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shortlisted problems being used in undergraduate competition
enter16180   0
6 minutes ago
Hello, I am posting here to let know ( clarified after a post in College Math forum) that Problem 10 of Open Mathematical Olympiad for University Students ( OMOUS-2025) held at Ashgabat, Turkmenistan on 13-18 April, 2025 is found to be A6 Shortlisted Problems IMO-2024.
Following is discussion on College Math Forum
https://artofproblemsolving.com/community/c7h3551018_omous2025_team_competition_p10


Image of problem from competition for reference below.
0 replies
enter16180
6 minutes ago
0 replies
Functional equation
shactal   1
N 21 minutes ago by Mathzeus1024
Source: Own
Hello, I found this functional equation that I can't solve, and I haven't got any hints. Could someone try and find the solution, it's actually quite difficult:
Find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that, for all $x, y \in \mathbb{R} $,
$$
f(x + f(y)) + f(y + f(x)) = f(x \, f(y) + y \, f(x)) + f(x + y)$$Thank you.
1 reply
shactal
Yesterday at 11:15 PM
Mathzeus1024
21 minutes ago
9 MathCounts prep
ericheathclifffry   13
N Today at 3:02 AM by Aniharry
Redacted
13 replies
ericheathclifffry
Monday at 11:06 PM
Aniharry
Today at 3:02 AM
AMC 8 DHR
PhoenixMathClub   15
N Today at 1:50 AM by K1mchi_
Hey y'all so I just started preparing for the 2026 AMC 8 and I was wondering if you could give me some suggestions on what I should do so I can get DHR. I barely practice math competition and I just started to get serious this year so my scores might be bad so here it is:

5th Grade: 12
6th Grade: 16
6th Grade Summer/7th Grade: Mocking 20-24 on AMC 8 Tests

I saw many recommendations about buying AoPS books but not many about the number theory so I got the Introduction to Algebra, Introduction to Counting and Probability, and Introduction to Geometry Books. I was wondering what other suggestions that you would have for me to get better at AMC 8.
15 replies
PhoenixMathClub
Jul 10, 2025
K1mchi_
Today at 1:50 AM
9 Was this year's AMC 8 hard or easy
ChuMath   69
N Today at 1:30 AM by a.zvezda
Question: Was this year's AMC 8 harder or easier than last year's (or the historical average)

69 replies
ChuMath
Jan 25, 2024
a.zvezda
Today at 1:30 AM
9 How many Math Olympiad concepts/theorems can you name?
a.zvezda   1
N Today at 1:15 AM by OWOW
I know these
1 reply
a.zvezda
Today at 1:09 AM
OWOW
Today at 1:15 AM
9 Worst math subject
a.zvezda   58
N Today at 1:13 AM by a.zvezda
Mine is geo and C&P because it's really annoying and last year, I got 16 on the AMC 8 when I sillied a few geo and C&P problems. :wallbash_red:
58 replies
a.zvezda
Jul 14, 2025
a.zvezda
Today at 1:13 AM
Random but useful theorems
booking   60
N Today at 1:12 AM by booking
There have been all these random but useful theorems
Please post any theorems you know, random or not, but please say whether they are random or not.
I'll start give an example:
Random
I am just looking for some theorems to study.
60 replies
booking
Jul 16, 2025
booking
Today at 1:12 AM
The 24 Game, but with a twist!
PikaPika999   416
N Today at 12:52 AM by Biglion
So many people know the 24 game, where you try to create the number 24 from using other numbers, but here's a twist:

You can only use the number 24 (up to 5 times) to try to make other numbers :)

the limit is 5 times because then people could just do $\frac{24}{24}+\frac{24}{24}+\frac{24}{24}+...$ and so on to create any number!

honestly, I feel like with only addition, subtraction, multiplication, and division, you can't get pretty far with this, so you can use any mathematical operations!

Banned functions
416 replies
PikaPika999
Jul 1, 2025
Biglion
Today at 12:52 AM
Troll Problem
giratina3   13
N Today at 12:14 AM by Ryanzzz
If $\frac{a}{a - 1} = \frac{b^2 + 2b - 1}{b^2 + 2b - 2}$, then what does $a$ equal in terms of $b$?

Hint 1
Hint 2
Hint 3
13 replies
giratina3
Jul 11, 2025
Ryanzzz
Today at 12:14 AM
counting problems
BlueAnglerfish42   8
N Today at 12:11 AM by Ryanzzz
1. How many ways are there for 5 skibidi toilets to swallow 30 different people if each toilet can swallow up to 5 people in one gulp? (A toilet can gulp more than once.)

2. Call a number "rizztastic" if it is divisible by more than 4 odd numbers. How many three-digit numbers are "rizztastic"?

3. 10 skibidi toilets don't have names, and 5 toilets have names. Each toilet name is three letters long. How many ways are there to give the rest of the toilets names?
8 replies
BlueAnglerfish42
Jul 17, 2025
Ryanzzz
Today at 12:11 AM
9 Easiest math competition
a.zvezda   51
N Today at 12:10 AM by Ryanzzz
MOEMS, Math Kangaroo, and Beestar are to me the easiest. :P
51 replies
a.zvezda
Jul 15, 2025
Ryanzzz
Today at 12:10 AM
Find all functions
WakeUp   21
N Apr 30, 2025 by CrazyInMath
Source: Baltic Way 2010
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
21 replies
WakeUp
Nov 19, 2010
CrazyInMath
Apr 30, 2025
Find all functions
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G H BBookmark kLocked kLocked NReply
Source: Baltic Way 2010
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WakeUp
1347 posts
#1 • 4 Y
Y by jhu08, Adventure10, Mango247, and 1 other user
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
This post has been edited 1 time. Last edited by WakeUp, Nov 19, 2010, 8:18 PM
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pco
23515 posts
#2 • 7 Y
Y by Abdollahpour, jhu08, Adventure10, Mango247, and 3 other users
WakeUp wrote:
Let $R$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]
for all $x,y\in\mathbb{R}$.
Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$

$P(0,x)$ $\implies$ $f(0)(f(x)+x-2)$

If $f(0)\ne 0$, this implies $f(x)=2-x$ which indeed is a solution.

Let us from know consider that $f(0)=0$

$P(x,0)$ $\implies$ $f(x^2)=xf(x)$
Then : $P(x,y)$ $\implies$ $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
Same : $P(y,x)$ $\implies$ $yf(y)+f(xy)=f(x)f((y)+xf(y)+yf(x+y)$
Subtracting implies $(x-y)f(x+y)-f(x)-f(y))=0$

and so $f(x+y)=f(x)+f(y)$ $\forall x\ne y$

Plugging this in $xf(x)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$, we get $f(xy)=f(x)f(y)+yf(x)+xf(y)$ $\forall x\ne y$

$\iff$ $f(xy)+xy=(f(x)+x)(f(y)+y)$

Let then $g(x)=f(x)+x$. We got :
$g(0)=0$
$g(x+y)=g(x)+g(y)$ $\forall x\ne y$
$g(xy)=g(x)g(y)$ $\forall x\ne y$

From the first, we get $g(-x)=-g(x)$ and so $g(2x+(-x))=g(2x)+g(-x)$ and so $g(2x)=2g(x)$ and so $g(x+y)=g(x)+g(y)$ $\forall x,y$
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

So :
$g(x+y)=g(x)+g(y)$ $\forall x,y$
$g(xy)=g(x)g(y)$ $\forall x,y$
And so, very classical, $g(x)=x$ and $f(x)=0$

Hence the two solutions :
$f(x)=2-x$
$f(x)=0$
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WakeUp
1347 posts
#3 • 3 Y
Y by jhu08, Adventure10, Mango247
pco wrote:
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution? Note also $f(x)=-x$ is a solution.
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Solving
185 posts
#4 • 3 Y
Y by jhu08, Adventure10, Mango247
SO im right?

x=y
2f(x^2)=f(x)^2+x(f(x)+f(2x))
2f(0)=f(0)^2
f(0)=0
or
f(0)=1/2
let x=x, y=0
f(x^2)+f(0)=f(x)f(0)+xf(x)
f(0)=1/2
f(x^2)+1/2=f(x)(1/2+x)
f(x)=ax+b
ax^2+b+1/2=(ax/2+ax^2+b/2+bx)
b/2=b+1/2
b=-1
a=2
f(x)=2x-1 is the solution
for
f(0)=0
f(x)=x
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pco
23515 posts
#5 • 3 Y
Y by jhu08, Adventure10, Mango247
WakeUp wrote:
pco wrote:
From the second, we get $g(x(x+1))=g(x)g(x+1)=g(x)^2+g(x)$
But also $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so $g(x^2)=g(x)^2$ and so $g(xy)=g(x)g(y)$ $\forall x,y$

Hi pco, could you please explain this part of the solution?
Yes, :oops:, I wrote too quickly !
First we can see that $g(x)=0$ is a solution (and so $f(x)=-x$ is indeed !
If $g(x)$ is not the all zero function, let then $u$ such that $g(u)\ne 0$. If $u=1$, choose instead $u=-1$.
Then the second equation gives us $g(u)(g(1)-1)=0$ and so $g(1)=1$

Then $g(x(x+1))=g(x)g(x+1)$ (using second equation since $x\ne x+1$) $=g(x)(g(x)+g(1))=g(x)^2+g(x)$
But $g(x(x+1))=g(x^2+x)=g(x^2)+g(x)$
And so, $g(x^2)=g(x)^2$ and so the second equation $g(xy)=g(x)g(y)$ is also true if $x=y$
...

WakeUp wrote:
Note also $f(x)=-x$ is a solution.
Yes, :oops: $g(x)=0$ is also a solution (I forgot it)

And so :
$f(x)=0$
$f(x)=-x$
$f(x)=2-x$
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borntobeweild
331 posts
#6 • 4 Y
Y by Jerry37284, jhu08, Adventure10, Mango247
This is just about as interesting as a FE can get while still dying to the standard strategies of plugging stuff in, taking cases, and testing. Nevertheless, it was a fun problem.

Reading this won't teach you anything except for what tricks you should have tried
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Onlygodcanjudgeme
18 posts
#7 • 3 Y
Y by jhu08, EHHSW, Adventure10
put $ (x,y) = (0,0) $ then we take that 1) $ f(0) =0 $ 2)$ f(0) =2 $
1) put $ (x,y) = (x,0) $ then we take that $ f(x^2) = x \cdot f(x) $ and this is odd function .
put $ (x,y) = (x,-x) $ then we take that f(-x) = x , f(x) =0
2)put $ (x,y) = (0,x) $ then we take that f(x) = 2-x
so answer is f(x)=0 , f(x) = -x ,f(x) = 2-x
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Jerry37284
46 posts
#8 • 3 Y
Y by jhu08, Adventure10, Mango247
@Onlygodcanjudgeme :I think you got "$\forall x$ : either $f(x)=0$, either $f(x)=-x$" and not "either $f(x)=0$ $\forall x$ , either $f(x)=-x$ $\forall x$ "
This post has been edited 1 time. Last edited by Jerry37284, Dec 10, 2018, 3:34 AM
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Keith50
464 posts
#9 • 3 Y
Y by jhu08, Mango247, Mango247
Let $P(x,y)$ be the given assertion, $P(0,0)\implies 2f(0)=f(0)^2\implies f(0)=0,2$.
If $f(0)=2$, then $P(0,x)\implies 4=2f(x)+2x\implies f(x)=2-x \ \ \forall x\in\mathbb{R}$.
So, if $f(0)=0,$ $P(x,0)\implies f(x^2)=xf(x) \ \ \ (1)$
$P(x,-x)\implies f(x^2)+f(-x^2)=f(x)f(-x)-xf(x) \ \ \ (2)$
Using $P(x,y)$ and $P(-x,-y)$, we can arrive at \[f(x)f(y)+yf(x)+xf(x+y)=f(-x)f(-y)-yf(-x)-xf(-x-y),\]letting $y=0$, we get $f(x)=-f(-x)$ where $x\ne0$, so $f$ is odd.
From $(2)$, using the fact that $f$ is odd, \[f(x)(f(x)+x)=0\implies f(x)=0,-x.\]Now, assume that there exists $a,b\in \mathbb{R}, a,b\ne 0$ such that $f(a)=0$ and $f(b)=-b$, using $(1)$, $f(a^2)=0, f(b^2)=-b^2$,
using $P(a,b)$ we get \[f(ab)=af(a+b).\]If $f(ab)=-ab$, \[-b=f(a+b).\]If $f(a+b)=0$, then $b=0$, a contradiction.
If $f(a+b)=-(a+b)$, then $a=0$, a contradiction. Thus, when $f(ab)=0$, $f(a+b)=0$ as $a=0$ is a contradiction.
Take $P(a,b)$ and $P(b,a)$, subtracting one from another gives \[f(a^2)-f(b^2)=bf(a)+af(a+b)-af(b)-bf(a+b)\]which simplifies to \[b^2=ab\implies b(a-b)=0\]and so $a=b$ since $b\ne 0$ but this means $f(b)=f(a)=0=-b$, a contradiction.
Hence, we have \[f(x)\equiv 0, 2-x, -x\]as solutions and plugging them into the equation, we see that they indeed satisfy. $\blacksquare$
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Kimchiks926
256 posts
#10 • 2 Y
Y by mkomisarova, jhu08
Solved together with @blastoor
Let $P(x,y)$ denote assertion of given functional equation.

Note that $P(0,0)$ gives us $2f(0) = f(0)^2$, which means that $f(0)=0$ or $f(0)=2$. If $f(0) =2$, then $P(0,x)$ gives us:
\begin{align*} 
 f(0) + f(0) = f(0)f(x) + xf(0) \\
4 = 2f(x) + 2x \\
f(x) = 2 -x 
\end{align*}It is easy to check that function $f(x) =2-x$ works.

From now we assume that $f(0) =0$. Note that $P(x,0)$ gives us:
$$ f(x^2) = xf(x) \qquad (1) $$In relation $(1)$ replacing $x$ by $-x$ yields:
$$ f(x^2) = xf(x) = -xf(-x) \implies -f(x) = f(-x) $$Also note that $P(x,-x)$ gives us:
\begin{align*}
f(x^2) + f(-x^2) = f(x)f(-x) -xf(x) \\
f(x^2) -f(x^2) = -f(x)^2 - xf(x) \\ 
f(x)^2 = -xf(x) 
\end{align*}We conclude that $f(x) = 0$ or $f(x) =-x$. Now we are left to escape pointwise trap. Assume that there exist nonzero real numbers $a, b$ such that $f(a) =-a$ and $f(b) = 0 $. Note that $f(b^2) = bf(b) = 0$ and that $P(b,a-b)$ gives us:
\begin{align*} 
f(b^2) +f(b(a-b))= f(b)f(a-b) + (a-b)f(b) + bf(a) \\ 
f(b(a-b)) = -ab 
\end{align*}If $f(b(a-b)) = b^2 -ab$, then $b^2 =0 $, which is contradiction since $b \ne 0$. On another hand id $f(b(a-b)=0=ab$, then one of the numbers $a,b$ is zero, which is again contradiction.

We conclude that $f(x)=0$, $f(x) = 2 -x$, $f(x) =-x$ are only solutions.
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508669
1040 posts
#11 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

Let $P(x, y)$ be the assertion. We claim that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$. It is not hard to see they work. Now we show that these are the only such functions.

We see that $P(x, 0) \implies f(0)(f(x) + x - 2) = 0$, so if $f(0) \neq 0$, then $\boxed{f(x) = 2-x}$ which is indeed a solution.

Otherwise, let $f(0) = 0 \dots (1)$. Then by $P(x, 0)$, we yield that $f(x^2) = xf(x) = -xf(-x)$ (by replacing $x$ by $-x$) and so $f$ is odd function. Now $P(x, y) - P(y, x)$ along with $f(x^2) = xf(x)$ gives that $(x-y)(f(x+y)-f(x)-f(y)) = 0$ and so if $x \neq y$, definitely $f(x+y) = f(x) + f(y)$ and so $f$ is additive. Now, we re-arrange few terms in $P(x, y)$.

$P(x, y) \implies f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y) \implies xf(x) + yf(x) = f(x)f(y) + yf(x) + xf(x) + xf(y) \implies 0 = f(x)f(y) + xf(y) = f(y)(f(x) + x)$ which means that $f(x) = -x$ or $f(x) = 0$ for all reals $x$.

Let us say that $A = \{ x \lvert f(x) = -x, x \neq 0 \}$ and $A = \{ x \lvert f(x) = 0, x \neq 0 \}$. Let $a \in A, b \in B$. We see that $f(ab) = bf(a) + af(b)$, and so here in this case, $f(ab) = b \times -a = a \times 0 = 0$, so either of $a$ or $b$ is $0$, a contradiction to definition of elements belonging to sets $A$ and $B$. Hence, $\lvert A \rvert = 0$ or $\lvert B \rvert = 0$. We see that $f(0) = 0 = -0$. Therefore, we see that all solutions to the given functional equation are of form $\boxed{f(x) = 2-x}$ for reals $x$, $\boxed{f(x) = -x}$ for all positive reals $x$ and $\boxed{f(x) = 0}$ for all reals $x$
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jasperE3
11445 posts
#13 • 1 Y
Y by jhu08
Hint

Let $P(x,y)$ be the assertion $f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)$
$P(0,0)\Rightarrow 2f(0)=f(0)^2$
If $f(0)=2$ then:
$P(0,x)\Rightarrow\boxed{f(x)=2-x}$, which works.
Now assume $f(0)=0$.
$P(x,0)\Rightarrow f(x^2)=xf(x)$
$P(1,x)\Rightarrow f(x+1)=f(x)+f(1)-f(x)f(1)-xf(1)$
We use this recurrence to find $f(1)$.
$P(1,1)\Rightarrow f(2)=f(1)-f(1)^2$
$P(1,2)\Rightarrow f(3)=f(2)-f(1)-f(2)f(1)=f(1)^3-2f(1)^2$
$P(1,3)\Rightarrow f(4)=f(3)-2f(1)-f(3)f(1)=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$
But $f(4)=2f(2)=2f(1)-2f(1)^2$, so we find that $2f(1)-2f(1)^2=-f(1)^4+3f(1)^3-2f(1)^2-2f(1)$. Solving, we have $f(1)\in\{-1,0,2\}$.

$\textbf{Case 1: }f(1)=0$
$P(1,x)\Rightarrow f(x)=f(x+1)$
$P\left(x,\frac yx+1\right)-P\left(x,\frac yx\right)\Rightarrow f(x+y)=f(x)+f(y)$ if $x\ne0$, but since it holds for $x=0$, $f$ is additive.
By USAMO 2002/4, since $f(x^2)=xf(x)$ and $f$ is additive, we must have $f(x)=xf(1)$, hence $\boxed{f(x)=0}$ which works.

$\textbf{Case 2: }f(1)=-1$
$P(1,x)\Rightarrow f(x+1)=2f(x)+x-1$
$P(x,1)\Rightarrow f(x^2)+f(x)=xf(x+1)\Rightarrow xf(x)+f(x)=2xf(x)+x^2-x\Rightarrow\boxed{f(x)=-x}$ since $f(1)=-1$, which works.

$\textbf{Case 3: }f(1)=2$
$P(1,x)\Rightarrow f(x+1)=-f(x)-2x+2$
$P(x,1)\Rightarrow(x-1)f(x)=-x^2+x\Rightarrow f(x)=\begin{cases}-x&\text{if }x\ne1\\2&\text{if }x=1\end{cases}$ which doesn't work.
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RopuToran
609 posts
#14 • 1 Y
Y by jhu08
Here is my way to solve the first case where $f(0)=0$ :D
$P(x,0)$ gives us $$f(x^2) = xf(x),  \forall x \quad (1)$$$P(y,x)$ gives us $$f(y^2) + f(xy)= f(x)f(y)+xf(y)+yf(x+y),  \forall x,y \quad (2)$$By pluging (1) into (2) then subtracting (2) and the original FE, we got $$(x-y) (f(x)+f(y)) = (x-y) f(x+y), \forall x,y$$, which implies $$ f(x) + f(y) = f(x+y), \forall x \neq y \quad (3)$$From $(1)$, we also have $f(x^2)= -x f(x)$ which leads to $$f(x)= f(-x), \forall x \quad (4)$$By $P(x,-x)$ and using $(4)$, we got $$f(x)^2 = -xf(x), \forall x (5)$$Using $(1)$ and $(3)$, from the origina FE, we have $$f(xy) = f(x)f(y)+xf(y) + yf(x), \forall x \neq y$$, which equivalent to $$ f(x)(f(y)+y) = f(xy) - xf(y), \forall x \neq y \quad (6)$$Case 1: There is a number $k \neq 0$ such that $f(k) = -k$. In $(6)$, let $y=k$, we have $f(kx)=-kx, \forall x \neq k$. Thus $f(x) = -x, \forall x \neq k^2$. In the other hand, $f(k^2) = kf(k) = -k^2$. So, $f(x)=-x, \forall x$.
Case 2: There is no number $k$ other than $0$ such that $f(k) = -k$, which means $f(x) \neq -x, \forall x \neq 0$. With $(5)$, we implies $f(x) = 0, \forall x \neq 0$. In the otherhand, $f(0)=0$, thus $f(x) = 0$.

P/S
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MathLuis
1597 posts
#15 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

Case 1: $f$ is constant.
We will set $f(x)=c$ where $c$ is some real constant. Plugging this on the F.E. we get:
$$2c=c^2+cy+cx \implies c=0 \implies f(x)=0$$Cade 2: $f$ is non-constant.
Let $P(x,y)$ the assertion of the given F.E.
$P(0,0)$
$$f(0)^2=2f(0) \implies f(0)=0 \; \text{or} \; f(0)=2$$Case 2.1: $f(0)=2$
$P(0,x)$
$$4=2f(x)+2x \implies f(x)=2-x$$Case 2.2: $f(0)=0$
$P(x,0)$
$$f(x^2)=xf(x) \implies f \; \text{odd}$$$P(x,-x)$ where $x$ is any non-cero real
$$f(x)^2+xf(x)=0 \implies f(x)=-x$$Since $f(0)=0$ we have that $f(x)=-x \; \forall x \in \mathbb R$
Thus the solutions are:

$\boxed{f(x)=0 \; \forall x \in \mathbb R}$

$\boxed{f(x)=2-x \; \forall x \in \mathbb R}$

$\boxed{f(x)=-x \; \forall x \in \mathbb R}$

Thus we are done :blush:
This post has been edited 1 time. Last edited by MathLuis, Aug 8, 2021, 6:26 PM
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rama1728
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#16 • 1 Y
Y by jhu08
WakeUp wrote:
Let $\mathbb{R}$ denote the set of real numbers. Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that
\[f(x^2)+f(xy)=f(x)f(y)+yf(x)+xf(x+y)\]for all $x,y\in\mathbb{R}$.

A good problem for oddness of a function and how to tackle pointwise traps. Other steps are natural.

Solution
This post has been edited 3 times. Last edited by rama1728, Aug 8, 2021, 7:14 PM
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JustKeepRunning
2958 posts
#17 • 1 Y
Y by jhu08
A nice exercise for pointwise trap!

The answers are $f\equiv 0, 2-x, -x$. These work.

Denote the assertion by $P(x,y)$. $P(0,0)$ gives that $f(0)=0,2$.

Case 1: $f(0)=2$.

$P(0,y)$ gives that $f(y)=2-y.$

Case 2: $f(0)=0$

$P(x,0)$ gives that $f(x^2)=xf(x),$ so we have that $f$ is odd. Then $P(x,-x)$ gives that $0=f(x)(f(x)+x),$ so $f(x)=0,-x$. To avoid pointwise trap, suppose that $f(x)=0$ and $f(y)=-y$ for some $x,y\neq 0$. Then from $P(x,y)$ and $P(y,x)$ and subtracting, we get that $xf(x)-yf(y)=yf(x)-xf(y)+(x-y)f(x+y)$. Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.
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RopuToran
609 posts
#18 • 1 Y
Y by jhu08
JustKeepRunning wrote:
Obviously, $x\neq y,$ and simplifying gives that $f(x+y)=f(y)=-y$. If $y=0,$ we are done, and if $x+y=y,$ then $x=0,$ and we are done as well.

How did you simplify the equation into this?
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jasperE3
11445 posts
#19 • 1 Y
Y by jhu08
They used the properties $f(x)=0$ and $f(y)=-y$.
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ZETA_in_olympiad
2211 posts
#20
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Let $P(x,y)$ denote the assertion. Quickly $P(0,x)$ gives $f(x)\equiv 2-x$ or $f(0)=0.$ The former works, so we explore the latter.

$P(x,0)$ gives $f(x^2)=xf(x).$ And so comparing $P(x,y)$ with $P(y,x)$ shows that $f(x+y)=f(x)+f(y)$ for all $x\neq y.$ To conclude $P(x,-x)$ implies $f(x)\in \{0,-x\}$ but since additive $f\equiv 0$ or $f\equiv -x$ and both satisfy.
This post has been edited 1 time. Last edited by ZETA_in_olympiad, Aug 1, 2022, 11:08 AM
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HamstPan38825
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#22
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This dies to basically anything.

Setting $x=y=0$ yields $2f(0) = f(0)^2$. If $f(0) = 2$, setting $x=0$ yields $f(x) = 2-x$ immediately.

If $f(0) = 0$, setting $y=0$ yields $f(x^2) = xf(x)$, implying $f$ is odd. Setting $y=-x$ in the original, \[0=f\left(x^2\right) + f\left(-x^2\right) = f(x)f(-x)-xf(x) = -f(x)^2 - xf(x).\]So for each $x$, either $f(x) = 0$ or $f(x) = -x$. There are many ways to resolve the pointwise trap, but here is a really stupid way. By setting $y=x$ we get $f(x)^2 + xf(2x) = xf(x)$, i.e. $xf(2x) = 2xf(x)$ or $f(2x) = f(x)$. Furthermore, by swapping $x$ and $y$, we get \[(x-y)f(x) - xf(x+y)=(y-x)f(y) - yf(x+y)\]so $f(x+y) = f(x)+f(y)$ for all $x \neq y$. Combining this with the previous equation yields that $f$ is Cauchy and bounded below on $x \geq 0$, thus $f$ is linear. We can check that only $f \equiv 0$ works here.
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math-olympiad-clown
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#23
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case 1. f is a constant function :
c+c=c^2+cy+xc x=y=1 plug in we get c=0
so f(x)=0

case 2. f is not a constant function :
P(0,y) :2f(0)=f(0)f(y)+yf(0)

2-1. if f(0) is not 0 then y=1 plug in we get f(1)=1
P(1,y): 1+f(y)=f(y)+y+f(y+1) and we know that f(y)=2-y

2-2.f(0)=0 : P(0,0): f(x^2)=xf(x) this imply f is a odd function
P(x,-x) : 0=-(f(x)^2)-xf(x) we get f(x)=-x

so the answer is f(x)=0 or 2-x or -x
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CrazyInMath
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#24
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$f(x)=0$ works, now assume $f$ is not constant
$P(0,0)$ gives $f(0)=0, 2$

If $f(0)=2$, $P(0, x)$ gives $f(x)=2-x$ which works
If $f(0)=0$, $P(x, 0)$ gives $f(x^2)=xf(x)$ so $f$ is odd
then $P(x, -x)$ gives $-f(x)^2-xf(x)=0$ so $f(x)(f(x)-x)=0$, so $f(x)=-x, 0$.
If $f(a)=0$, $f(b)=-b$, by $P(a, b)$ and $P(b, a)$ we have contradiction.
So $f(x)=-x$ or $f(x)=0$ in this case.
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