Also Indian TST 4, Problem 2

The answer is $f(x) = \lambda_1 x + \lambda_2 x ^{-1}$ for any real constants $\lambda_1$ and $\lambda_2$. Indeed, $f(x) \equiv x$ and $f(x) \equiv 1/x$ work and the set of solutions is an ${\mathbb R}$-vector space. To prove they are the only ones, we may by shifting assume that \[ f(1) = f(2) = 0 \]and try to show $f$ is the zero function. Let $P(x,y)$ denote the given assertion. First, note that $P(x,x)$ gives we get \[ f(x^2) = (x+1/x) f(x) \qquad \forall x. \]By replacing $xy$ and $y/x$ with $a^2$ and $b^2$ in the given we get the more symmetric (and square-free) assertion \begin{align*} Q(a,b) : \qquad \left( \frac ab + \frac ba \right) f(ab) &= f(a^2) + f(b^2) \\ &= \left( a+\frac1a \right) f(a) + \left( b+\frac1b \right) f(b). \end{align*}The idea is to now compute $f(4x)$ in terms of $f(x)$ in two different ways: because the way the equations are linear, we should get some constant multiple. To begin $Q(2,2) \implies f(4) = 0$. Next, for any $x > 0$, note that \begin{align*} Q(x,4) \implies \left( \frac x4 + \frac 4x \right) f(4x) &= \left( x + \frac 1x \right) f(x) \\ Q(x,2) \implies \left( \frac x2 + \frac 2x \right) f(2x) &= \left( x+\frac1x \right) f(x) \\ Q(2x,2) \implies \left( x + \frac 1x \right) f(4x) &= \left( 2x+\frac1{2x} \right) f(2x) = \left( 2x + \frac{1}{2x} \right) \left( x+ \frac 1x \right) f(x). \end{align*}Thus $f(4x)$ is a multiple of $f(x)$ in two different ways! If $f(x) \neq 0$, then we would have to have \[ 2x + \frac{1}{2x} = \left( x + \frac{1}{x} \right) \left( \frac x4 + \frac 4x \right) = \frac{x^2}{4} + \frac{4}{x^2} + \frac{17}{4}. \]But by AM-GM $\frac{x^2}{4} + 4 \ge 2x$, and $\frac{4}{x^2} + \frac 14 \ge \frac{2}{x} > \frac{1}{2x}$, so this is a contradiction.

This problem is quite interesting! It seems like it's very easy to solve if one found the solution space and noticed it is a real vector space, and very difficult to approach otherwise.

Since the set of all solutions form a real vector space $V$ and $f(x)\equiv x$, $f(x)\equiv\frac{1}{x}$ are two linearly independent solutions. It suffices to show that $\dim V = 2$. Pick one solution $g$ such that $g(1)=g(2)=0$. Thus $$\left(x+\frac{1}{x}\right)g(y)=g(xy)+g\left(\frac{y}{x}\right)$$ Call the last equation $P(x,y)$. We get \begin{align*} P(x,1)&\implies g(x)+g\left(\frac{1}{x}\right)=0\quad\forall\, x>0 \\ P(2x,2)&\implies g(4x)+g\left(\frac{1}{x}\right)=0\quad\forall\, x>0. \end{align*}Hence, $g(4x)=g(x)$ for every $x\in (0,\infty)$. Using this we get $$P(4,y)\implies \frac{15}{4}g(y)=g(4y)+g\left(\frac{y}{4}\right)=2g(y)$$Hence, $g(x)\equiv 0\implies \dim V=2$. Hence $ f(x)\equiv ax+\frac{b}{x}$ are the only possible solutions, which clearly satisfies the equation.

The solutions are $ax+\frac{b}{x}$ for any constants $a,b\in\mathbb{R}$. These work because $x,\frac{1}{x}$ are solutions and the solution set is a vector space over $\mathbb{R}$. Observe that \begin{align*} \left(x+\frac{1}{x}\right)f(x)&=f(x^2)+f(1)\\ &=\left(\frac{x}{2}+\frac{2}{x}\right)f(2x)-f(4)+f(1)\\ &=\frac{\frac{x}{2}+\frac{2}{x}}{2x+\frac{1}{2x}}\left[f(4x^2)+f(1)\right]-f(4)+f(1)\\ &=\frac{\frac{x}{2}+\frac{2}{x}}{2x+\frac{1}{2x}}\left[\left(4x+\frac{1}{4x}\right)f(x)-f(\frac{1}{4})+f(1)\right]-f(4)+f(1) \end{align*}where we used the substitutions $(x,x)$, $(x/2,2x)$, $(2x,2x)$, and $(4x,x)$. So \[\left(f(\frac{1}{4})-f(1)\right)\cdot\frac{\frac{x}{2}+\frac{2}{x}}{2x+\frac{1}{2x}}+f(4)-f(1)=\left[\frac{\left(\frac{x}{2}+\frac{2}{x}\right)\left(4x+\frac{1}{4x}\right)}{2x+\frac{1}{2x}}-\left(x+\frac{1}{x}\right)\right]f(x)=\frac{\frac{45}{8}}{2x+\frac{1}{2x}}\cdot f(x)\]and thus \begin{align*} f(x)&=\frac{8}{45}\left(f(\frac{1}{4})-f(1)\right)\left(\frac{x}{2}+\frac{2}{x}\right)+\frac{8}{45}\left(f(4)-f(1)\right)\left(2x+\frac{1}{2x}\right)\\ &=\frac{4f(\frac{1}{4})-20f(1)+16f(4)}{45}\cdot x+\frac{16f(\frac{1}{4})-20f(1)+4f(4)}{45}\cdot\frac{1}{x} \end{align*}as desired.

Ignore this.

Another idea. Let $P(x,y)$ denote the given assertion. $P(e^x,e^y)$ leads us to let $g:\mathbb {R}\to \mathbb {R}$ such that $g(x)=f(e^x)$. Then $P(x,y)$ becomes: $g(x+y)+g(x-y)=2\cosh (y) g(x)$ which is a well known FE. We can work either with complex numbers or like pco here (https://artofproblemsolving.com/community/c6h467179p2616006). We prove that: $g(x)=a\sinh (x)+ b\cosh (y)$ and as a result $f(x)=ax+\frac{b}{x} $.

BOBTHEGR8 wrote: juckter wrote: This problem is quite interesting! It seems like it's very easy to solve if one found the solution space and noticed it is a real vector space, and very difficult to approach otherwise. I had no idea about space, but only noticed the solution $Ax+Bx^{-1}$,after that the $g(x)=f(x)-Ax-Bx^{-1}$ trick is standard. I think these are the same idea!

The answer is $\boxed{f(x)=ax+\tfrac bx}$ for real numbers $a$ and $b$. It is easy to check that this works; so it suffices to prove that these are the only solutions. Let $P(x,y)$ denote the assertion. Notice that if $f$ and $g$ are solutions to the functional equation, $f+g$ and $f-g$ are also solutions. Given $f(1)$ and $f(2)$, we can easily find $a$ and $b$ such that $f(1)=a+b$ and $f(2)=2a+\tfrac b2$. Shift down to give $f(1)=f(2)=0$. We will show that $f\equiv 0$. $P(2,2)$ gives $$\frac52f(2)=f(4)+f(1)\implies f(4)=0.$$Now, $P(2x,2)$ and $P(x,1)$ give $$f(4x)=-f\left(\frac1x\right)=f(x)$$for all $x$, so $P(2,2x)$ gives $$\frac52f(2x)=f(4x)+f(x)=2f(x),$$and thus $f(2x)=\tfrac45f(x)$ and $f(x)=f(4x)=\tfrac{16}{25}f(x)$, which immediately yields $f\equiv 0$, as desired. $\square$

Functional wrote: Determine all functions $f:(0,\infty)\to\mathbb{R}$ satisfying $$\left(x+\frac{1}{x}\right)f(y)=f(xy)+f\left(\frac{y}{x}\right)$$for all $x,y>0$. Here it is I belive a new solution that didn't see yet and actually the first time I did try to solve it I did solve it this way and it did take $1$ and half hour to completed. First in the condition lets take $x\rightarrow \frac{x}{y}$ and condition transform to: $$\left(\frac{x^2+y^2}{xy}\right)f(y)=f(x)+f\left(\frac{y^2}{x}\right)$$ I'm gonna write this with $P(x,y)$. Taking $P(1,x)$ we have: $$f(x^2)=\left(\frac{x^2+1}{x}\right)f(x)-f(1) \hspace{2cm}(1)$$ Now taking $P(x^2,y^2)$ and using relation $(1)$ we have: $$\left(\frac{x^4+y^4}{x^2y^2}\right)f(y^2)=f(x^2)+f\left(\frac{y^4}{x^2}\right)$$$$\Rightarrow \left(\frac{x^4+y^4}{x^2y^2}\right)\left(\frac{y^2+1}{y}\right)f(y)- \left(\frac{x^4+y^4}{x^2y^2}\right)f(1)=\left(\frac{x^2+1}{x}\right)f(x)-f(1)+\left(\frac{x^2+y^4}{xy^2}\right)f\left(\frac{y^2}{x}\right)-f(1)$$$$\Rightarrow \left(\frac{x^4+y^4}{x^2y^2}\right)\left(\frac{y^2+1}{y}\right)f(y)- \frac{(x^2-y^2)^2}{x^2y^2}f(1)=\left(\frac{x^2+1}{x}\right)f(x)+\left(\frac{x^2+y^4}{xy^2}\right)f\left(\frac{y^2}{x}\right)$$ Now we use $f\left(\frac{y^2}{x}\right)=\left(\frac{x^2+y^2}{xy}\right)f(y)-f(x)$ in last relation we have: $$\left(\frac{x^4+y^4}{x^2y^2}\right)\left(\frac{y^2+1}{y}\right)f(y)- \frac{(x^2-y^2)^2}{x^2y^2}f(1)=\left(\frac{x^2+1}{x}\right)f(x)+\left(\frac{x^2+y^4}{xy^2}\right)\left(\left(\frac{x^2+y^2}{xy}\right)f(y)-f(x)\right)$$ After some manipulation we find it is equivalent to: $$\left(\frac{(x^2-1)(x^2-y^2)}{x^2y}\right)f(y)-\frac{(x^2-y^2)^2}{x^2y^2}f(1)=\left(\frac{(y^2-1)(x^2-y^2)}{xy^2}\right)f(x)$$ For $x\neq y$ last relation is equivalent to: $$y(x^2-1)f(y)-(x^2-y^2)f(1)=x(y^2-1)f(x)\Rightarrow (x^2-1)(yf(y)-f(1))=(y^2-1)f(xf(x)-f(1))$$$$\Rightarrow \frac{xf(x)-f(1)}{x^2-1}=\frac{yf(y)-f(1)}{y^2-1}$$ Last relation it is true for all postive real numbers $x,y$ different from each other and different from $1$ hence this means that: $$\frac{xf(x)-f(1)}{x^2-1}$$ is constant for any $x\in\mathbb{R}^+\setminus\{1\}$, so exist constant $c\in\mathbb{R}$ such that: $$\frac{xf(x)-f(1)}{x^2-1}=c\Rightarrow f(x)=\frac{c(x^2-1)+f(1)}{x}$$ for all $x\in\mathbb{R}^+\setminus\{1\}$, still true for $x=1$ hence: $$f(x)=cx+\frac{f(1)-c}{x}=cx+\frac{d}{x}$$ for all $x\in\mathbb{R}^+$. Substituting in first relation we can see that it is true for any constant $c,d\in\mathbb{R}$.

I think A4 is more difficult

juckter wrote: This problem is quite interesting! It seems like it's very easy to solve if one found the solution space and noticed it is a real vector space, and very difficult to approach otherwise. In general how do you know that the solution is going to be a vector space?

hellomath010118 wrote: In general how do you know that the solution is going to be a vector space? Because if $f$ and $g$ are solutions to the functional equation, then so are $f+g$, $100g$, etc. You can see this directly from the given equation.

v_Enhance wrote: hellomath010118 wrote: In general how do you know that the solution is going to be a vector space? Because if $f$ and $g$ are solutions to the functional equation, then so are $f+g$, $100g$, etc. You can see this directly from the given equation. And how do you know there are just ‘2’ f,g?

hellomath010118 wrote: And how do you know there are just ‘2’ f,g? I don't understand the question. The space $\mathbb R^{\mathbb R}$ of all functions $\mathbb R \to \mathbb R$ is a real vector space. Consider the subset $V$ of functions satisfying the given functional equation. Then as I mentioned, $V$ is closed under addition and scaling. Therefore, $V$ is a subspace of $\mathbb R^{\mathbb R}$.

I am asking its dimension;for example how do you know that the dimension is 2 in this case

hellomath010118 wrote: I am asking its dimension;for example how do you know that the dimension is 2 in this case By solving the problem

After analysing I have 1 thing to say: Like you made a handout on guessing properties of mixtilinears , you should also make 1 on guessing solutions to functional equations

He did: http://web.evanchen.cc/handouts/FuncEq-Intro/FuncEq-Intro.pdf

Here is my solution during the TST in Greece (lul). Let $P(x,y)$ denote $\left(x+\frac{1}{x}\right)f(y)=f(xy)+f\left(\frac{y}{x}\right)$ I first note that if $f$ satisfies then $cf$ satisfies too with $c$ constant. Hence, we may discern $2$ cases: A. $f(1)=0$. $P(x,1): f(x)+f(\dfrac{1}{x})=0$ $P(x,y)+P(y,x): (x+\dfrac{1}{x})f(y)+(y+\dfrac{1}{y})f(x)=2f(xy)+f(\dfrac{x}{y})+f(\dfrac{y}{x})=2f(xy) (1)$ Setting $y=yz$ we get $2(x+\dfrac{1}{x})f(yz)+2(yz+\dfrac{1}{yz})f(x)=4f(xyz)$ , which due to $(1)$ becomes $((y+\dfrac{1}{y})f(z)+(z+\dfrac{1}{z})f(y))(x+\dfrac{1}{x})+2(yz+\dfrac{1}{yz})f(x)=4f(xyz)$. Now swapping $z$ with $x$ and substracting leads to $(y+\dfrac{1}{y})(x+\dfrac{1}{x})f(z)+2(yz+\dfrac{1}{yz})f(x)=(y+\dfrac{1}{y})(z+\dfrac{1}{z})f(x)+2(yx+\dfrac{1}{yx})f(z)$ $\Leftrightarrow$ (just simplify) $\dfrac{f(x)x}{x^2-1}=\dfrac{f(z)z}{z^2-1}$ $\Rightarrow$ $f(x)=\dfrac{k(x^2-1)}{x}$ which satisfies. B. $f(1)=1$ The same stuff apply here (I am too lazy to write it all up but once I get the chance (never) I will). Only the last part of the swap gets harder but is easily done with some slick algebraic manipulations. And since we had assumed that $f(1)=1$ the solution set will be $c \times$ (what we find) ( which if I recall correctly should be something in the form of $x+(x-\dfrac{1}{x})s$ where $s$ is an arbitary constant and eventually leads to the solution set $ax+\dfrac{b}{x}$.)

We claim that I am high or the set of all solutions are $f(x) = ux + \frac{v}{x} (u,v \in \mathbb R)$. Note that the set of solutions $f$ form a group so it is easy to see that all these $f$ satisfy the functional equation. For any $f$, replace $f$ by $f - ax - \frac{b}{x}$ with $a,b$ chosen so that $f(1) = f(2) = 0$. Then $f(x) + f(\frac{1}{x}) = 0, f(2x) + f(\frac{2}{x}) = 0$ by letting $y = 1,2$ in the above FE. But from the first relation $f(2x) + f(\frac{1}{2x}) = 0 \implies f(\frac{2}{x}) = f(\frac{1}{2x}) \forall x \implies f(4u) = f(u) \forall u$. Now observe that $\forall n \in \mathbb Z$, $f(x(x^ny)) + f(\frac{x^ny}{x}) = f(x^ny)(x+\frac{1}{x})$ so there exists $\alpha_y, \beta_y$ such that $f(x^ny) = \alpha_y x^n + \beta_y\cdot \frac{1}{x^n}$. Letting $x = 2$ gives that the sequence $\alpha_y x^n + \beta_y\cdot \frac{1}{x^n}$ is bounded as $n$ ranges over $\mathbb Z$ (recall that $f(4u) = f(u)$). This can only happen if $\alpha_y = \beta_y = 0$, i.e. $f(y) = 0$. ་༺ཉༀ༔ཨཱིY͓̽U͓̽H͓̽.༃ༀདྷ༻་

With awang11: The answer is $\boxed{f(x)=px+\frac{q}{x}}$ for any reals $p,q$; this can be checked to work. Now we show these are the only solutions. Note that if some pair of functions $g(x),h(x)$ are both solutions to the given functional equation, then $g(x)+h(x)$ is also a solution. Thus we can shift $f$ by $ax+\frac{b}{x}$ for a suitable choice of $a,b$ so that $f(1)=f(2)=0$; we will show that $f(x)=0$ for all $x$. Plugging $y=1$ into the original functional equation gives $f(x)+f(\frac{1}{x})=0$, and plugging $y=2$ gives $f(2x)+f(\frac{2}{x})=0$. Comibining these gives $f(x)=f(4x)$. Now plugging $x=4$ into the original functional equation gives $(4+\frac{1}{4} )f(y) = f(4y)+f(y/4) = 2f(y)$, so $f(y)=0$ for all $y$ as desired.

ignore this

Define $g : (0, \infty) \to \mathbb{R}$ by $g(x) = f(x) - xf(1)$. Note that $g(1) = 0$ and $g$ satisfies \[ \left( x + \frac{1}{x} \right) g(y) = g(xy) + g \left( \frac{y}{x} \right) \]Let $P(x, y)$ denote this assertion. Also define $k : (0, \infty) \setminus \{1\} \to \mathbb{R}$ by \[ k(x) = \frac{g(x)}{x - \frac{1}{x}} .\]We will show that $k(x)$ is constant. Now for some substitutions: \[ P(x, 1) \implies g(x) = -g\left( \frac{1}{x} \right) \]\[ P(x, x) \implies g(x^2) = \left( x + \frac{1}{x} \right) g(x) \implies k(x) = k(x^2) .\]\[ P(a, b) + P(b, a) \implies \left( a + \frac{1}{a} \right) g(b) + \left( b + \frac{1}{b} \right) g(a) = 2g(ab) \]So we can use $P(x, x)$ and $P(a, b) + P(b, a)$ to get an expression for $g(a^2b^2)$, and we can get another by using $P(a^2, b^2) + P(b^2, a^2)$. Equating these we get: \begin{align*} \left( ab + \frac{1}{ab} \right) \left( \left( a + \frac{1}{a} \right) g(b) + \left( b + \frac{1}{b} \right) g(a) \right) &= \left( a^2 + \frac{1}{a^2} \right) g(b^2) + \left( b^2 + \frac{1}{b^2} \right) g(a^2) \\ \implies (a^2b^2 + 1) ((a^2 + 1)(b^2 - 1)k(b) + (b^2 + 1)(a^2 - 1)k(a)) &= (a^4 + 1)(b^4 - 1)k(b) + (b^4 + 1)(a^4 - 1)k(a) \\ \implies (a^4b^4 - a^4b^2 + a^2b^4 - a^2 + b^2 - 1)k(b) + (a^4b^4 + a^4b^2 - a^2b^4 + a^2 - b^2 - 1)k(a) &= (a^4b^4 - a^4 + b^4 - 1)k(b) + (a^4b^4 + a^4 - b^4 - 1)k(a) \\ \implies (a^2b^4 - a^4b^2 + a^4 - b^4 - a^2 + b^2)k(b) &= (a^2b^4 - a^4b^2 + a^4 - b^4 - a^2 + b^2)k(a) \end{align*}So as long as $(a^2b^4 - a^4b^2 + a^4 - b^4 - a^2 + b^2) \neq 0$ then we can deduce that $k(a) = k(b)$. For any fixed $a \neq 1$, this is a degree 4 non constant polynomial in $b$, so there are at most 4 values of $b$ for which we can't immediately deduce $k(a) = k(b)$. But we know that $k(x) = k(x^{2^k})$ so it's easy to also deduce it for these values. Hence $k(x)$ is constant, and therefore $g$ must be of the form $Ax - \frac{A}{x}$ for some constant $A$, and hence $f$ must be of the form $Ax + \frac{B}{x}$. It is easy to see that any solution of this form satisfies the original equation.

dame dame

Let $a=\frac{2f(2)-f(1)}3$ and $b=\frac{4f(1)-2f(2)}3$. Also, define $g(x)=f(x)-ax-\frac bx$. Then $g(1)=g(2)=0$ and we have the assertion $P(x,y):\left(x+\frac1x\right)g(y)=g(xy)+g\left(\frac yx\right)$. $P(x,1)\Rightarrow g(x)+g\left(\frac1x\right)=0$ $P(2x,2)\Rightarrow g(4x)+g\left(\frac1x\right)=0\Rightarrow g(4x)=g(x)$ $P(2,2x)\Rightarrow g(2x)=\frac45g(x)\Rightarrow g(4x)=\frac{16}{25}g(x)$ Then we must have $g(x)=0$, so $\boxed{f(x)=ax+\frac bx}$ for any real constants $a,b\in\mathbb R$. This works since $x$ and $\frac1x$ are solutions and whenever $f,g$ satisfy the FE, $f+g$ and $cf$ also do for any $c\in\mathbb R$.

Functional wrote: Determine all functions $f:(0,\infty)\to\mathbb{R}$ satisfying $$\left(x+\frac{1}{x}\right)f(y)=f(xy)+f\left(\frac{y}{x}\right)$$for all $x,y>0$. Cute and easy problem. Firstly, note that the set of all solutions forms a vector space, and since \(x\), \(\frac{1}{x}\) are indeed solutions, it suffices to show that the dimension of the vector space is \(2\). To show this, let there exist two distinct elements \(a\) and \(b\) whose image is \(0\). Since we can always transform, we may assume that \(f(1)=f(2)=0\) where \(f\) is a function in the vector space. First of all, \(P(x,1)\) and \(P(x,2)\) give \(f(x)=-f\left(\frac{1}{x}\right)\) and \(f(2x)=-f\left(\frac{2}{x}\right)\). From these, \(P\left(2,\frac{1}{x}\right)\) gives us that \(f(2y)=\frac{-5}{8}f(y)\). Now \(P(2,2y)\) gives us that \[\frac{-25}{8}f(y)=f(4y)+f(y)\]or \(f(y)=-\frac{225}{64}f(y)\) implying that \(f\) is identically \(0\) as desired. Therefore, the dimension of our vector space is \(2\) and since \(x\) and \(\frac{1}{x}\) are solutions, we conclude that the solution set is \(f(x)=ax+\frac{b}{x}\) for all \(a,b\in\mathbb{R}\) and \(x\in\mathbb{R}^+\).

We claim the answer is $\boxed{f(x)=ax+\frac{b}{x}}$ for real constants $a$ and $b$. This clearly works (we have $ayx+\frac{ay}{x}+\frac{bx}{y}+\frac{b}{xy}=axy+\frac{b}{xy}+\frac{ay}{x}+\frac{bx}{y}$, which is true). Thus it suffices to show that these are the only solutions. Let $a=\frac{2f(2)-f(1)}{3}$ and let $b=\frac{4f(1)-2f(2)}{3}$ and $g(x)=f(x)-ax-\frac{b}{x}$. We have $g(1)=f(1)-a-b=0$ and $g(2)=f(2)-2a-\frac{b}{2}=0$. Claim: $(x+\frac{1}{x})g(y)=g(xy)+g(\frac{y}{x})\forall x,y>0$. Proof: We have $(x+\frac{1}{x})(g(y)+ay+\frac{b}{y})=(x+\frac{1}{x})g(y)+axy+\frac{bx}{y}+\frac{ay}{x}+\frac{b}{xy}=g(xy)+axy+\frac{b}{xy}+g(\frac{y}{x})+\frac{ay}{x}+\frac{bx}{y}$, which after cancelling terms proves our claim. Let $P(x,y)$ denote the assertion that \[(x+\frac{1}{x})g(y)=g(xy)+g(\frac{y}{x})\] $P(x,1): g(x)+g(\frac{1}{x})=0$. $P(2x,2): g(4x)+g(\frac{1}{x})=0$. So $g(4x)=g(x)$. $P(2,2x): 2.5g(2x)=g(4x)+g(x)=2g(x)\implies g(2x)=\frac{4}{5}g(x)$. Thus, $g(4x)=\frac{16}{25}g(x)=g(x)\implies g\equiv 0$. This proves that $f(x)=ax+b$, for real constants $a$ and $b$.

Let $P(x,y)$ be the assertion given in the statement. WLOG $f(1)=f(2)=0.$ Note that it is possible by shifting. Compare $P(x,1)$ and $P(x,2)$ to get $f(x)=f(4x).$ Now $P(2,2y)$ readily gives $f(x)=ax+bx^{-1}~~\forall a,b \in \mathbb{R},$ which obviously works. $\blacksquare$

ZETA_in_olympiad wrote: Claim: No solution exists except $f(x)=ax+bx^{-1}~~\forall a,b \in \mathbb{R},$ which obviously fits. Proof. WLOG $f(1)=f(2)=f(0).$ Note that it is possible by shifting. Let $P(x,y)$ be the assertion. Compare $P(x,1)$ and $P(x,2)$ to get $f(x)=f(4x).$ Now $P(2,2y)$ readily yields the claim. $\blacksquare$ Um? "It is possible by shifting" Seems okay, but how do you shift the domain as well. $f(0)$ isn't even defined in the domain.

Let $P(x,y)$ denote the assertion, then $P(x,1)$ and $P(2x,2)$ give \begin{align*} \left(x+\frac1x\right)f(1) &=f(x)+f\left(\frac{1}{x}\right) \\ \left(2x+\frac1{2x}\right)f(2) &=f(4x)+f\left(\frac{1}{x}\right) \end{align*}What we get is that \[f(x)-f(1)\left(x+\frac1x\right)=f(4x)-f(2)\left(2x+\frac1{2x}\right)\]but also $P(2,2x)$ gives us \[\frac{5}{2}f(2x)=f(4x)+f(x)\]Note that in the original equation, if $f_1$ and $f_2$ work then $af_1+bf_2$ also works. Note that $f(x)=x$ and $\tfrac1x$ work, so we can assume $f(1)=f(2)=0$ which solves the problem since $f(x)=f(4x)$ so $f(2x)=\tfrac54 f(x)$ so $f(4x)=\tfrac{25}{16}f(x)$ which implies $f=0$.

They mean that $f(1)=f(2)=0$ I believe.

Am I clowning I will credit Jorzdan Lethobsquik for the idea of substituting $h(r) = g(e^r),$ and RazzMath for the limit idea Let $f(x)$ be any solution. Let $g(x)=f(x)+Ax+\frac{B}{x}$ where $A $ and $B$ are chosen such that $g(1)=g(e)=0$. So $g(x)$ satisfies the given. Now we try to show $g$ is $0$ everywhere. First, $g(x) = -g(1/x)$ and $g(x) = -g(e/x).$ We can also find $g(e^{a}) = 0$ for any positive integer $a.$ So now $g(x) = -g(e^a/x)$ for any positive integer $a.$ Now define $h(r) = g(e^{r})$ for any real $r.$ So now $h(-x) = -h(1-x)$ and $h(-x) = -h(x)$, and $h(a) = 0$ for any positive integer $a.$ So basically now $h(x) = -h(a-x)$ for any $x.$ So actually $h$ is odd and is periodic mod $2.$ FTSOC it's not zero everywhere, so some $h(r)$ is nonzero. Then $$\left(\frac{1}{e^{a}} + e^a \right) h(r) = h(r+a) + h(r-a).$$Take positive integers $a$ going to infinity, $h(r+a) + h(r-a)$ goes to infinity or negative infinity. But $h(r+a+1)+h(r-a-1)$ and $h(r-a)+h(r+a)$ have opposite signs. This is bad.

Note that the solution set of this function is an ${\mathbb R}$-vector space. Then, since $f(x) = x$ and $f(x) = \frac1x$ are both solutions, any linear combinations of them is still a solution. Claim: If $f(1) = f(2) = 0$, then $f(x) = 0$. Proof. By $P(x, 1)$ and $P(2x, 2)$ it follows that \[ f(x) + f\left(\frac1x\right) = f(4x) + f\left(\frac1x\right) = 0. \]and thus $f(x) = f(4x)$. Thus, by $P(4, y)$ it follows that $2f(y) = \frac{17}{4}f(y)$ and thus $f(y) = 0$. $\blacksquare$ As such, the vector space has dimension $2$ and $f(x) = ax + \frac{b}{x}$ is the solution set.

I claim the answer is $f(x)=ax+\frac{b}{x}$ for constants $a$ and $b$. To begin, shift by such a function to get that it is sufficient to show $f(x)=0$ given $f(1)=f(2)=0$. Plugging in $y=1$ yields $f(x)+f(\frac{1}{x})=0$. Plugging in $y=2$ yields $f(2x)+f(\frac{2}{x})=0$. These two combine to give $f(4x)=f(x)$. This means that we can plug $x=4$ into our equation to get that $\frac{17}{4}f(y)=f(4y)+f(\frac{y}{4})=2f(y)$, which implies $f(y)=0$ for all $y$.

Prove that $f(x)$ is continuity. Let $P(x,y)$ denote the given assertion. Let's say $F_{x}$ the set of continuous functions by $x$. In first $f(ax)+f\left(\frac{b}{x}\right)$ for all $a,b > 0$ by $P(x \sqrt{\frac{a}{b}},\sqrt{ab})$ In second $f(x)-f(ax) \in F_{x}$ for all $a>0$(difference of two equations the line above) Note that $f(\frac{y}{x})+f(\frac{x}{y}) \in F_{x}(a=\frac{1}{y}, b=y)$ and $f(\frac{x}{y})+f(\frac{y}{x})=(x+\frac{1}{x})f(y)+(y+\frac{1}{y})f(x)-2f(xy)$ By subtraction $2f(x)-2f(xy)+(x+\frac{1}{x})f(y)$ сontinuity is not lost and $(y+\frac{1}{y}-2)f(x) \in F_{x}$ It means $f(x)$ continuity. After this seeing that they can consider $f(1)=f(2)=0$ and if $f(x)=f(y)=0$ then $f(\frac{x^{2}}{y})=f(\sqrt{xy})=0$. It follows from this that for all $x$ having finite binary notation $f(x)=0$. Therefore $f(x)=0$ for all $x$ by continuetly.