# For IMO

$\large{ f(xf(x + y)) = f(yf(x)) + x^2}$

Find all functions $$f$$ from the set of real numbers into the set of real numbers which satisfy for all real numbers $$x,y$$ and the equation above.

Note by Department 8
2 years, 8 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:

Put $$y=0$$ in the given functional equation.

we get $\boxed{f(xf(x))=f(0)+x^{2}}...........(1)$ Now put $$x=-x$$ in the given functional equation .

We get $\boxed{f(-xf(-x)=f(0)+x^{2}}...........(2)$

Equating $$(1),(2)$$ we get

$f(-xf(-x))=f(xf(x))$ $-xf(-x)=xf(x)$ $\boxed{f(x)+f(-x)=0}..........(3)$

Now by putting $$x=0$$ we get $$\boxed{f(0)=0}........(4)$$. Now equations $$(1),(2)$$ transform into

$\boxed{f(xf(x))=x^{2}}...........(I)$

$\boxed{f(xf(x))=x^{2}}...........(II)$

Our main functional equation is $$f(xf(x+y))=f(yf(x))+x^{2}$$

This implies that degree of $$f(yf(x)),f(xf(x+y))$$ is $$2$$.

This implies that degree of $$f(x),f(x+y)=1$$.

Therefore we can write $$f(x)=ax+b$$ where $$a,b$$ are real numbers.

But $$f(0)=0$$ therefore $$b=0$$ . therefore $$f(x)=ax$$ . But for $$f(x)=ax$$ to satisfy $$(I)$$ $$a=1$$ is a must condition .

Therefore our final function reduces to $$f(x)=x$$.

Therefore the required function is $$\boxed{f(x)=x}$$.

- 2 years, 8 months ago

How can u say that in the line above (3) ? I dont think that it is obvious that the given functional equation is injective. If yes then please prove it Also I dont see how do u get f(0)=0. Also arguments like degree of so and so .... are not allowed. What is the function is not a polynomial ? If it is something like exponential then argument fails. Please check the solution again.

- 2 years, 7 months ago

nice but typo.. if you put y=0 in the first line you would get $f(xf(x))=f(0)+x^2$

- 2 years, 8 months ago

Nice observations but it did not affect the solution greatly.

- 2 years, 8 months ago

Yeah just saw that now

- 2 years, 8 months ago