## FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm
FE Marathon!
আমি যতদূর সম্ভব সার্চ দিয়ে কোনো FE ম্যারাথন খুঁজে পাই নি। তাই এই ম্যারাথন চালু করতেসি।সবাই এগিয়ে আসলে হয়তো এটা চালু হবে। হ্যাপি প্রব্লেম সলভিং!

Problem 1:
Find all functions $\ f$ such that
$\ f: \mathbb{R} \rightarrow \mathbb{R}$ and $\ f(xf(x)+f(y))=f(x)^2+y$ for all real $x,\ y$.

Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm

### Re: FE Marathon!

Is $f(x)^2=f(x) \times f(x)$?
or $f(x)^2=f(f(x))$?
Last edited by Mehrab4226 on Fri Dec 11, 2020 12:02 am, edited 1 time in total.

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm

### Re: FE Marathon!

Dustan wrote:
Mon Oct 05, 2020 4:32 pm
আমি যতদূর সম্ভব সার্চ দিয়ে কোনো FE ম্যারাথন খুঁজে পাই নি। তাই এই ম্যারাথন চালু করতেসি।সবাই এগিয়ে আসলে হয়তো এটা চালু হবে। হ্যাপি প্রব্লেম সলভিং!

Problem 1: Find all functions $\ f$ such that
$\ f: \mathbb{R} \rightarrow \mathbb{R}$ and $\ f(xf(x)+f(y))=f(x)^2+y$ for all real $x,\ y$.
Good initiative. Use ${\large LaTeX}$ properly. I have edited this one.
"Questions we can't answer are far better than answers we can't question"

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm

### Re: FE Marathon!

Mehrab4226 wrote:
Thu Dec 10, 2020 9:49 pm
Is $f(x)^2=f(x) \times f(x)$
or $f(x)^2=f(f(x))$
$\ f(x)^2 = f(x) \times f(x)$
"Questions we can't answer are far better than answers we can't question"

Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm

### Re: FE Marathon!

If we can prove $f(f(x))$ is onto can we prove $f(x)$ is onto too? If we can than how?
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm

### Re: FE Marathon!

[N.B I am not sure the solution is correct. If somebody would kindly check it, I will be grateful.]
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

tanmoy
Posts:312
Joined:Fri Oct 18, 2013 11:56 pm

### Re: FE Marathon!

Mehrab4226 wrote:
Fri Dec 11, 2020 11:26 pm
[N.B I am not sure the solution is correct. If somebody would kindly check it, I will be grateful.]
The solution is okay, well done! Just a few remarks:
(1) Hide your solution. I have hidden it; check the latex code.
(2) Use proper spaces between your solutions so that they become reader friendly.
(3) You have latexed your solution, that's great. Though there were some typos. I have edited them; check.
(4) After solving a problem in a marathon, try to post a new problem.
Good luck!
"Questions we can't answer are far better than answers we can't question"

Mehrab4226
Posts:230
Joined:Sat Jan 11, 2020 1:38 pm

### Re: FE Marathon!

tanmoy wrote:
Sun Dec 13, 2020 11:39 pm
Mehrab4226 wrote:
Fri Dec 11, 2020 11:26 pm
[N.B I am not sure the solution is correct. If somebody would kindly check it, I will be grateful.]
The solution is okay, well done! Just a few remarks:
(1) Hide your solution. I have hidden it; check the latex code.
(2) Use proper spaces between your solutions so that they become reader friendly.
(3) You have latexed your solution, that's great. Though there were some typos. I have edited them; check.
(4) After solving a problem in a marathon, try to post a new problem.
Good luck!
Thank you Bhaya. I will keep those in mind. I am very new in LaTeX and in the forum so sorry for this time.
$\text{Problem 2:}$

Find all the function of $f : {\Bbb R} \to {\Bbb R}, satisfying$

$$f(x^2+y)=f(f(x)-y)+4f(x)y$$

for all real numbers x and y.Good Luck!
Source:
The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.
-Henri Poincaré

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

It took 2.30 hours

Dustan
Posts:71
Joined:Mon Aug 17, 2020 10:02 pm

### Re: FE Marathon!

$Problem 3$:

Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all $x$, $y$ the identity $f\left(xf(x+y)\right) = f\left(yf(x)\right) +x^2$