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$f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$

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

$f(x+1)=f(x)+1$

Can this functional equation be solved ?

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    $\begingroup$ Yes, it can be solved by you. $\endgroup$ – Watson Jun 12 '18 at 14:18
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    $\begingroup$ Or at the very least, attempted by you. $\endgroup$ – Rhys Hughes Jun 12 '18 at 14:21
  • $\begingroup$ For $f$ continuous this has been solved on MSE here. $\endgroup$ – Dietrich Burde Jun 12 '18 at 14:34
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Perhaps obvious but what about $$f(x)=x?$$

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    $\begingroup$ What about it? Do you think this is an answer? $\endgroup$ – Aqua Jun 12 '18 at 14:25
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    $\begingroup$ Yes. It works but just seemed too obvious. $\endgroup$ – Rhys Hughes Jun 12 '18 at 14:27
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    $\begingroup$ The point is to find, with proof, all solutions. No? $\endgroup$ – the_fox Jun 12 '18 at 14:29
  • $\begingroup$ More often than not, solving functional equations involves finding all solutions, not just one. So yes, it's probably too obvious. $\endgroup$ – Arthur Jun 12 '18 at 14:29

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