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I am considering the problem of finding all functions $f:(0,\infty)\to(0,\infty)$ satisfying the functional equation:

$$f\big(xf(y)+f(x)\big) = 2f(x)+xy\text.$$

I have been able to prove the following three results/properties:

  1. $f$ is not surjective.
  2. $f$ does not have any fixed points.
  3. $f(x)=x+1$ is a solution.

My intuition tells me that $x\mapsto x+1$ is the only solution, but I have not been successful in proving or disproving this claim.

Any ideas on how I can make further progress is appreciated.

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  • $\begingroup$ You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $g\left(xg\left(y\right)+g\left(x\right)+2x+xy+1\right)+xg\left(y\right)=g\left(x\right)$ and the question is now: is $g(x)=0$ the only solution? I really don't know whether that will help, though. $\endgroup$ – drhab Sep 3 '18 at 8:04
  • $\begingroup$ Thanks, the idea is great. But this new equation for g(x) seems to be almost as complicated as the original. $\endgroup$ – Wuberdall Sep 3 '18 at 9:34
  • $\begingroup$ Injectivity can be easily established also. $\endgroup$ – Sil Apr 7 '19 at 10:19
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    $\begingroup$ In this form it has a solution for example this: artofproblemsolving.com/community/c6h611705p3637387. $\endgroup$ – Sil Jul 18 '20 at 15:32
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    $\begingroup$ @MohsenShahriari I prefer to add some additional info / thoughts, but I haven't checked this one too thoroughly. Anyway added as a wiki post with the reference at least, anyone can also improve it as necessary. $\endgroup$ – Sil Oct 12 '20 at 10:57
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As suggested in comments, here is a solution by user pco in https://artofproblemsolving.com/community/c6h611705p3637387:

Let $P(x,y)$ be the assertion $f\big(xf(y)+f(x)\big)=2f(x)+xy$. Let $a=f(1)$.

$P(1,x) \implies f\big(f(x)+a\big)=x+2a$ and so $(2a, +\infty) \subseteq f(\mathbb{R})$

$P\big(1,f(x)+a\big) \implies f(x+3a)=f(x)+3a$ and so $f(x+3ka)=f(x)+3ka$ $\forall x >0, \forall k \in \mathbb{Z}_{\ge 0}$.

Let $x,y>0$ and $n \in \mathbb{N}$ such that $y+3na>2ax+f(x)$. Then there is $t>0$ such that $y+3na=xf(t)+f(x)$. Comparing $P(x,t)$ with $P(x,t+3a)$, we get $f(y+3na+3ax)=f(y+3na)+3ax$ and so $f(y+3ax)=f(y)+3ax$. So $f(x+y)=x+f(y)$ $\forall x,y >0$ and so $f(x)=x+c$ $\forall x$ and for some $c \in \mathbb{R}$.

Plugging this back in original equation, we get $c=1$ and so the unique solution is $f(x)=x+1$ $\forall x$.

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