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The following is the final problem from this page:

Find all the functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $$f(x+y)=f(x)+f(y) \; \; \; \forall \,x,y\in \mathbb{R}$$ and also (this is the important part) $$f(x^{2019})=f(x)^{2019}\tag{$*$}$$

My idea is to prove that $f(x)=x \; \; \forall x \in \mathbb{R}$, $f(x)=-x \; \; \forall x \in \mathbb{R}$ or $f\equiv 0$.

If we change $2019$ for an even number this is easy because it implies that the image of a positive number is positive and from there $f$ is linear and hence the identity or zero.

If we change $2019$ by $3$ then this is related (although I don't know how to deal with the case $f(1)=0$ or $f(1)=-1$)

But in this case I don't know how to prove any type of regularity from $(*)$ to conclude that $f$ must be linear.

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    $\begingroup$ You have tagged this as "contest-math". Can you please provide a link to the contest that this problem comes from? $\endgroup$ – Xander Henderson Aug 16 '19 at 19:14
  • $\begingroup$ @XanderHenderson Oh sorry, but I don't know if it comes from a contest. My source is that it is problem 10 of the proposed problems of this blog entry how-did-i-get-here.com/61 $\endgroup$ – A123 Aug 16 '19 at 19:51
  • $\begingroup$ I have edited your question to include the link (in the future, you should provide such references yourself---these provide valuable context). Given that this problem is not from a contest, is there any particular reason that you have chosen to tag it as a contest problem? $\endgroup$ – Xander Henderson Aug 16 '19 at 23:29
  • $\begingroup$ @XanderHenderson Similar problems (e.g. replacing 2019) have appeared as training exercises before. I can't link one off the top of my head, but I think the tag is perfectly fine here, even if it does not come from an explicit contest. $\endgroup$ – user574848 Aug 17 '19 at 0:35
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Let $f$ satisfy the premises. Then $f(ax)=af(x)$ for any $x\in\mathbb{R}$ and $a\in\mathbb{Q}$. Now $$f\big((a+x)^{2019}\big)=f(a+x)^{2019}$$ (with both sides expanded using the binomial formula and the above), being a polynomial identity in $a\in\mathbb{Q}$, implies $$f(x^k)=f(1)^{2019-k}f(x)^k\qquad(0\leqslant k\leqslant 2019).$$ Taking $k=2$, we get $f(x^2)=f(1)f(x)^2$. This reduces to the case you have worked out (after replacing $f$ by $-f$ if needed).

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    $\begingroup$ That's really cool, thanks!! $\endgroup$ – A123 Aug 16 '19 at 21:33
  • $\begingroup$ @metamorphy I still seem to be missing something. Why cannot $f$ satisfy something like e.g., $f(e)=e^2$? [where $e$ is the base of the natural logarithm]. $\endgroup$ – Mike Aug 17 '19 at 1:53
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    $\begingroup$ @Mike If $f(x+y)=f(x)+f(y)$ and $x\geqslant 0\implies f(x)\geqslant 0$, then $f$ is (necessarily) linear. The Wikipedia article suggests this as an easy consequence of the fact that nonlinear solutions must have graphs (everywhere) dense in $\mathbb{R}^2$. $\endgroup$ – metamorphy Aug 17 '19 at 9:07
  • $\begingroup$ I think I see it now: the condition $f(x^2)=f(1)(f(x)^2$ implies that for all positive $y$, the sign of $f(y)$ has to be the same as the sign of $f(1)$. Which implies that there is at least one quadrant of $\mathbb{R}^2$ not covered by the graph of $f$, which implies (via the reasoning in the link you posted) that $f$ has to be linear. Thank you for explaining @metamorphy ! $\endgroup$ – Mike Aug 17 '19 at 16:01
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The first equation is Cauchy's functional equation, and hence the existence of nonlinear solutions depends on the axiom of choice.

If we assume that $f$ is linear, then it must be of the form $f(x)=ax$ for some $a \in \mathbb{R}$. The second equation then says that $\forall x \in \mathbb{R} (ax)^{2019} = ax^{2019}$. Setting $x=1$, it follows that $a$ must be its own 2019th power. The only real numbers that are their own 2019th (or nth for any odd $n>1$) powers are $0$, $1$, and $-1$. Hence, the three linear solutions are $f(x)=0$, $f(x)=x$, and $f(x)=-x$.

If instead of considering solutions over $\mathbb{R}$, we had considered solutions over $\mathbb{C}$, then there would be 2016 more linear solutions, with one corresponding to each non-real 2018th root of unity.

For nonlinear solutions (assuming AC), we don't know.

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We can prove that $f$ is continuous at $\mathbb{R}$. Furthermore $f$ has the first derivative on $(0, 0)$. Using these properties leads to show that $f(x) = x$ for every $x \in \mathbb{R}$ or $f \equiv 0$.

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    $\begingroup$ Can you please show the continuity? $\endgroup$ – Virtuoz Aug 16 '19 at 18:49
  • $\begingroup$ I agree that there is more work to be done. If $f$ were a function on $\mathbb{Q}$ instead of $\mathbb{R}$ then this would be much easier. But e.g., just because one has $f(e)$ does not mean that one has $f(e^2)$. $\endgroup$ – Mike Aug 16 '19 at 19:48
  • $\begingroup$ For example, let $X$ be the subset of $\mathbb{R}$ of the form $\{a +be+ce^2\}$; $a,b,c \in \mathbb{Q}$. Consider $f$ defined as follows: $f(1)=1$; $f(e)=0$; $f(e^2)=100000$. Then this could be extended to an additive real-valued function on $X$. Now there may very well be "something else" happening for the entire real line $\mathbb{R}$ but if so what is it. $\endgroup$ – Mike Aug 16 '19 at 19:51

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