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.

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    $\begingroup$ f(x)=0 is also a solution $\endgroup$ – Loreno Heer Feb 1 '15 at 16:03
  • $\begingroup$ There is of course other solutions because not only linear function are solutions. $\endgroup$ – idm Feb 1 '15 at 16:09
  • $\begingroup$ Where did this problem come from? Do you have a background for solving such problems? $\endgroup$ – Mhenni Benghorbal Feb 1 '15 at 16:17
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    $\begingroup$ yes i have but having problem in this one' $\endgroup$ – user8795 Feb 1 '15 at 16:18
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    $\begingroup$ no a friend of mine gave me $\endgroup$ – user8795 Feb 1 '15 at 16:22

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

Suppose $f(a)=0$ for some $a\ne 0$, 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 $x\ne0$.

Now for $x\ne 0$ 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 $x\ne 0$.

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

  • $\begingroup$ Once you show that $f(x\neq 0$ for $x \neq 0$ you can conclude by just putting $y=-x$ which gives $f(x^{2}-xf(x))=xf(x-x)=0$ so $x^{2}-xf(x) \equiv 0$ and so $f(x) \equiv x$. $\endgroup$ – Kabo Murphy Jul 18 '18 at 6:24

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

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

If there exists $a\neq 0$ 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 $$x^2-xf(x)=0\Rightarrow f(x)=x$$

$f(x)$ 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 $y\in\mathbb{R}$ 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 $y\in\mathbb{R}$. 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 $x\in\mathbb{R}$, $$f(x^2-xf(x))=0$$ and so $$x^2-xf(x)=0 $$ or $$f(x)=x.$$


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