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Find all function $f:\mathbb Q \rightarrow \mathbb Q $ $$f(xf(y)+f(x))=yf(x)+x , \forall x,y \in \mathbb Q$$

So , for $x = y = 0$ we get that $f(f(0))=0$ . For $x = 0$ $\implies f(f(0))=yf(0) , \forall y \in \mathbb Q \iff yf(0)=0 ,\forall y \in \mathbb Q \implies f(0)=0$. Now if we make $y=0 \implies f(f(x))=x ,\forall x \in \mathbb Q \implies$ f bijective(don't really used it but it will probably come in handy at some point).It seems that the function that checks is $f(x)=x$ and the fact that the function is defined on rational numbers, we will probably have to arrive at a Cauchy functional equation. If you have any idea or a solution, I'm here to listen. Thanks!

Unknowduck
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  • What are the chances this is one of those where you rely on $ \mathbb Q $ being dense in $ \mathbb R $ and continuity to arrive at the only solution being $ f(x) = 0 $ ? – 3dguy Mar 07 '24 at 16:36
  • ah no, $ f(x) = x $ is also a solution – 3dguy Mar 07 '24 at 16:43
  • I don't really master dense sets very well, but this problem is from a book where dense sets have not yet been presented, so probably the solution would be using injectivity, surjectivity, additive or other tricks. In the meantime, I managed to find f(1) and f(-1) and proved that f(x)=x for any integer x (basically showed that f(x+1)=f(x)+1 ) – Unknowduck Mar 07 '24 at 17:02
  • Or this https://math.stackexchange.com/questions/4872237/find-all-functions-satisfying-fxfy-fx-yfx-x – Aig Mar 07 '24 at 17:11
  • Yes thank you. Looking at the solution , I was relatively close. – Unknowduck Mar 07 '24 at 17:12

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