2
$\begingroup$

Let $\mathbb{R}^+$ denote the set of the positive real numbers. Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ satisfying $$\dfrac{f(f(y)^2+xy)}{f(y)}=f(x)+y$$ for all $x,y \in\mathbb{R}^+$.

I am very thankful for any solution, please help!

I tried to set $x=y=1, x=y=2, x=1, y=2$, so on, but this problem is more difficult.

$\endgroup$
  • $\begingroup$ Where is this problem from? $\endgroup$ – the_fox May 11 '18 at 14:55
0
$\begingroup$

Put $y=0$. Then the equation shows that $f$ has to be constant. Say $f(x)=c$ for all $x$. Using this in the equation results in $1=c+y$ for all $y>0$ which is impossible. Thus the equation has no solution.

EDIT: As pointed out below the function is defined on $\mathbb{R}_+$. Thus the arguments are not valid.

$\endgroup$
  • 3
    $\begingroup$ $0 \notin \mathbb R^+$ $\endgroup$ – Fred May 11 '18 at 7:40
  • $\begingroup$ I will edit my answer. $\endgroup$ – Jens Schwaiger May 11 '18 at 13:57
  • $\begingroup$ $f(x)=x$ is an example, Jens. $\endgroup$ – Piquito May 11 '18 at 15:07
0
$\begingroup$

Proving with $f(x)=kx^n$ we have$$f(k^2y^{2n}+xy)=k(k^2y^{2n}+xy))^n\\f(y)(f(x)+y)=k(f(y)f(x)+yf(y))=k(k^2y^nx^n+ky^{n+1})$$ which is valid only for $k=n=1$

We have an example with $$f(x)=x$$ I hope to find other examples or prove that there are not

$\endgroup$
  • $\begingroup$ So in the class of monomials there are very few solutions. Nice result. $\endgroup$ – Jens Schwaiger May 11 '18 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.