Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(x)f(x+y)=xf(x)+f\big(f(x)\big)f(y)$$ for all $x,y\in\mathbb R$.

Source: Problem by me.

  • $\begingroup$ This is more strictly a math olympiad question rather than a puzzle. There is no appropriate site for such creations as of yet, so it is better suited somewhere else. $\endgroup$ Jun 6 '20 at 23:14
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    $\begingroup$ @Nij, these are not math textbook style. $\endgroup$ Jun 7 '20 at 3:09
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    $\begingroup$ @Nij, this is a Math Olympiad style problem, way far than math textbook kind of problem. $\endgroup$ Jun 7 '20 at 3:12
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    $\begingroup$ @Nij This question is very clearly on topic. Textbook problems are mostly those where you can just follow an algorithm to solve it. This is a maths puzzle that requires you to think outside the box. I'd argue that most Math Olympiad questions are more puzzles than problems. puzzling.meta.stackexchange.com/questions/2783/… $\endgroup$
    – Hjulle
    Jun 7 '20 at 9:39
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    $\begingroup$ @Nij: Not trying to defend the quality of my solution, but I can say for sure that what I wrote in my answer only included half the things I tried. I agree that no particular step requires anything beyond basic algebra, but finding the correct sequence of steps was non-trivial, at least for me. Maybe I should have written it better, but I absolutely did have a bit of "aha" once I struck on the correct path. $\endgroup$ Jun 7 '20 at 22:00

The set of functions which satisfy the functional equation is:

the all zero function and the identity function. Let $g$ be a function satisfying the relationship, and first let $x=y=0$. Then we have $$g(0)^2 = 0 + g(g(0)) \times g(0)$$ This implies either $g(0) = 0$ or $g(g(0)) = g(0)$.

In either case:

$\lambda = g(0)$ is a fixed point of $g$, or in other words $g(\lambda) = \lambda$. Let $x=\lambda$ and $y=0$, in which case we have $$g(\lambda)g(\lambda) = \lambda g(\lambda) + g(g(\lambda))g(0)$$ This gives the equality $$\lambda^2 = 2\lambda^2$$ which forces $\lambda = 0$.

With this knowledge:

let $x$ be any real and let $y=0$. This forces $$g(x)^2 = xg(x)$$ for all real $x$, which implies either $g(x) = 0$ or $g(x) = x$ for all real $x$. This implies either $g$ is the zero function, which one can easily check satisfies the condition above, or there exists $\mu \neq 0$ such that $g(\mu) = \mu$.


Let $y$ be any real such that $g(y) = 0$. Then we have $$g(\mu)g(\mu+y) = \mu g(\mu) + g(g(\mu))g(y)$$ which implies $\mu g(\mu+y) = \mu^2$ and thus $g(\mu+y) = \mu$ since $\mu \neq 0$. Since $\mu \neq 0$ this forces $\mu + y = \mu$, which forces $y=0$. Thus $g(x) = x$ for all real $x$ and $g$ is the identity function. Again it is easy to check this satisfies the condition above.


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