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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
About old Inequality
perfect_square   0
4 minutes ago
Source: Arqady
This is: $a,b,c>0$ which satisfy $abc=1$
Prove that: $ \frac{a+b+c}{3} \ge \sqrt[10]{\frac{a^3+b^3+c^3}{3}}$
By $  uvw $ method, I can assum $b=c=x,a=\frac{1}{x^2}$
But I can't prove:
$ \frac{2x+\frac{1}{x^2}}{3} \ge \sqrt[10]{ \frac{2x^3+ \frac{1}{x^6}}{3}} $
Is there an another way?
0 replies
1 viewing
perfect_square
4 minutes ago
0 replies
inquality
karasuno   1
N 27 minutes ago by sqing
The real numbers $x,y,z \ge \frac{1}{2}$ are given such that $x^{2}+y^{2}+z^{2}=1$. Prove the inequality $$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2 .$$
1 reply
1 viewing
karasuno
an hour ago
sqing
27 minutes ago
Number Theory
karasuno   0
an hour ago
Solve the equation $$n!+10^{2014}=m^{4}$$in natural numbers m and n.
0 replies
karasuno
an hour ago
0 replies
Minimum number of values in the union of sets
bnumbertheory   5
N an hour ago by Rohit-2006
Source: Simon Marais Mathematics Competition 2023 Paper A Problem 3
For each positive integer $n$, let $f(n)$ denote the smallest possible value of $$|A_1 \cup A_2 \cup \dots \cup A_n|$$where $A_1, A_2, A_3 \dots A_n$ are sets such that $A_i \not\subseteq A_j$ and $|A_i| \neq |A_j|$ whenever $i \neq j$. Determine $f(n)$ for each positive integer $n$.
5 replies
bnumbertheory
Oct 14, 2023
Rohit-2006
an hour ago
No more topics!
functional equation
Raja Oktovin   18
N Jun 19, 2023 by hectorleo123
Source: Indonesia IMO 2010 TST, Stage 1, Test 2, Problem 2
Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(x^3+y^3)=xf(x^2)+yf(y^2)\] for all real numbers $ x$ and $ y$.
Hery Susanto, Malang
18 replies
Raja Oktovin
Nov 12, 2009
hectorleo123
Jun 19, 2023
functional equation
G H J
Source: Indonesia IMO 2010 TST, Stage 1, Test 2, Problem 2
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Raja Oktovin
277 posts
#1 • 3 Y
Y by GeorgeRP, Adventure10, Mango247
Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(x^3+y^3)=xf(x^2)+yf(y^2)\] for all real numbers $ x$ and $ y$.
Hery Susanto, Malang
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matrix41
68 posts
#2 • 1 Y
Y by Adventure10
Set $ y=0$ then

$ f(x^3)=xf(x^2)$ , so $ f(y^3)=yf(y^2)$ , then

$ f(x^3+y^3)=xf(x^2)+yf(y^2)=f(x^3)+f(y^3)$ $ \rightarrow$ $ f(x)+f(y)=f(x+y)$

simple induction gives : $ f(x_1+x_2+...+x_n)=f(x_1)+f(x_2)+f(x_3)+...+f(x_n)$ so $ \forall \ n\in\mathbb{R}$

then $ f(nx)=nf(x)$ and thus by setting $ x=1$ we have $ f(n)=nf(1)$, since $ f(1)$ is a constant then by assuming $ f(1)=a$ , gives $ f(n)=an$ with $ a\in\mathbb{R}$

done
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math10
478 posts
#3 • 1 Y
Y by Adventure10
matrix41 wrote:
Set $ y = 0$ then

$ f(x^3) = xf(x^2)$ , so $ f(y^3) = yf(y^2)$ , then

$ f(x^3 + y^3) = xf(x^2) + yf(y^2) = f(x^3) + f(y^3)$ $ \rightarrow$ $ f(x) + f(y) = f(x + y)$

simple induction gives : $ f(x_1 + x_2 + ... + x_n) = f(x_1) + f(x_2) + f(x_3) + ... + f(x_n)$ so $ \forall \ n\in\mathbb{R}$ then $ f(nx) = nf(x)$
Why for all $ n \in R$ :?:
matrix41 wrote:
Set $ y = 0$ then

$ f(x^3) = xf(x^2)$ , so $ f(y^3) = yf(y^2)$ , then

$ f(x^3 + y^3) = xf(x^2) + yf(y^2) = f(x^3) + f(y^3)$ $ \rightarrow$ $ f(x) + f(y) = f(x + y)$

simple induction gives : $ f(x_1 + x_2 + ... + x_n) = f(x_1) + f(x_2) + f(x_3) + ... + f(x_n)$ so $ \forall \ n\in\mathbb{R}$
then $ f(nx) = nf(x)$ and thus by setting $ x = 1$ we have $ f(n) = nf(1)$, since $ f(1)$ is a constant then by assuming $ f(1) = a$ , gives $ f(n) = an$ with $ a\in\mathbb{R}$

done
you only solve equation on $ N$ not $ R$
We have:$ f(x + y) = f(x) + f(y)$ and $ f(x^3) = xf(x^2)$,$ (x) = - f( - x)$,$ f(kx) = kf(x)$ for all $ k \in N$
We have:
$ f((x + 1)^3 + (x - 1)^3) = (x + 1)f(x^2 + 2x + 1) + (x - 1)f(x^2 - 2x + 1) = 2xf(x^2) + 2xf(1) + 4f(x)$
and $ f((x + 1)^3 + (x - 1)^3) = f(2x^3 + 6x) = 2xf(x^2) + 6f(x)$
So $ f(x) = xf(1)$ for all $ x \in R$
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matrix41
68 posts
#4 • 2 Y
Y by Adventure10, Mango247
Sorry there are some mistakes in my solution

Let me write my full solution

First from my first post, we have

$ f(x_1)+f(x_2)+...+f(x_n)=f(x_1+x_2+x_3+...+x_n)$ with $ n\in\mathbb{N}$ so $ f(nx)=nf(x)$ $ \forall n\in\mathbb{N}$

and then by setting $ x=m+1$ and $ y=m-1$ it gives

$ f((m+1)^3+(m-1)^3)=f((m^3+3m^3+3m+1)+(m^3-3m^3+3m-1))=f(2m^3+6m)=f(2m^3)+f(6m)=2f(m^3)+6f(m)$

and

$ f((m+1)^3+(m-1)^3)=(m+1)f(m^2+2x+1)+(m-1)f(x^2-2x+1)=(m+1)\left(f(m^2)+f(2m)+f(1)\right)+(m-1)\left(f(m^2)+f(-2m)+f(1)\right)=2mf(m^2)+4f(m)+2mf(1)$

since $ 2mf(m^2)=2f(m^3)$ , so $ f((m+1)^3+(m-1)^3)=2mf(m^2)+4f(m)+2mf(1)=2f(m^3)+4f(m)+2mf(1)$

$ 2f(m^3)+6f(m)=f((m+1)^3+(m-1)^3)=2f(m^3)+4f(m)+2mf(1)$

$ 2f(m)=2mf(1)$ $ \rightarrow$ $ f(m)=mf(1)$ $ \forall m\in\mathbb{R}$

please check....

EDITED : Sorry I didn't know that my solution is similar to math10's solution :oops:
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littletush
761 posts
#5 • 1 Y
Y by Adventure10
not so hard.
let $y=0$,then
$f(x^3)=xf(x^2)$
so $f(x^3+y^3)=f(x^3)+f(y^3)$
hence f satisfies Cauchy's function,so for rational x,$f(x)=x$
hence $f(1)=1$
by letting $x=x+1$ we get
$f((x+1)^3)=(x+1)f((x+1)^2)$
hence $2f(x^2)=(2x-1)f(x)+x$
let $x=x+1$
we get $2(f(x^2)+2f(x)+1)=(2x+1)(f(x)+1)+x+1$
hence $f(x)=x$.
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pco
23437 posts
#6 • 2 Y
Y by Adventure10, Mango247
littletush wrote:
... hence f satisfies Cauchy's function,
Right
littletush wrote:
so for rational x,$f(x)=x$
Wrong : $f(x)=ax$
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littletush
761 posts
#7 • 2 Y
Y by Adventure10, Mango247
pco wrote:
littletush wrote:
... hence f satisfies Cauchy's function,
Right
littletush wrote:
so for rational x,$f(x)=x$
Wrong : $f(x)=ax$
oh gush!such a terrible mistake!
a can be any real number.
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nawaites
204 posts
#8 • 1 Y
Y by Adventure10
So which solution is correct?????
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Takeya.O
769 posts
#9 • 1 Y
Y by Adventure10
It is easy to show that $a\in \mathbb R$ s.t. $f(x)=ax(\forall x\in \mathbb Q)$.

If we show that $f$ is
-continuous at some point
-monotonous
-either upperbounded or lowerbounded on some open interval

$f(x)=ax(\forall x\in \mathbb R)$ :P

Anyone has idea? :?
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pco
23437 posts
#10 • 4 Y
Y by Takeya.O, bgn, Wizard_32, Adventure10
Raja Oktovin wrote:
Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(x^3+y^3)=xf(x^2)+yf(y^2)\]for all real numbers $ x$ and $ y$.
Hery Susanto, Malang

Let $P(x,y)$ be the assertion $f(x^3+y^3)=xf(x^2)+yf(y^2)$
Let $a=f(1)$

$P(0,0)$ $\implies$ $f(0)=0$
$P(x,0)$ $\implies$ $f(x^3)=xf(x^2)$
And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive.

So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$

Let $x\in\mathbb R$ and $k\in\mathbb Q$
$P(x+k,0)$ $\implies$ $f(x^3+3kx^2+3k^2x+k^3)=(x+k)f(x^2+2kx+k^2)$

Which may be written $k^2(f(x)-ax)+2k(f(x^2)-xf(x))=0$

Considering this as a polynomial in $k$ with infinitely many roots (any rational), we get thet it must be the zero polynomial and so, looking at coefficient of $k^2$ :

$\boxed{f(x)=ax\text{  }\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$
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anhtaitran
363 posts
#11 • 3 Y
Y by Takeya.O, Wizard_32, Adventure10
My solution:Plug x=y=0 so f(0)=0.
Plug x=0 so f(x^3)=xf(x^2).(1)
So f(x^3+y^3)=f(x^3)+f(y^3) for every x;y real.
So f is addictive.
Now we will caculate f( (x+1)^3+(x-1)^3) in 2 ways.
f( (x+1)^3+(x-1)^3) = (x+1)f((x+1)^2)+(x-1)f((x-1)^2).
=(x+1)[f(x^2)+2f(x)+f(1)]+(x-1)(f(x^2)-2f(x)+f(1)].
=2xf(x^2)+2xf(1)+4f(x).(2)
On the other hand,
f( (x+1)^3+(x-1)^3) =f(2x^3+6x)=2f(x^3)+6f(x).(3)
By (1);(2) and (3) we have :
f(x)=cx(c=f(1)).
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Takeya.O
769 posts
#12 • 2 Y
Y by Adventure10, Mango247
@pco,@anhtaitran
What a nice solution! :w00t: Are you a FE master? :lol:
This post has been edited 1 time. Last edited by Takeya.O, Sep 5, 2016, 10:03 AM
Reason: Latex miss
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Evenprime123
104 posts
#13 • 2 Y
Y by Adventure10, Mango247
pco wrote:
And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive.

So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$

How do I conclude this? Induction would work on \(\mathbb N \) but what can I do here?
This post has been edited 1 time. Last edited by Evenprime123, Feb 2, 2018, 4:32 PM
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pco
23437 posts
#14 • 2 Y
Y by Adventure10, Mango247
Evenprime123 wrote:
pco wrote:
And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive.

So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$

How do I conclude this? Induction would work on \(\mathbb N \) but what can I do here?
You should consider this as a well-known property of additive functions.

If you want to prove it :
$f(2x)=f(x)+f(x)=2f(x)$
$f(3x)=f(2x)+f(x)=3f(x)$
And, with induction : $f(nx)=nf(x)$ $\forall x$, $\forall n\in\mathbb N$

So $f(x)=f(q\frac xq)=qf(\frac xq)$ and so $f(\frac xq)=\frac 1qf(x)$ $\forall x$, $\forall q\in\mathbb N$

So $f(\frac pqx)=pf(\frac xq)=\frac pqf(x)$ and so $f(px)=pf(x)$ $\forall x$, $\forall p\in\mathbb Q^+$

It remains to remember that $f(-x)=-f(x)$ and $f(0)=0$ and so
$f(px)=pf(x)$ $\forall x$, $\forall p\in\mathbb Q$
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pco
23437 posts
#16 • 1 Y
Y by Adventure10
GeronimoStilton wrote:
Some progress:
...
... so I can't finish the problem.
Sorry, but what is the interest of this post ?
Just publicly claim that you are working on this problem ? And what about your progress about your sister's garden ? and what is the last film you liked ?
Just request for some help ? : read the thread ! there is already a full solution (see post #10)
Just bring some new informations to the community ? but this contribution already has been given at least thrice in the current thread (and btw is the beginning of the full solution in the post #10)
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Keith50
464 posts
#17
Y by
This is a folklore:
$\bigstar \color{blue}{\textit{\textbf{ANS:}}}$ $f(x)=cx \quad x\in \mathbb{R}$ where $c\in \mathbb{R}$.
$\spadesuit \color{red}{\textit{\textbf{Proof:}}}$ It's easy to see that this is a solution to the given FE. Let $P(x,y)$ denote the given assertion, \[P(x,-x): f(0)=0\]\[P(x,0): f(x^3)=xf(x^2).\]So, let $u=x^3, v=y^3$, $f(u+v)=f(x^3+y^3)=xf(x^2)+yf(y^2)=f(x^3)+f(y^3)=f(u)+f(v)$ and $f$ is additive. We use a standard trick: \[P(x+1,x-1): \boxed{f(x)=f(1)x \quad \forall x\in \mathbb{R}}\]as many terms cancel nicely due to additivity and it yields the desired result. $\quad \blacksquare$
This post has been edited 3 times. Last edited by Keith50, Mar 23, 2021, 8:30 AM
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hectorleo123
338 posts
#18
Y by
Raja Oktovin wrote:
Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying \[ f(x^3+y^3)=xf(x^2)+yf(y^2)\]for all real numbers $ x$ and $ y$.
Hery Susanto, Malang
$\color{blue} \boxed{\textbf{Answer: f(x)=cx}}$
$\color{blue} \boxed{\textbf{Proof:}}$
$\color{blue} \rule{24cm}{0.3pt}$
$$f(x^3+y^3)=xf(x^2)+yf(y^2)...(\alpha)$$In $(\alpha) y=0:$
$$\Rightarrow f(x^3)=xf(x^2)...(\beta)$$$(\beta)$ in $(\alpha):$
$$\Rightarrow f(x^3+y^3)=f(x^3)+f(y^3)$$$$\Rightarrow f(x+y)=f(x)+f(y)...(I)$$In $(\beta) x=x+1:$
$$\Rightarrow f(x^3+3x^2+3x+1)=(x+1)f(x^2+2x+1)$$By $(I):$
$$\Rightarrow f(x^3)+f(3x^2)+f(3x)+f(1)=xf(x^2)+xf(2x)+xf(1)+f(x^2)+f(2x)+f(1)$$$$\Rightarrow f(x)=xf(1)$$$$\Rightarrow \boxed{\textbf{f(x)=cx}}_\blacksquare$$$\color{blue} \rule{24cm}{0.3pt}$
This post has been edited 1 time. Last edited by hectorleo123, Jun 19, 2023, 10:49 PM
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F10tothepowerof34
195 posts
#19
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This problem is very similar to Macedonian Mathematical Olympiad 2007 P4
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hectorleo123
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#20
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F10tothepowerof34 wrote:
This problem is very similar to Macedonian Mathematical Olympiad 2007 P4
It's the same problem...
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