# Solving $f(yf(x)+x/y)=xyf(x^2+y^2)$ over the reals

Find all functions $$f:\mathbb{R}\to \mathbb{R}$$ such that $$f(1)=1$$ and for all real numbers $$x$$ and $$y$$ with $$y \neq 0$$, $$f\Bigg (yf(x)+\frac{x}{y}\Bigg)=xyf(x^2+y^2)$$

This seems quite hard. $$f(x)=\begin{cases}\frac{1}{x}, x\neq 0 \\ 0, x=0 \end{cases}$$ works by inspection, as does $$f(x)=0$$ (though I'm not sure if this is legitimate. If we set $$y=1$$, we have $$f(f(x)+x)=xf(x^2+1)$$. If we set $$x=1$$, we have $$f(\frac{y^2+1}{y})=yf(y^2+1)$$. which seems to imply $$f(f(x)+x)=f(\frac{x^2+1}{x})$$. If I could show that $$f$$ is injective I could go further, but I'm stuck - subbing in other values doesn't really seem to lead anywhere either.

I believe this problem came from an Olympiad camp.

## 4 Answers

Here is one approach,

the equation $$y f(x)+ \frac{x}{y}=x^2 + y^2$$ for $$x \neq 0$$ does have at least one real root, since the equation is equivalent to a cubic equation. Denote this root by $$\lambda$$.

inputting $$\lambda$$ into the equation we find $$f(x^2+y^2)=x \lambda f(x^2+y^2)$$. Now, notice that $$x^2+y^2 \neq 0$$, if you can show that for some $$\sigma$$ that $$f(\sigma)=0\, \iff \sigma =0$$ then the result folllows since, then $$\lambda = \frac{1}{x}$$ and form the initial functional equation you would obtain $$f(x)\lambda + \frac{x}{\lambda}=x^2+\lambda^2\, \implies f(x)=\frac{1}{x}$$

Thus, the solution to the problem is found by asserting that you can prove that

$$f(\sigma)=0 \, \iff \sigma = 0$$

The road to glory I belive requires analysis of the following relation

$$f(f(x)+x) = xf(x^2+1)$$

Allows us to show that for $$x \neq 0$$ then $$f(x)=0 \implies f(x^2+1)=0$$ which implies $$\exists\,$$ a sequence $$S_n \to \infty$$ such that $$f(S_n)=0$$. For the following, assume $$f$$ is continuous.

Now, take $$x^2+y^2=S_n$$, the initial relation implies that $$f\left(f(x)y+ \frac{x}{y}\right)=0\, \forall \,x^2+y^2=S_n$$

Now, for $$y \to 0^+$$ and $$x>0$$ one finds that $$f(x)y+\frac{x}{y} \to \infty$$ as $$x \to \sqrt{S_n}$$, thus $$f$$ takes zero values in the neighbourhood of $$\infty$$.

Since $$f(y+\frac{1}{y}) = y f(1+y^2)\, \implies f(x)=-f(-x)\, \forall\, |x| \geq 2$$ we find the same result in the neighbourhood of $$- \infty$$. Notice now that for $$y>1$$ that $$y+\frac{1}{y} > 1+y^2$$ one can translate the zeroes in the neighbourhood of $$\infty$$ until you reach $$2$$. So that

$$f(x)=0$$ on $$(-\infty, -2]\, \cup \, [2, \infty)$$

Not sure where else to go from here, but this may provide a useful aid to a full solution.

Summary: This solution shows that if a function $$f:\mathbb{R}\to\mathbb{R}$$ satisfies $$f(1)=1$$ and $$f\left(yf(x)+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$$ for all $$x,y\in\mathbb{R},y\neq0$$, then $$f(0)=0$$ and $$f(x)=\frac{1}{x}$$ for all $$x\neq0$$. No additional assumptions on $$f$$ are necessary!

Thanks @Sil for giving all these references! I meanwhile came up with a solution myself, and last but not least because not many solutions of this problem seem to be around, I would like to share mine.

Suppose $$f$$ is a solution to this functional equation and for $$x,y\in\mathbb{R},\ y\neq0$$ write $$P(x,y)$$ for the assertion $$f\left(yf(x)+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$$. Furthermore, we define $$\mathcal{N}:=\{x>0\ |\ f(x)=0\}$$, and assume that $$\mathcal{N}\neq\emptyset$$.

As @Kevin already pointed out, we have $$\alpha\in\mathcal{N}\implies\alpha^2+1\in\mathcal{N}$$, and in particular $$\mathcal{N}$$ is unbounded. Furthermore, $$P(1,\alpha)$$ gives also $$\alpha\in\mathcal{N}\implies\alpha+\frac{1}{\alpha}\in\mathcal{N}$$. Now, if $$\alpha,\beta\in\mathcal{N}$$ then $$P(\alpha,\beta):\quad f\left(\frac{\alpha}{\beta}\right)=\alpha\beta f\left(\alpha^2+\beta^2\right)\\ P(\beta,\alpha):\quad f\left(\frac{\beta}{\alpha}\right)=\beta\alpha f\left(\beta^2+\alpha^2\right)$$ and thus $$f\left(\frac{\alpha}{\beta}\right)=f\left(\frac{\beta}{\alpha}\right)$$. Therefore, if $$\alpha\in\mathcal{N}$$ then $$f\left(\frac{1}{\alpha}\right)=f\left(\frac{\alpha+\frac{1}{\alpha}}{\alpha^2+1}\right)=f\left(\frac{\alpha^2+1}{\alpha+\frac{1}{\alpha}}\right)=f(\alpha)=0$$ and thus also $$\frac{1}{\alpha}\in\mathcal{N}$$. This gives, together with the unboundedness of $$\mathcal{N}$$, the existence of $$(\alpha_n)_{n\in\mathbb{N}}\in\mathcal{N}^\mathbb{N}$$ with $$\lim_{n\to\infty}\alpha_n=0$$.

Now notice that for $$x\neq 0$$ and $$\alpha\in\mathcal{N}$$ $$P(\alpha,\alpha^2 x):\quad f\left(\frac{1}{\alpha x}\right)=\alpha^3 x f\left(\alpha^4\left(x^2+\frac{1}{\alpha^2}\right)\right)\\ P(\frac{1}{\alpha}, x):\quad f\left(\frac{1}{\alpha x}\right)=\frac{x}{\alpha} f\left(x^2+\frac{1}{\alpha^2}\right)$$ and thus $$\alpha^4 f\left(\alpha^4\left(x^2+\frac{1}{\alpha^2}\right)\right)=f\left(x^2+\frac{1}{\alpha^2}\right)$$ for all $$x\neq 0$$, or in more simple terms $$(*)\qquad\alpha^4 f(\alpha^4 z) = f(z) \quad\forall \alpha\in\mathcal{N},\ z>\frac{1}{\alpha^2}$$ Now for a fixed $$\alpha\in\mathcal{N}$$ and $$y>0$$ let $$n\in\mathbb{N}$$ be such that $$y>\max\{\frac{\alpha_n^2}{\alpha^4},\frac{\alpha_n^4}{\alpha^2},\alpha_n^2\}$$, which exists as $$(\alpha_n)\to 0$$. Then $$\alpha^4 f(\alpha^4 y)=\frac{\alpha^4}{\alpha_n^4} \alpha_n^4 f\left(\alpha_n^4\frac{\alpha^4 y}{\alpha_n^4}\right)\overset{\frac{\alpha^4 y}{\alpha_n^4}>\frac{1}{\alpha_n^2}}{=}\frac{1}{\alpha_n^4}\alpha^4 f\left(\alpha^4\frac{y}{\alpha_n^4}\right)\overset{\frac{y}{\alpha_n^4}>\frac{1}{\alpha^2}}{=}\frac{1}{\alpha_n^4}f\left(\frac{y}{\alpha_n^4}\right)\overset{y>\alpha_n^2}{=}f(y).$$ Thus we can strengthen $$(*)$$ and actually have $$(**)\qquad\alpha^4 f(\alpha^4 z) = f(z) \quad\forall \alpha\in\mathcal{N},\ z>0.$$ In particular, for $$z=\frac{1}{\alpha^2}$$ we get $$\alpha^4f(\alpha^2)=f\left(\frac{1}{\alpha^2}\right)$$. On the other hand, we see that as $$1+\alpha^2,1+\frac{1}{\alpha^2}\in\mathcal{N}$$, we have by again combining $$P(1+\alpha^2,1+\frac{1}{\alpha^2})$$ and $$P(1+\frac{1}{\alpha^2},1+\alpha^2)$$ that $$f(\alpha^2)=f\left(\frac{1+\alpha^2}{1+\frac{1}{\alpha^2}}\right)=f\left(\frac{1+\frac{1}{\alpha^2}}{1+\alpha^2}\right)=f\left(\frac{1}{\alpha^2}\right)$$ and thus, as $$\alpha>0$$ and $$\alpha\neq 1$$, we have $$\alpha^4f(\alpha^2)=f\left(\frac{1}{\alpha^2}\right)=f(\alpha^2)\implies f(\alpha^2)=0$$ so $$\alpha^2\in\mathcal{N}$$. But then also $$\alpha^4\in\mathcal{N}$$ which gives by $$(**)$$ $$0=\alpha^4 f(\alpha^4)\overset{(**)}{=} f(1)=1,$$ contradiction!

Therefore we conclude that $$\mathcal{N}$$ is empty, and as @Kevin already saw this gives $$f(x)=\frac{1}{x}$$ for all $$x\neq 0$$. As $$P(0,y)$$ gives $$f(0)=0$$, we have uniquely determined $$f$$, and we can verify easily that $$f$$ is indeed a solution of the equation at hand.

$$\color{brown}{\textbf{Some forms of the equation.}}$$

If $$\underline{x\not=0},$$ then unknowns can be swapped. So $$f\left(yf(x)+\dfrac xy\right) = xyf(x^2+y^2) = f\left(xf(y)+\dfrac yx\right),\tag1$$ with the partial cases $$\begin{cases} y=1,\quad f(f(x)+x) = xf(x^2+1) = f\left(x+\dfrac1x\right) \hspace{226mu}(2.1)\\[4pt] y=x,\quad f(xf(x)+1) = x^2f(2x^2) \hspace{350mu}(2.2)\\[4pt] \end{cases}$$ Denote $$g(x) = xf(x)\tag3,$$ then from $$(2)$$ should $$\begin{cases} g(1+x^2) = g\left(x+\dfrac1x\right) \hspace{432mu}(4.1)\\[4pt] g(g(x)+1) = \frac12g(2x^2)(g(x)+1)\hspace{370mu}(4.2)\\[4pt] \end{cases}$$ Assume $$g(x)$$ continuous function.

$$\color{brown}{\textbf{Corollaries from the formula (4.1).}}$$

Using the relationships between the arguments in $$(4.1)$$ in the form of $$\begin{cases} L_{1,2}=1+x^2=\dfrac 12R(R\pm\sqrt{R^2-4})\\[4pt] R_{1,2}=x+\dfrac1x = \pm\left(\sqrt {L-1}+\dfrac1{\sqrt{L-1}}\right), \end{cases}$$ one can present equation $$(4.1)$$ in the forms of $$\begin{cases} g(x) = g\left(\dfrac 12x(x\pm\sqrt{x^2-4})\right)\hspace{382mu}(5.1)\\[4pt] g(x) = g\left(\pm\left(\sqrt{x-1} + \dfrac1{\sqrt{x-1}}\right)\right).\hspace{330mu}(5.2) \end{cases}$$ From $$(5.2)$$ should $$g(1)=g(\pm\infty),$$ $$g(\pm\infty)=1.$$ Also, formula $$(5.2)$$ allows to assign to each point of the interval $$(1,2)$$ the point of the interval $$(2,\infty)$$ with the same value of the function $$g.$$

At the same time, repeated recursive application of formula $$(5.1)$$ allows to prove that $$g(x)=1\quad \forall \quad x\in((-\infty,-2]\cup[2,\infty)).\tag6$$

$$\color{brown}{\textbf{Corollaries from the formula (4.2).}}$$

Let us consider such neighbour of the point $$x=1,$$ where $$g(x)>0.$$ Then the right part of the system $$(4.2)$$ can be presented in the form of $$g(2x^2)(1+g(x)) = 2.\tag7$$ Applying $$(7)$$ for $$x$$ from $$1$$ to $$+0$$ and from $$-2$$ to $$-0,$$ easy to see that $$g(x)=1.\tag8$$

The value in the singular point $$x=0$$ can be defined immediately from the equaion $$(1)$$ and equals to zero.

Theerefore, the OP solution $$f(x) = \begin{cases} 0,\quad\text{if}\quad x=0\\[4pt] \dfrac1x,\quad\text{otherwize} \end{cases}$$ is the single non-trivial solution.

• How do you conclude that $f(-y)=f(y)$ from $(1)$? – Servaes Mar 29 at 21:49
• Also, how do you account for the solution $f=0$? – Servaes Mar 29 at 21:54
• In fact, if $f$ is monotonic it is injective and then the fact that $$f(yf(x)+\tfrac{x}{y})=xyf(x^2+y^2)=(-x)(-y)f((-x)^2+(-y)^2)=f(-yf(-x)+\tfrac{-x}{-y}),$$ shows that $f(-x)=-f(x)$ for all $x\in\Bbb{R}$, directly contradicting $(2)$ unless $f=0$. – Servaes Mar 29 at 22:00
• Also, I don't see immediately how you prove that $f$ is monotonic on these intervals; your observation partitions $(1,\infty)$ into infinite sequences on which $f$ is monotonic, but I don't see how this implies that $f$ is monotonic on the entire interval. The same for the interval $(0,1)$. – Servaes Mar 29 at 22:06
• @Servaes Thanks! Fixed all. – Yuri Negometyanov Apr 1 at 19:17

Let $$f:\textbf{R}\rightarrow\textbf{R}$$, such that $$f(1)=1$$ and $$f\left(yf(x)+\frac{x}{y}\right)=xyf(x^2+y^2)\textrm{, }\forall (x,y)\in\textbf{R}\times\textbf{R}^{*}\tag 1$$ Set $$x=0$$, $$y=1$$ in (1), then easily $$f(0)=0$$ (the case $$f(0)\neq 0$$ is trivial). For $$y=1$$ in (1) we get $$xf(x^2+1)=f(f(x)+x)\textrm{, }x\neq 0.\tag 2$$ Also using the symmetry we get $$f\left(yf(x)+\frac{x}{y}\right)=f\left(xf(y)+\frac{y}{x}\right)\textrm{, }x,y\neq 0$$ For $$y=1$$, we get $$f\left(x+\frac{1}{x}\right)=f(f(x)+x)\textrm{, }x\neq 0\tag 3$$ Assume that $$h(x,y)$$ is a surface (function) such that $$f(y+h(x,y))=f(x).\tag 4$$ Note.

One choise of $$h(x,y)$$ is $$h(x,y)=-y+x$$ but may be there others. For example if $$f(x)=x^2-x+1$$, then $$h(x,y)=x-y$$ or $$h(x,y)=1-x-y$$.

We assume here that $$f(x)=h\left(x+\frac{1}{x},x\right)\tag{4.1}$$
From (4) with $$x\rightarrow x+\frac{1}{x}$$ we get $$f\left(x+\frac{1}{x}\right)=f\left(y+h\left(x+\frac{1}{x},y\right)\right).$$ Hence for $$y=x$$ in the above identity we get $$f\left(x+\frac{1}{x}\right)=f\left(x+h\left(x+\frac{1}{x},x\right)\right)=f(x+f(x))$$ Hence we can get (3). From (2) and (3) we get also $$xf(x^2+1)=f\left(x+\frac{1}{x}\right)\tag 6$$ Setting $$x\rightarrow x^{-1}$$ in (6) $$\frac{1}{x}f\left(1+\frac{1}{x^2}\right)=f\left(x+\frac{1}{x}\right)=xf(x^2+1).\tag 7$$ Hence if we set $$x^2\rightarrow x>0$$ in (7), then $$f\left(1+\frac{1}{x}\right)=xf(x+1)\textrm{, for all }x>0.\tag 8$$ Set also $$x=y-1>0$$ in (8), then $$f\left(\frac{y}{y-1}\right)=(y-1)f(y)\textrm{, }y>1.$$ With $$y=1/w>1$$ we get $$\frac{1}{1-w}f\left(\frac{1}{1-w}\right)=\frac{1}{w}f\left(\frac{1}{w}\right).$$ Hence if $$0 and $$g(w):=\frac{1}{w}f\left(\frac{1}{w}\right)$$, then $$f(x)=x^{-1}g\left(x^{-1}\right)\textrm{, where }x>1\textrm{ and }g(1-w)=g(w)\textrm{, }0 The solution (9) satisfies (8),(7),(6), but for to holds (3) we get plus a new functional equation for $$f(x)$$ and this is (4.1).

Hence the general solution is $$f(x)=h\left(x+\frac{1}{x},x\right)\tag{11}$$ with $$\frac{1}{x-1}h\left(x-1+\frac{1}{x-1},\frac{1}{x-1}\right)=\frac{1}{x}h\left(x+\frac{1}{x},\frac{1}{x}\right)\tag{12}$$ and $$h(x,y)$$ solution of $$f(y+h(x,y))=f(x).\tag{13}$$ An obvious solution of $$(13)$$ is $$h(x,y)=x-y$$, but it may exist and other solutions.