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I'm looking for a method to solve:

$$f(f(x))=x$$

Where $f$ is defined for $x \in R$

So far by inverting both sides I have:

$f(x)=f^{-1}(x)$

Which means that my function should be symmetrical over $y=x$. I may "guess" the functions:

$y=x$

$y=c-x$

However I'm wondering is there a way to solve this without "guessing".

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  • $\begingroup$ $f(x)=c-x$ is also a solution. $\endgroup$ – Bernard Sep 27 '15 at 15:04
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    $\begingroup$ @AhmedS.Attaalla It's not - point $(0,1)$ belongs to this line, but $(1,0)$ doesn't. $\endgroup$ – Wojowu Sep 27 '15 at 15:07
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    $\begingroup$ Related: math.stackexchange.com/questions/46635/… $\endgroup$ – A.Sh Sep 27 '15 at 15:08
  • $\begingroup$ Wow how did I miss that. I was thinking parrellel $\endgroup$ – Ahmed S. Attaalla Sep 27 '15 at 15:08
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Functions satisfying this property are known as involutions. See the article for many examples in various fields. You are correct about the symmetry condition.

Edit: I should add that there are unaccountably infinitely many of such functions (even restricting to continuous functions).

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