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Find all functions $f:mathbbRto mathbbR$ such that $f(1)=1$ and for all real numbers $x$ and $y$ with $y neq 0$, $$fBigg (yf(x)+fracxyBigg)=xyf(x^2+y^2)$$

This seems quite hard. $f(x)=begincasesfrac1x, xneq 0 \ 0, x=0 endcases$ 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 $f(fracy^2+1y)=yf(y^2+1)$. which seems to imply $f(f(x)+x)=f(fracx^2+1x)$. 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 $y f(x)+ fracxy=x^2 + y^2$ 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 $lambda = frac1x$ and form the initial functional equation you would obtain $f(x)lambda + fracxlambda=x^2+lambda^2, implies f(x)=frac1x$

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
$$fleft(f(x)y+ fracxyright)=0, forall ,x^2+y^2=S_n$$

Now, for $y to 0^+$ and $x>0$ one finds that $f(x)y+fracxy to infty$ as $x to sqrtS_n$, thus $f$ takes zero values in the neighbourhood of $infty$.

Since $f(y+frac1y) = y f(1+y^2), implies f(x)=-f(-x), forall, |x| geq 2$ we find the same result in the neighbourhood of $- infty$. Notice now that for $y>1$ that $y+frac1y > 1+y^2$ 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 $f:mathbbRtomathbbR$ satisfies $f(1)=1$ and $fleft(yf(x)+fracxyright)=xyfleft(x^2+y^2right)$ for all $x,yinmathbbR,yneq0$, then $f(0)=0$ and $f(x)=frac1x$ 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 $x,yinmathbbR, yneq0$ write $P(x,y)$ for the assertion $fleft(yf(x)+fracxyright)=xyfleft(x^2+y^2right)$. Furthermore, we define $mathcalN:=x>0 $, and assume that $mathcalNneqemptyset$.

As @Kevin already pointed out, we have $alphainmathcalNimpliesalpha^2+1inmathcalN$, and in particular $mathcalN$ is unbounded. Furthermore, $P(1,alpha)$ gives also $alphainmathcalNimpliesalpha+frac1alphainmathcalN$. Now, if $alpha,betainmathcalN$ then
$$
P(alpha,beta):quad fleft(fracalphabetaright)=alphabeta fleft(alpha^2+beta^2right)\
P(beta,alpha):quad fleft(fracbetaalpharight)=betaalpha fleft(beta^2+alpha^2right)
$$
and thus $fleft(fracalphabetaright)=fleft(fracbetaalpharight)$. Therefore, if $alphainmathcalN$ then
$$
fleft(frac1alpharight)=fleft(fracalpha+frac1alphaalpha^2+1right)=fleft(fracalpha^2+1alpha+frac1alpharight)=f(alpha)=0
$$
and thus also $frac1alphainmathcalN$. This gives, together with the unboundedness of $mathcalN$, the existence of $(alpha_n)_ninmathbbNinmathcalN^mathbbN$ with $lim_ntoinftyalpha_n=0$.

Now notice that for $xneq 0$ and $alphainmathcalN$
$$
P(alpha,alpha^2 x):quad fleft(frac1alpha xright)=alpha^3 x fleft(alpha^4left(x^2+frac1alpha^2right)right)\
P(frac1alpha, x):quad fleft(frac1alpha xright)=fracxalpha fleft(x^2+frac1alpha^2right)
$$
and thus $alpha^4 fleft(alpha^4left(x^2+frac1alpha^2right)right)=fleft(x^2+frac1alpha^2right)$ for all $xneq 0$, or in more simple terms
$$
(*)qquadalpha^4 f(alpha^4 z) = f(z) quadforall alphainmathcalN, z>frac1alpha^2
$$
Now for a fixed $alphainmathcalN$ and $y>0$ let $ninmathbbN$ be such that $y>maxfracalpha_n^2alpha^4,fracalpha_n^4alpha^2,alpha_n^2$, which exists as $(alpha_n)to 0$. Then
$$
alpha^4 f(alpha^4 y)=fracalpha^4alpha_n^4 alpha_n^4 fleft(alpha_n^4fracalpha^4 yalpha_n^4right)oversetfracalpha^4 yalpha_n^4>frac1alpha_n^2=frac1alpha_n^4alpha^4 fleft(alpha^4fracyalpha_n^4right)oversetfracyalpha_n^4>frac1alpha^2=frac1alpha_n^4fleft(fracyalpha_n^4right)oversety>alpha_n^2=f(y).
$$
Thus we can strengthen $(*)$ and actually have
$$
(**)qquadalpha^4 f(alpha^4 z) = f(z) quadforall alphainmathcalN, z>0.
$$
In particular, for $z=frac1alpha^2$ we get $alpha^4f(alpha^2)=fleft(frac1alpha^2right)$. On the other hand, we see that as $1+alpha^2,1+frac1alpha^2inmathcalN$, we have by again combining $P(1+alpha^2,1+frac1alpha^2)$ and $P(1+frac1alpha^2,1+alpha^2)$ that
$$
f(alpha^2)=fleft(frac1+alpha^21+frac1alpha^2right)=fleft(frac1+frac1alpha^21+alpha^2right)=fleft(frac1alpha^2right)
$$
and thus, as $alpha>0$ and $alphaneq 1$, we have $alpha^4f(alpha^2)=fleft(frac1alpha^2right)=f(alpha^2)implies f(alpha^2)=0$ so $alpha^2inmathcalN$. But then also $alpha^4inmathcalN$ which gives by $(**)$
$$
0=alpha^4 f(alpha^4)overset(**)= f(1)=1,
$$
contradiction!

Therefore we conclude that $mathcalN$ is empty, and as @Kevin already saw this gives $f(x)=frac1x$ 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.

$colorbrowntextbfSome forms of the equation.$

If $underlinexnot=0,$ 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
$$begincases
y=1,quad f(f(x)+x) = xf(x^2+1) = fleft(x+dfrac1xright) hspace226mu(2.1)\[4pt]
y=x,quad f(xf(x)+1) = x^2f(2x^2) hspace350mu(2.2)\[4pt]
endcases$$
Denote
$$g(x) = xf(x)tag3,$$
then from $(2)$ should
$$begincases
g(1+x^2) = gleft(x+dfrac1xright) hspace432mu(4.1)\[4pt]
g(g(x)+1) = frac12g(2x^2)(g(x)+1)hspace370mu(4.2)\[4pt]
endcases$$
Assume $g(x)$ continuous function.

$colorbrowntextbfCorollaries from the formula (4.1).$

Using the relationships between the arguments in $(4.1)$ in the form of
begincases
L_1,2=1+x^2=dfrac 12R(RpmsqrtR^2-4)\[4pt]
R_1,2=x+dfrac1x = pmleft(sqrt L-1+dfrac1sqrtL-1right),
endcases
one can present equation $(4.1)$ in the forms of
begincases
g(x) = gleft(dfrac 12x(xpmsqrtx^2-4)right)hspace382mu(5.1)\[4pt]
g(x) = gleft(pmleft(sqrtx-1 + dfrac1sqrtx-1right)right).hspace330mu(5.2)
endcases
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
$$g(x)=1quad forall quad xin((-infty,-2]cup[2,infty)).tag6$$

$colorbrowntextbfCorollaries from the formula (4.2).$

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) =
begincases
0,quadtextifquad x=0\[4pt]
dfrac1x,quadtextotherwize
endcases$$
is the single non-trivial solution.

Let $f:textbfRrightarrowtextbfR$, such that $f(1)=1$ and
$$
fleft(yf(x)+fracxyright)=xyf(x^2+y^2)textrm, forall (x,y)intextbfRtimestextbfR^*tag 1
$$
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
$$
fleft(yf(x)+fracxyright)=fleft(xf(y)+fracyxright)textrm, x,yneq 0
$$
For $y=1$, we get
$$
fleft(x+frac1xright)=f(f(x)+x)textrm, xneq 0tag 3
$$
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
$$
f(x)=hleft(x+frac1x,xright)tag4.1
$$

From (4) with $xrightarrow x+frac1x$ we get
$$
fleft(x+frac1xright)=fleft(y+hleft(x+frac1x,yright)right).
$$
Hence for $y=x$ in the above identity we get
$$
fleft(x+frac1xright)=fleft(x+hleft(x+frac1x,xright)right)=f(x+f(x))
$$
Hence we can get (3). From (2) and (3) we get also
$$
xf(x^2+1)=fleft(x+frac1xright)tag 6
$$
Setting $xrightarrow x^-1$ in (6)
$$
frac1xfleft(1+frac1x^2right)=fleft(x+frac1xright)=xf(x^2+1).tag 7
$$
Hence if we set $x^2rightarrow x>0$ in (7), then
$$
fleft(1+frac1xright)=xf(x+1)textrm, for all x>0.tag 8
$$
Set also $x=y-1>0$ in (8), then
$$
fleft(fracyy-1right)=(y-1)f(y)textrm, y>1.
$$
With $y=1/w>1$ we get
$$
frac11-wfleft(frac11-wright)=frac1wfleft(frac1wright).
$$
Hence if $0<w<1$ and $g(w):=frac1wfleft(frac1wright)$, then
$$
f(x)=x^-1gleft(x^-1right)textrm, where x>1textrm and g(1-w)=g(w)textrm, 0<w<1tag 9
$$
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
$$
f(x)=hleft(x+frac1x,xright)tag11
$$
with
$$
frac1x-1hleft(x-1+frac1x-1,frac1x-1right)=frac1xhleft(x+frac1x,frac1xright)tag12
$$
and $h(x,y)$ solution of
$$
f(y+h(x,y))=f(x).tag13
$$
An obvious solution of $(13)$ is $h(x,y)=x-y$, but it may exist and other solutions.