Dtjytyk

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

This seems quite hard. $f(x)=begin{cases}frac{1}{x}, xneq 0 \ 0, x=0 end{cases}$ 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.

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
$$fleft(f(x)y+ frac{x}{y}right)=0, forall ,x^2+y^2=S_n$$

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 $f:mathbb{R}tomathbb{R}$ satisfies $f(1)=1$ and $fleft(yf(x)+frac{x}{y}right)=xyfleft(x^2+y^2right)$ for all $x,yinmathbb{R},yneq0$, then $f(0)=0$ and $f(x)=frac{1}{x}$ for all $xneq0$. 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 $x,yinmathbb{R}, yneq0$ write $P(x,y)$ for the assertion $fleft(yf(x)+frac{x}{y}right)=xyfleft(x^2+y^2right)$. Furthermore, we define $mathcal{N}:={x>0 | f(x)=0}$, and assume that $mathcal{N}neqemptyset$.

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

Now notice that for $xneq 0$ and $alphainmathcal{N}$
$$
P(alpha,alpha^2 x):quad fleft(frac{1}{alpha x}right)=alpha^3 x fleft(alpha^4left(x^2+frac{1}{alpha^2}right)right)\
P(frac{1}{alpha}, x):quad fleft(frac{1}{alpha x}right)=frac{x}{alpha} fleft(x^2+frac{1}{alpha^2}right)
$$
and thus $alpha^4 fleft(alpha^4left(x^2+frac{1}{alpha^2}right)right)=fleft(x^2+frac{1}{alpha^2}right)$ for all $xneq 0$, or in more simple terms
$$
(*)qquadalpha^4 f(alpha^4 z) = f(z) quadforall alphainmathcal{N}, z>frac{1}{alpha^2}
$$
Now for a fixed $alphainmathcal{N}$ and $y>0$ let $ninmathbb{N}$ 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
$$
alpha^4 f(alpha^4 y)=frac{alpha^4}{alpha_n^4} alpha_n^4 fleft(alpha_n^4frac{alpha^4 y}{alpha_n^4}right)overset{frac{alpha^4 y}{alpha_n^4}>frac{1}{alpha_n^2}}{=}frac{1}{alpha_n^4}alpha^4 fleft(alpha^4frac{y}{alpha_n^4}right)overset{frac{y}{alpha_n^4}>frac{1}{alpha^2}}{=}frac{1}{alpha_n^4}fleft(frac{y}{alpha_n^4}right)overset{y>alpha_n^2}{=}f(y).
$$
Thus we can strengthen $(*)$ and actually have
$$
(**)qquadalpha^4 f(alpha^4 z) = f(z) quadforall alphainmathcal{N}, z>0.
$$
In particular, for $z=frac{1}{alpha^2}$ we get $alpha^4f(alpha^2)=fleft(frac{1}{alpha^2}right)$. On the other hand, we see that as $1+alpha^2,1+frac{1}{alpha^2}inmathcal{N}$, 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
$$
f(alpha^2)=fleft(frac{1+alpha^2}{1+frac{1}{alpha^2}}right)=fleft(frac{1+frac{1}{alpha^2}}{1+alpha^2}right)=fleft(frac{1}{alpha^2}right)
$$
and thus, as $alpha>0$ and $alphaneq 1$, we have $alpha^4f(alpha^2)=fleft(frac{1}{alpha^2}right)=f(alpha^2)implies f(alpha^2)=0$ so $alpha^2inmathcal{N}$. But then also $alpha^4inmathcal{N}$ 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 $xneq 0$. 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 $underline{xnot=0},$ then unknowns can be swapped. So
$$fleft(yf(x)+dfrac xyright) = xyf(x^2+y^2) = fleft(xf(y)+dfrac yxright),tag1$$
with the partial cases
$$begin{cases}
y=1,quad f(f(x)+x) = xf(x^2+1) = fleft(x+dfrac1xright) 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) = gleft(x+dfrac1xright) 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(Rpmsqrt{R^2-4})\[4pt]
R_{1,2}=x+dfrac1x = pmleft(sqrt {L-1}+dfrac1{sqrt{L-1}}right),
end{cases}
one can present equation $(4.1)$ in the forms of
begin{cases}
g(x) = gleft(dfrac 12x(xpmsqrt{x^2-4})right)hspace{382mu}(5.1)\[4pt]
g(x) = gleft(pmleft(sqrt{x-1} + dfrac1{sqrt{x-1}}right)right).hspace{330mu}(5.2)
end{cases}
From $(5.2)$ should $g(1)=g(pminfty),$
$$g(pminfty)=1.$$
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
$$g(x)=1quad forall quad xin((-infty,-2]cup[2,infty)).tag6$$

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

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,quadtext{if}quad x=0\[4pt]
dfrac1x,quadtext{otherwize}
end{cases}$$
is the single non-trivial solution.

Let $f:textbf{R}rightarrowtextbf{R}$, such that $f(1)=1$ and
$$
fleft(yf(x)+frac{x}{y}right)=xyf(x^2+y^2)textrm{, }forall (x,y)intextbf{R}timestextbf{R}^{*}tag 1
$$
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
$$
xf(x^2+1)=f(f(x)+x)textrm{, }xneq 0.tag 2
$$
Also using the symmetry we get
$$
fleft(yf(x)+frac{x}{y}right)=fleft(xf(y)+frac{y}{x}right)textrm{, }x,yneq 0
$$
For $y=1$, we get
$$
fleft(x+frac{1}{x}right)=f(f(x)+x)textrm{, }xneq 0tag 3
$$
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
$$
f(x)=hleft(x+frac{1}{x},xright)tag{4.1}
$$

From (4) with $xrightarrow x+frac{1}{x}$ we get
$$
fleft(x+frac{1}{x}right)=fleft(y+hleft(x+frac{1}{x},yright)right).
$$
Hence for $y=x$ in the above identity we get
$$
fleft(x+frac{1}{x}right)=fleft(x+hleft(x+frac{1}{x},xright)right)=f(x+f(x))
$$
Hence we can get (3). From (2) and (3) we get also
$$
xf(x^2+1)=fleft(x+frac{1}{x}right)tag 6
$$
Setting $xrightarrow x^{-1}$ in (6)
$$
frac{1}{x}fleft(1+frac{1}{x^2}right)=fleft(x+frac{1}{x}right)=xf(x^2+1).tag 7
$$
Hence if we set $x^2rightarrow x>0$ in (7), then
$$
fleft(1+frac{1}{x}right)=xf(x+1)textrm{, for all }x>0.tag 8
$$
Set also $x=y-1>0$ in (8), then
$$
fleft(frac{y}{y-1}right)=(y-1)f(y)textrm{, }y>1.
$$
With $y=1/w>1$ we get
$$
frac{1}{1-w}fleft(frac{1}{1-w}right)=frac{1}{w}fleft(frac{1}{w}right).
$$
Hence if $0<w<1$ and $g(w):=frac{1}{w}fleft(frac{1}{w}right)$, then
$$
f(x)=x^{-1}gleft(x^{-1}right)textrm{, where }x>1textrm{ and }g(1-w)=g(w)textrm{, }0<w<1tag 9
$$
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
$$
f(x)=hleft(x+frac{1}{x},xright)tag{11}
$$
with
$$
frac{1}{x-1}hleft(x-1+frac{1}{x-1},frac{1}{x-1}right)=frac{1}{x}hleft(x+frac{1}{x},frac{1}{x}right)tag{12}
$$
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.