×

# Cool functions

Find all functions $$f:\mathbb{R} \rightarrow \mathbb{R}$$ such that

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

for all real numbers $$x$$ and $$y$$.

Note by Sharky Kesa
2 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Let $$P(x,y)$$ be the statement $$f(f(x)+y) = x+f(f(y))$$.

$$P(x,0)$$ implies $$f(f(x)) = x + f(f(0))$$. Applying this to $$P$$ gives $$f(f(x)+y) = x+y+f(f(0))$$.

$$P(x,f(0))$$ implies $$f(f(x)+f(0)) = x+f(0)+f(f(0))$$. $$P(0,f(x))$$ implies $$f(f(0)+f(x)) = f(x)+f(f(0))$$. Equating the two gives $$x+f(0)+f(f(0)) = f(x)+f(f(0))$$, or $$f(x) = x + c$$ for some fixed real number $$c$$ for all real $$x$$. It can be easily verified to satisfy the equation.

- 2 years, 2 months ago