of $x$, defined as For example, $\{2,7\}=0,7$ .
Problem
Source: Turkey National MO 2024 P5
Tags: function
starchan
17.12.2024 13:53
solved with rawlat_vanak, Siddharth03, mathsmania, mxlcv I don't know about the proposer, but the solvers are definitely handsome.
Let $P(x, y, z)$ denote the given assertion. From setting $y =z$, we obtain the following for all positive reals $x, y$: \[\biggl\{\frac{f(x)}{f(y)}\biggl\} + \biggl\{\frac{f(y)}{f(x)}\biggl\} = \biggl\{\frac{x}{y}\biggl\} + \biggl\{\frac{y}{x}\biggl\}\]Call the above $Q(x, y)$.
Claim 1: $f$ is injective.
Proof: Suppose $f(a) = f(b)$. Use $P(a, b, b)$. $\square$
Claim 2: Let $\alpha$ satisfy $\frac{7+\sqrt{13}}{6} < \alpha < 2$. Suppose that $\beta$ is a positive real number that satisfies \[\biggl\{\frac{1}{\beta}\biggl\} + \biggl\{\beta\biggl\} = \biggl\{\frac{1}{\alpha}\biggl\} + \biggl\{\alpha\biggl\}\]Then $\alpha = \beta$ or $\alpha \beta = 1$.
Proof: From some simple algebra, we deduce that the right hand side of the above equality is larger than $4/3$. At the same time, if we WLOG assume $\beta > 1$, so that $\beta = n+f$ for $n \geq 1$. If $n \geq 2$, we can see that one has \[\frac{1}{n+f} + f < \frac{4}{3},\]impossible and thus $n = 1$ from where $\alpha = \beta$ follows, proving the claim. $\square$
Claim 3: Let $\alpha$ satisfy $\frac{7+\sqrt{13}}{6} < \alpha < 2$. Then $f(\alpha x) = \alpha f(x)$ for all positive real $x$.
Proof: First, from $Q(\alpha x, x)$, we see that $f(\alpha x) = \alpha f(x)$ or $f(x) = \alpha f(\alpha x)$. Suppose that the latter is true. Then, from $Q(\alpha^2x, \alpha x)$ and injectivity, we obtain $f(x) = \alpha^2 f(\alpha^2 x)$. Now plug $P(x, \alpha x, \alpha^2 x)$ to get a contradiction. $\square$
Claim 4: Let $r, sr$ be positive reals, where $s > 1$. Then, $f(sr) = sf(r)$.
Proof: Pick $N$ huge so that $s^{1/n}$ is small. Since this is very small, we can express it as the ratio $\beta/\alpha$ so that $\alpha, \beta$ are both in \[\left(\frac{7+\sqrt{13}}{6}, 2\right).\]Now we get $f(sr) = f\left(r \cdot \frac{\beta^n}{\alpha^n}\right) = \frac{\beta^n}{\alpha^n} f(r) = sf(r)$, proving the claim. $\square$.
From Claim 4, we immediately see that $f(x) = cx$ for all $x > 0$ (where $c = f(1)$) and it may be easily seen that all such functions work as well, so they are the only ones. $\blacksquare$
BarisKoyuncu
17.12.2024 13:57
If a function $f(x)$ satisfies the given equation, then so does $cf(x)$. Hence, we can assume that $f(1)=1$. Take $x=ny$ where $z, y$ are fixed and $n$ is very large positive integer. Then, $\left\{\dfrac{z}{ny}\right\}=\dfrac{z}{ny}$ so we get $$\left\{\dfrac{y}{z}\right\}+ \dfrac{z}{ny} = \left\{\dfrac{f(ny)}{f(y)}\right\}+ \left\{\dfrac{f(y)}{f(z)}\right\}+ \left\{\dfrac{f(z)}{f(ny)}\right\}>\left\{\dfrac{f(y)}{f(z)}\right\}$$ Now since we can choose $n$ as large as we wish, we can make the term $ \dfrac{z}{ny}$ as small as we want. Therefore, $\left\{\dfrac{y}{z}\right\}\ge \left\{\dfrac{f(y)}{f(z)}\right\}$ for all $y,z\in\mathbb{R^+}$. Then, from the given equation, we get $\left\{\dfrac{y}{z}\right\}=\left\{\dfrac{f(y)}{f(z)}\right\}$ for all $y,z\in\mathbb{R^+}$ (Note that this equation implies the original equation so we can just focus on solving this). Now fix $z$. Choosing $y=nz$ for a positive integer $n\ge 2$ gives us $\dfrac{f(nz)}{f(z)}=m$ for some integer $m$. Then, choosing $y=z$ and $z=nz$, gives us $\dfrac 1n=\dfrac 1m\Rightarrow n=m$. Hence, $f(nz)=nf(z)$. Then, for a positive rational number $q=\dfrac ab$ (where $a,b\in\mathbb{Z^+}$), we get $$f(qx)=f\left(a\cdot\dfrac 1b\cdot x\right)=af\left(\dfrac 1b\cdot x\right)=a\cdot\dfrac 1b\cdot f(x)=qf(x)$$. In particular, $f(q)=q$. Now fix $z$ and take $y=q$. We get $\left\{\dfrac{q}{z}\right\}=\left\{\dfrac{q}{f(z)}\right\}\Rightarrow q\left(\dfrac 1z-\dfrac 1{f(z)}\right)\in\mathbb{Z}$. The only real number $r$ for which $qr\in\mathbb{Z}$ is $r=0$ so we get $f(z)=z$. Then, the general solution is $f(x)=cx$ for some positive real number $c$.
After proving $\left\{\dfrac{y}{z}\right\}=\left\{\dfrac{f(y)}{f(z)}\right\}$, choosing $y=1$ and then $z=1$ gives us that $f(x)=x+n_x$ and $\dfrac 1{f(x)}=\dfrac 1x+m_x$ for some integers $n_x,m_x$. From these equations, we can derive the quadratic equation $m_xx^2+m_xn_xx+n_x=0$. Unless $m_x=n_x=0$, this equation has at most $2$ solutions. Since $m_x$ and $n_x$ are both integers, the numbers of pairs $(n_x,m_x)$ is countably infinite. For each pair (except for $(0,0)$) there are at most two real numbers $x$ so for uncountably many real numbers $x$, we must have $m_x=n_x=0$.
Now fix $y$ and let $z$ be any number such that $f(z)=z$. As $\left\{\dfrac{y}{z}\right\}=\left\{\dfrac{f(y)}{f(z)}\right\}$, we find that $\dfrac{f(y)-y}z$ is an integer (here, $z$ can take uncountably many values). The only real numbers $r$ for which there are uncountably many real numbers $z$ such that $\dfrac rz\in\mathbb{Z}$ is $r=0$ so we find that $f(y)=y$ for all $y$.
GuvercinciHoca
18.12.2024 15:17
The only functions that satisfies the condition are in the form of $f(x) = cx$ where $c\in \mathbb{R^+} $.
Let $x$, $y$ be positive reel numbers and take $z = Nx$ where $N$ is a positive integer. Then we have,
$$\biggl\{\frac{f(x)}{f(y)}\biggl\}+\biggl\{\frac{f(y)}{f(z)}\biggl\}+ \biggl\{\frac{f(z)}{f(x)}\biggl\}= \bigl\{ \frac{x}{y} \bigr\} + \bigl\{ \frac{y}{Nx} \bigr\} \rightarrow \bigl\{ \frac{f(x)}{f(y)} \bigr\} \leq \bigl\{ \frac{x}{y} \bigr\} + \bigl\{ \frac{y}{Nx} \bigr\} $$
When $N$ goes to infinity, we get $\bigl\{ \frac{f(x)}{f(y)} \bigr\} \leq \bigl\{ \frac{x}{y} \bigr\}$. When plug this inequality in original equation we get $\bigl\{ \frac{f(x)}{f(y)} \bigr\} = \bigl\{ \frac{x}{y} \bigr\}$.
Since if $f(x)$ is a solution then $g(x)=c.f(x)$ is also a solution, WLOG we can take $f(1)=1$. Then we have, $\{ f(x) \} = \{ x \}$ for every positive reel $x$, hence, $f(x) - x \in \mathbb{Z}$.
Now lets take $x \notin \mathbb{Q}$ and $y \in \mathbb{Q}$ positive reel numbers. Let $f(x)=x+m$ and $f(y)=y+n$. Then there exist an integer $k$ such that $\frac{x+m}{y+n} = \frac{x}{y} + k = \frac{x + yk}{y}$.
From here we get,
$$xy + my = xy + y^2 k + xn + ykn \rightarrow my = y^2 k + xn + ykn$$Since, $x$ is an irrational number and the other numbers are rationals, we get $n=0$. Thus, $f(x)=x$ for all positive rational $x$. On the other hand, if $m\neq 0$, then for all $y$ we have $m=yk$. Take $x$ as a constant (then $m$ is also a constant integer), then take $y$ as an integer such that $y \nmid m$. Contradiction since $k$ is also an integer and $m=yk$. Thus, $m=0$ which means $f(x)=x$ for all positive reel $x$.
bin_sherlo
18.12.2024 21:28
Answer is $f(x)=cx$. Let $P(x,y,z)$ be the assertion. If $f$ holds the conditions, then $cf$ also satisfies thus we can assume that $f(1)=1$. $P(nm,n,1)$ where $n,m\in Z^+$ gives $\{f(n)\}\leq\{\frac{f(nm)}{f(n)}\}+\{f(n)\}+\{\frac{1}{f(nm)}\}=\{\frac{1}{nm}\}$ We pick $m$ sufficienlty large which implies $\{f(n)\}=0\iff f(n)\in Z^+$. $P(n,1,1)$ yields $\{\frac{1}{f(n)}\}=\{f(n)\}+\{1\}+\{\frac{1}{f(n)}\}=\{n\}+\{1\}+\{\frac{1}{n}\}=\frac{1}{n}$ Since both $f(n),n$ are positive integers, we conclude that $f(n)=n$. $P(x,n,1)$ implies $\{\frac{f(x)}{n}\}+\{\frac{1}{f(x)}\}=\{\frac{x}{n}\}+\{\frac{1}{x}\}\iff \{\frac{f(x)}{n}\}-\{\frac{x}{n}\}=\{\frac{1}{x}\}-\{\frac{1}{f(x)}\}$ Right hand side is independent from $n$ so if we pick $n>>f(x),x$ then $\frac{f(x)-x}{n}$ is independent from $n$ which gives $f(x)=x$ as desired.$\blacksquare$
ezpotd
19.12.2024 02:49
The answer is $f(x) = kx$ for all positive real $k$, it is obvious to see that this works. Claim 1: If $\frac ba \in \mathbb{N}$, then $\frac{f(b)}{f(a)} \in \mathbb{N}$. Take $x = b, y = a, \frac az \in \mathbb{N}$. As $z$ approaches $0$, so does the right hand side, so $\frac{f(b)}{f(a)}$ cannot have a fractional part. Claim 2: If $\frac ba \in \mathbb{N}$, then $\frac{f(a)}{f(b)} = \frac ab $. Take $z \mid y \mid x$, then we have $\frac{f(x)}{f(z)} \in \mathbb{N}, $ so $\frac{f(z)}{f(x)} \le 1$. Thus the equality becomes $\frac{f(z)}{f(x)} = \frac zx$ as desired. Claim 3: If $f(1) = k$, then $f(\frac pq) = k\frac pq$ for all $p,q \in \mathbb{N}$. Obviously follows from Claim 2. We now show $f(x) = kx$ for all $x$. Let $a,b$ be rationals satisfying $\frac ba < 1.1$. We show $a < r < b$ implies $f(a) < f(r) < f(b)$. Take $x = a, y = r, z = b$, then we have $\left \{ \frac{ka}{f(r)} \right \} + \left \{ \frac{f(r)}{kb} \right \} + \left \{ \frac ba \right \}= \frac{a}{r} + \frac{r}{b} + \left \{ \frac{b}{a} \right \}$. If $f(r) < \frac{f(a)}{2}$, then $\left \{ \frac{ka}{f(r) } \right \} < 1, \left \{ \frac{f(r)}{kb} \right \} < \frac{a}{2b}$, so $\left \{ \frac{ka}{f(r)} \right \} + \left \{ \frac{f(r)}{kb} \right \} < 1 + \frac{a}{2b} < \min(\frac ar + \frac rb) = 2\sqrt{\frac ab}$, as $1 > t = \frac ab > 0.9$ guarantees $t^2 - 4t + 2 < 0$, so we cannot have $f(r) < \frac{f(a)}{2}$. If $\frac 12 f(a) < f(r) < f(a)$, observe that $\left \{ \frac{ka}{f(r)} \right \} + \left \{ \frac{f(r)}{kb} \right \} = \frac{ka}{f(r)} + \frac{f(r)}{kb} - 1$, taking the derivative of $\frac{ka}{x} + \frac{x}{kb}$ (which is $-\frac{ka}{x^2} + \frac{1}{kb}$) which is negative for all $f(r) < ka$, we see the maximum over the desired interval is as $f(r)$ approaches $\frac 12 f(a)$, which is less than the minimum value of $\frac ar + \frac rb$ as previously determined. We can make literally the exact same arguments in the cases where $f(b) <f(r) < 2f(b), f(r) > 2f(b)$, the expressions evaluate the exact same way. To finish, assume $f(r) = ks < kr$ (we can make the exact same argument if $ks > kr$) for some $r$. Then by the density of the rationals we can find some rationals $a,b$ with $s < a < r < b$, which forces $f(r) > ka > ks $, contradiction.
grupyorum
19.12.2024 04:11
Interesting. I think there's a shorter way to conclude—let me know if I am missing anything below. As above, deduce $\{f(x)/f(y)\} = \{x/y\}$ for all $x,y\in\mathbb{R}^+$ and assume wlog that $f(1)=1$. Now, consider $x=n$ and $y=1$ where $n\in\mathbb{N}$ is arbitrary, we find $\{f(n)\} = \{n\}=0$. So, $f(n)$ is a positive integer for all $n$. Now, taking $x=1$ and $y=n$, we find $\{1/f(n)\} = \{1/n\}$. For $n>1$, we have $\{1/n\} = 1/n = \{1/f(n)\} = 1/f(n)$ since $f(n)\in\mathbb{N}$. So, $f(n)=n,\forall n\in\mathbb{N}$. Lastly, fix $x>0$ and let $y>\max\{x,f(x)\}$ be a positive integer. Then, $\{f(x)/f(y)\} = \{f(x)/y\} = f(x)/y$ is equal to $x/y$, yielding $f(x)=x$ for all $x$.