FE Marathon!

For discussing Olympiad Level Algebra (and Inequality) problems
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Mehrab4226
Posts: 212
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Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Fri Dec 18, 2020 9:37 pm

I am giving up :( :( :( :( :(
I tried a lot for the last 2 days but couldn't solve it :( :( :(

My progress:
Let us denote the functional equation with (1)
Let $x=0 $in (1)
Then,
$f(0)=f(yf(0)) \cdots (2) $

let $f(0) \neq 0$ then $f(0)=c$ for some real constant c

Using (2) we can change y as whatever we want and get all real numbers in the input of ther R.H.S but the L.H.S remains the same. Then $f(x)$ is a constant function that obviously not true.CONTRADICTION

$\therefore f(0)=0$

Now putting $y=0$ in (1) we get,

$f(xf(x))=f(0)+x^2$

Or,$f(xf(x))=x^2 \cdots (3)$

Or,$f(f(1))=1[x=1]$

Putting x=f(1) in (3) we get,

$f(f(1)f(f(1)))=f(1)^2$

Or,$f(f(1))=f(1)^2[f(f(1))=1]$

Or$f(1)^2=1$

Or,$f(1)= \pm 1$

Finally putting $x=1$ in (1) we get,
$f(f(y+1))=f(\pm y)+1$
-----------------------------------------------
From here I got no progress :( :(
But if we can get $f(f(x))=x$[I couldn't get it so it might be a wrong way to go] we can use it in(3) when x=f(t),
and get $f(t)^2=t^2$ and then get $f(t)=\pm t $ which is indeed the answer.
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: 65
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Sat Dec 19, 2020 11:04 am

Mehrab4226 wrote:
Fri Dec 18, 2020 9:37 pm
I am giving up :( :( :( :( :(
I tried a lot for the last 2 days but couldn't solve it :( :( :(

My progress:
Let us denote the functional equation with (1)
Let $x=0 $in (1)
Then,
$f(0)=f(yf(0)) \cdots (2) $

let $f(0) \neq 0$ then $f(0)=c$ for some real constant c

Using (2) we can change y as whatever we want and get all real numbers in the input of ther R.H.S but the L.H.S remains the same. Then $f(x)$ is a constant function that obviously not true.CONTRADICTION

$\therefore f(0)=0$

Now putting $y=0$ in (1) we get,

$f(xf(x))=f(0)+x^2$

Or,$f(xf(x))=x^2 \cdots (3)$

Or,$f(f(1))=1[x=1]$

Putting x=f(1) in (3) we get,

$f(f(1)f(f(1)))=f(1)^2$

Or,$f(f(1))=f(1)^2[f(f(1))=1]$

Or$f(1)^2=1$

Or,$f(1)= \pm 1$

Finally putting $x=1$ in (1) we get,
$f(f(y+1))=f(\pm y)+1$
-----------------------------------------------
From here I got no progress :( :(
But if we can get $f(f(x))=x$[I couldn't get it so it might be a wrong way to go] we can use it in(3) when x=f(t),
and get $f(t)^2=t^2$ and then get $f(t)=\pm t $ which is indeed the answer.
But i won't give you the solution, you can find the solution of this problem. Its IMO SL A7 2009.
You can give a new problem to continue this marathon.

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Mehrab4226
Posts: 212
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Sun Dec 20, 2020 3:28 pm

Find all functions $f$ $:$ $\Bbb Q \to \Bbb Q$ that satisfies,

\[f(x+y)=f(x)+f(y)+xy\]
Source FE notes
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: 65
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Mon Dec 21, 2020 11:50 am

Solution 4:Let $g(x)=f(x)+xy$
So,$f(x)=g(x)-xy$
Plugging this back to the given equation,
$g(x+y)-xy$=$g(x)-xy+g(y)-xy+xy$
Or,$g(x+y)=g(x)+g(y)$
Now this is well known cauchy function.
When f is continuous or bounded.

$f(x)$=$cx-xy$ is the solution where x,c,y belongs to R.

Dustan
Posts: 65
Joined: Mon Aug 17, 2020 10:02 pm

Re: FE Marathon!

Unread post by Dustan » Mon Dec 21, 2020 1:07 pm

Problem 5: $f:R\rightarrow R$
Find all the function such that for all real valued x,y
$f(x^4+ y) $= $x^3\cdot f(x) + f(f(y))$

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Mehrab4226
Posts: 212
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Tue Dec 22, 2020 11:33 pm

I am not sure this solution is correct. Bhaya plz check it if right then I will post a new problem if not then I will probably try more to solve it.
Ifx=0 in the given equation,
$f(y)=f(f(y))$
Let, $f(t)=q$ for some real t,q
Putting y=t in (1) we get,
$f(t)=f(f(t))$
Or, $q=f(q)$
Thus,
$f(x)=x$ is the required function which indeed works fine!
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: 305
Joined: Fri Oct 18, 2013 11:56 pm
Location: Rangpur,Bangladesh

Re: FE Marathon!

Unread post by tanmoy » Tue Dec 22, 2020 11:46 pm

Mehrab4226 wrote:
Tue Dec 22, 2020 11:33 pm
I am not sure this solution is correct. Bhaya plz check it if right then I will post a new problem if not then I will probably try more to solve it.
Ifx=0 in the given equation,
$f(y)=f(f(y))$
Let, $f(t)=q$ for some real t,q
Putting y=t in (1) we get,
$f(t)=f(f(t))$
Or, $q=f(q)$
Thus,
$f(x)=x$ is the required function which indeed works fine!
The solution is not correct :( . You haven't proved that the function is injective, so, you can't say that the values of $q$ are different for different values of $t$. For example, suppose, $f(10) = f(15) = 15$, then $15 = f(10) = f(15) = f(f(10))$. It still holds the problem's conditions but $f(10) \neq 10$.
"Questions we can't answer are far better than answers we can't question"

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Mehrab4226
Posts: 212
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Tue Dec 22, 2020 11:52 pm

tanmoy wrote:
Tue Dec 22, 2020 11:46 pm
Mehrab4226 wrote:
Tue Dec 22, 2020 11:33 pm
I am not sure this solution is correct. Bhaya plz check it if right then I will post a new problem if not then I will probably try more to solve it.
Ifx=0 in the given equation,
$f(y)=f(f(y))$
Let, $f(t)=q$ for some real t,q
Putting y=t in (1) we get,
$f(t)=f(f(t))$
Or, $q=f(q)$
Thus,
$f(x)=x$ is the required function which indeed works fine!
The solution is not correct :( . You haven't proved that the function is injective, so, you can't say that the values of $q$ are different for different values of $t$. For example, suppose, $f(10) = f(15) = 15$, then $15 = f(10) = f(15) = f(f(10))$. It still holds the problem's conditions but $f(10) \neq 10$.
Ok. It looked too good to be true anyway.
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

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Mehrab4226
Posts: 212
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Wed Dec 23, 2020 7:15 pm

This one kinda looks like an :| okay solution.[This looks big because of the spacings]
$f(x^4+y)=x^3f(x)+f(f(y))$

Putting x=0 we get,

$f(f(y))=f(y)\cdots (1)$

Putting y= 0 we get,

$f(x^4)=x^3f(x) $

Or, $f(x^4)=f(t)=x^3f(x)\cdots (2)$[Let $x^4=t$]

Now from the question we get,

$f(x^4+y)=x^3f(x)+f(f(y))$

Or, $f(t+y)=f(t)+f(y)$[Using (1) and (2)]

Which is the cauchy function,

$\therefore f(x)=cx$

Using it in the question's equation and solving it we get c=1

$\therefore f(x)=x $ which is indeed the answer.
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

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Mehrab4226
Posts: 212
Joined: Sat Jan 11, 2020 1:38 pm
Location: Dhaka, Bangladesh

Re: FE Marathon!

Unread post by Mehrab4226 » Wed Dec 23, 2020 8:24 pm

If my solution is correct then the next problem in the marathon, if my solution is not correct :cry: then ignore this message and plz tell me where I am wrong.


Problem 6:
Find all functions $f : \Bbb R \to \Bbb R$ that satisfies,

\[f((x-y)^2)=f(x)^2-2xf(y)+y^2\]
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

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