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I was trying to solve the following functional equation:

$f(f(x-y))=f(x)-f(y)$.

And I concluded that $f$ must be additive and bijective. The question is:

Let $f:\mathbb{R} \to \mathbb{R}$ be an additive bijective function. Is it true that $f$ has to be of the form $f(x)=cx$, $c\in\mathbb{R}-\{0\}$?

What if we impose the condition that $f(f(x))=f(x)$. Is the identity map the only solution?

Edit: if $f$ is surjective and $f(f(x))=f(x)$, then $f$ is the identity map (almost by definition)

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  • $\begingroup$ Constant functions $F$ also satisfy $F\circ F=F$. $\endgroup$ – lulu Apr 24 at 20:00
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    $\begingroup$ For what concerns the additive and bijective function, you'd like to read this: en.m.wikipedia.org/wiki/Cauchy%27s_functional_equation. For it to be bijective, just do this: take a basis of $\Bbb R$ over $\Bbb Q$ and make the function swap two elements of the basis, while being the identity on the others. $\endgroup$ – Marco Vergamini Apr 24 at 20:05

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