# Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $5f(x+y)+y^5=f(x)+(x+y)^5$?

I am a total beginner at functional equations; I am an undergrad studying them for my Putnam seminar, so I don't really know/understand what I am doing. I just got some results from symbolic manipulations and I am trying to interpret them correctly. If possible, a more heuristic answer would be appreciated. I got the question from Brilliant:

$f:\mathbb{R} \to \mathbb{R}$

$5f(x+y)+y^5=f(x)+(x+y)^5$

I plugged in $x=y=0$, and concluded that $f(0)=0$

I set $y=0$, and I got $f(x) = \dfrac {x^5}{4} \tag 1$

I set $x=0$, and I got $f(y) = 0 \tag 2$

I plugged $(1)$ into the original functional equation, and of course it only worked when $y$ was $0$.

I think from $(2)$ we can conclude that $0$ is the only solution. I guess it's because I think we can maybe come up with more functions such as $(1)$ which satisfy the original equation under some nieche conditions, but once we get $f$ is $0$ for all values in its domain, that feels like the end of the line. However I can't really come up with a cohesive argument.

• It's a good idea to explicitly write the conditions under which your equations hold. (1) holds for all real numbers $x$. (2) holds for all real numbers $y$. You could rewrite (2) as "$f(x)=0$ for all real numbers $x$". Can (1) and (2) both hold? – stewbasic Sep 23 '16 at 0:08

You're right that $f(x) = \frac{x^5}{4}$ for every $x$; and also that $f(y) = 0$ for every $y$ (since $f(0) = 0$). This is a contradiction: for example, $f(1) = \frac{1}{4}$ and also $0$.
To be concrete: you have already noted that $x=y=0$ yields $f(0) = 0$.
Now let $x=1, y=0$ to obtain $5 f(1) = f(1) + 1$ or $f(1) = \frac{1}{4}$.
On the other hand, let $x=0, y=1$ to obtain $5 f(1) + 1 = f(0) + 1$, or $5 f(1) = f(0)$, or $f(1) = 0$.
I think you will agree that $0 \not = \frac{1}{4}$; so we have a contradiction.