Krdytkyu

To find all functions
$f$ which is a real function from $Bbb R to Bbb R$ satisfying the relation
$$f(x^2 + yf(x)) = xf(x+y)$$
It can be easily seen that the identity function $i.e.$ $f(x)=x$ and $f(x)=0$ (verified just now) satisfies the above relation!! And putting $y=0$ I have got $f(x^2)=xf(x)$.

Help needed to find other functions satisfying the relation.

Substitute $x=y=0$, we have $f(0)=0$.

Suppose $f(a)=0$ for some $ane 0$, then substitute $x=a$, gives
$$f(a^2)=af(a+y) hspace{1cm}forall y$$
Hence we have the trivial solution $$f(y)=text{constant}=f(a)=f(0)=0$$ for all $y$.

Therefore, if other solutions exist, they must satisfy $f(x)ne0$ for all $xne0$.

Now for $xne 0$ let $$y=-x^2/f(x)$$ then $$f(0)=xf(x-x^2/(f(x)))=0$$

Hence we have $$x-x^2/f(x)=0$$ which implies $$f(x)=x$$ for all $xne 0$.

Combined with $f(0)=0$, the non-trivial solution is $$f(x)=x$$ for all $x$.

Putting $x=y=0$: $$f(0)=0$$

Putting $y=-x$: $$ f(x^2-xf(x))=0$$

If there exists $aneq 0$ such that $f(a)=0$, then put $x=a$: $$f(a^2)=af(a+y)$$

Then $f(x)$ is a constant, and easily find that $f(x)=0$.

If $f(x)=0$ iff $x=0$, then $x^2-xf(x)=0$ for all $x$, then $f(x)=x$ for all $x$.

Hint: Note that the relation is true for all $x,y$. Therefore

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

this can be solved as:

$$f(x) = C x$$.

Note also that

$$f(x^2 -xf(x)) = xf(0) $$

An incomplete solution. We present two results on this problem.

We show that if $f(x)$ is injective then the only non-trivial solution to the functional equation is $f(x)=x$.

Proof: Apart from the trivial solution $f(x)=0$ let us assume $f(x)$ is not identically zero and it is injective. Set $x=y=0$ to get
$$f(0^2+0f(0))=0f(0)Rightarrow f(0)=0$$
Then let $x=-y$ one would obtain
$$f(x^2-xf(x))=xf(x-x)=xf(0)Rightarrow f(x^2-xf(x))=0$$
Using the assumption that $f(x)$ is injective then $$x^2-xf(x)=0Rightarrow f(x)=x$$

$f(x)$ is an odd function.

Set $y=0$ then
$$f(x^2)=xf(x)$$
on the other hand
$$f((-x)^2)=-xf(-x)Rightarrow f(x^2)=-xf(-x)Rightarrow f(x)=-f(-x)$$
and hence $f(x)$ would be an odd function.

Putting $ x=0 $ gives $$f(yf(0))=0 $$ for any $yinmathbb{R}$ so $f(0)=0$. If $f$ is identically zero, then it satisfies the equation. So for non-constant solution we may assume $f$ is not zero at some point. Setting $y=0$, $$ f(x^2)=xf(x). $$ Let $x_0$ be a real number for which $f(x_0)=0$. Then $$f(x_0 ^2)=x_0 f(x_0 +y)$$ and so $$ x_0 f(x_0 +y) =0 $$ for all $yinmathbb{R}$. Thus $x_0 =0$ since f is not identically zero. Indeed,
$$f(x)=0 iff x=0.$$
Now setting $ y=-x $ we have for any $xinmathbb{R}$,
$$f(x^2-xf(x))=0$$ and so
$$x^2-xf(x)=0 $$ or
$$f(x)=x.$$