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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Conditional geo with centroid
a_507_bc   6
N 27 minutes ago by LeYohan
Source: Singapore Open MO Round 2 2023 P1
In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$, the extension of $AG$ meets $\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A, O, S$ are collinear. Prove that $\angle AGO = 90^{o}$.
6 replies
a_507_bc
Jul 1, 2023
LeYohan
27 minutes ago
Channel name changed
Plane_geometry_youtuber   0
28 minutes ago
Hi,

Due to the search handle issue in youtube. My channel is renamed to Olympiad Geometry Club. And the new link is as following:

https://www.youtube.com/@OlympiadGeometryClub

Recently I introduced the concept of harmonic bundle. I will move on to the conjugate median soon. In the future, I will discuss more than a thousand theorems on plane geometry and hopefully it can help to the students preparing for the Olympiad competition.

Please share this to the people may need it.

Thank you!
0 replies
Plane_geometry_youtuber
28 minutes ago
0 replies
IMO Shortlist 2010 - Problem G1
Amir Hossein   134
N an hour ago by happypi31415
Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$

Proposed by Christopher Bradley, United Kingdom
134 replies
Amir Hossein
Jul 17, 2011
happypi31415
an hour ago
Divisors on number
RagvaloD   34
N an hour ago by cubres
Source: All Russian Olympiad 2017,Day1,grade 10,P5
$n$ is composite. $1<a_1<a_2<...<a_k<n$ - all divisors of $n$. It is known, that $a_1+1,...,a_k+1$ are all divisors for some $m$ (except $1,m$). Find all such $n$.
34 replies
RagvaloD
May 3, 2017
cubres
an hour ago
IMO ShortList 2002, number theory problem 2
orl   59
N an hour ago by cubres
Source: IMO ShortList 2002, number theory problem 2
Let $n\geq2$ be a positive integer, with divisors $1=d_1<d_2<\,\ldots<d_k=n$. Prove that $d_1d_2+d_2d_3+\,\ldots\,+d_{k-1}d_k$ is always less than $n^2$, and determine when it is a divisor of $n^2$.
59 replies
orl
Sep 28, 2004
cubres
an hour ago
None of the circles contains the pentagon - ILL 1970, P34
Amir Hossein   1
N an hour ago by legogubbe
In connection with a convex pentagon $ABCDE$ we consider the set of ten circles, each of which contains three of the vertices of the pentagon on its circumference. Is it possible that none of these circles contains the pentagon? Prove your answer.
1 reply
Amir Hossein
May 21, 2011
legogubbe
an hour ago
interesting incenter/tangent circle config
LeYohan   0
2 hours ago
Source: 2022 St. Mary's Canossian College F4 Final Exam Mathematics Paper 1, Q 18d of 18 (modified)
$BC$ is tangent to the circle $AFDE$ at $D$. $AB$ and $AC$ cut the circle at $F$ and $E$ respectively. $I$ is the in-centre of $\triangle ABC$, and $D$ is on the line $AI$. $CI$ and $DE$ intersect at $G$, while $BI$ and $FD$ intersect at $P$. Prove that the points $P, F, G, E$ lie on a circle.
0 replies
LeYohan
2 hours ago
0 replies
interesting geo config (2/3)
Royal_mhyasd   5
N 2 hours ago by Royal_mhyasd
Source: own
Let $\triangle ABC$ be an acute triangle and $H$ its orthocenter. Let $P$ be a point on the parallel through $A$ to $BC$ such that $\angle APH = |\angle ABC-\angle ACB|$. Define $Q$ and $R$ as points on the parallels through $B$ to $AC$ and through $C$ to $AB$ similarly. If $P,Q,R$ are positioned around the sides of $\triangle ABC$ as in the given configuration, prove that $P,Q,R$ are collinear.
5 replies
Royal_mhyasd
Yesterday at 11:36 PM
Royal_mhyasd
2 hours ago
interesting geometry config (3/3)
Royal_mhyasd   2
N 2 hours ago by Royal_mhyasd
Let $\triangle ABC$ be an acute triangle, $H$ its orthocenter and $E$ the center of its nine point circle. Let $P$ be a point on the parallel through $C$ to $AB$ such that $\angle CPH = |\angle BAC-\angle ABC|$ and $P$ and $A$ are on different sides of $BC$ and $Q$ a point on the parallel through $B$ to $AC$ such that $\angle BQH = |\angle BAC - \angle ACB|$ and $C$ and $Q$ are on different sides of $AB$. If $B'$ and $C'$ are the reflections of $H$ over $AC$ and $AB$ respectively, $S$ and $T$ are the intersections of $B'Q$ and $C'P$ respectively with the circumcircle of $\triangle ABC$, prove that the intersection of lines $CT$ and $BS$ lies on $HE$.

final problem for this "points on parallels forming strange angles with the orthocenter" config, for now. personally i think its pretty cool :D
2 replies
Royal_mhyasd
Today at 7:06 AM
Royal_mhyasd
2 hours ago
Convex Quadrilateral with Bisector Diagonal
matinyousefi   8
N 3 hours ago by lpieleanu
Source: Germany TST 2017
In a convex quadrilateral $ABCD$, $BD$ is the angle bisector of $\angle{ABC}$. The circumcircle of $ABC$ intersects $CD,AD$ in $P,Q$ respectively and the line through $D$ parallel to $AC$ cuts $AB,AC$ in $R,S$ respectively. Prove that point $P,Q,R,S$ lie on a circle.
8 replies
matinyousefi
Apr 11, 2020
lpieleanu
3 hours ago
Kids in clubs
atdaotlohbh   0
3 hours ago
There are $6k-3$ kids in a class. Is it true that for all positive integers $k$ it is possible to create several clubs each with 3 kids such that any pair of kids are both present in exactly one club?
0 replies
atdaotlohbh
3 hours ago
0 replies
Turbo's en route to visit each cell of the board
Lukaluce   22
N 3 hours ago by HamstPan38825
Source: EGMO 2025 P5
Let $n > 1$ be an integer. In a configuration of an $n \times n$ board, each of the $n^2$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.

Proposed by Melek Güngör, Turkey
22 replies
Lukaluce
Apr 14, 2025
HamstPan38825
3 hours ago
n lamps
pohoatza   47
N 3 hours ago by yayyayyay
Source: IMO Shortlist 2006, Combinatorics 1, AIMO 2007, TST 2, P1
We have $ n \geq 2$ lamps $ L_{1}, . . . ,L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows: if the lamp $ L_{i}$ and its neighbours (only one neighbour for $ i = 1$ or $ i = n$, two neighbours for other $ i$) are in the same state, then $ L_{i}$ is switched off; – otherwise, $ L_{i}$ is switched on.
Initially all the lamps are off except the leftmost one which is on.

$ (a)$ Prove that there are infinitely many integers $ n$ for which all the lamps will eventually be off.
$ (b)$ Prove that there are infinitely many integers $ n$ for which the lamps will never be all off.
47 replies
pohoatza
Jun 28, 2007
yayyayyay
3 hours ago
prove that at least one of them is divisible by some other member of the set.
Martin.s   0
3 hours ago
Given \( n + 1 \) integers \( a_1, a_2, \ldots, a_{n+1} \), each less than or equal to \( 2n \), prove that at least one of them is divisible by some other member of the set.
0 replies
Martin.s
3 hours ago
0 replies
USAMO 2002 Problem 4
MithsApprentice   90
N Apr 25, 2025 by Ilikeminecraft
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y)  \] for all pairs of real numbers $x$ and $y$.
90 replies
MithsApprentice
Sep 30, 2005
Ilikeminecraft
Apr 25, 2025
USAMO 2002 Problem 4
G H J
G H BBookmark kLocked kLocked NReply
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KevinYang2.71
428 posts
#80 • 1 Y
Y by Orthogonal.
Solved with Orthogonal..

We claim the only functions are $\boxed{f(x)\equiv cx}$ for $c\in\mathbb{R}$. It is easy to check that these work.

Let $P(x,y)$ denote the given assertion. $P(0,0)$ gives $f(0)=0$ and $P(x,0)$ gives $f(x^2)=xf(x)$. From $P(x,x)$ we get $f(x)=-f(-x)$. Thus $P(x,y)$ becomes $f(x^2-y^2)=f(x^2)+f(-y^2)$. It follows that
\[
f(x+y)=f(x)+f(y)\ \ \ \ \ \ \ \forall(x,y)\in\mathbb{R}^+\times\mathbb{R}^-.\tag{*}
\]
Claim. We have $f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{R}$.

Proof. If $x$ and $y$ are both positive, $f(x+y)-f(x)=f(x+y)+f(-x)=f(y)$ so we are done. If exactly one of $x$ and $y$ are positive, we are done by $(*)$. If $x$ and $y$ are both negative,
\[
f(x+y)=-f(-x-y)=-f(-x)-f(-y)=f(x)+f(y)
\]by case $1$ so we are done. $\square$

Now
\begin{align*}
xf(x)+xf(1)+f(x)+f(1)&=(x+1)f(x+1)\\
&=f((x+1)^2)\\
&=f(x^2+2x+1)\\
&=xf(x)+2f(x)+f(1)
\end{align*}so $f(x)=f(1)x$, as desired. $\square$
This post has been edited 1 time. Last edited by KevinYang2.71, May 1, 2024, 6:01 PM
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Jndd
1417 posts
#81
Y by
We claim that the solutions are $f(x)=cx$ for any constant $c$, and it's easy to see that this satisfies the equation.

Plugging in $y=0$, we get $f(x^2)=xf(x)$, and plugging in $x=0$, we get $f(-y^2)=-yf(y)$. From this, we get $f(-x^2)=-xf(x)=-f(x^2)$, giving $-f(x)=f(-x)$.

Now, using the first two equations we got by plugging in $x=0$ and then $y=0$, we get $f(x^2-y^2)=f(x^2)-f(y^2)$, which further becomes $f(x^2-y^2)=f(x^2)+f(-y^2)$ since $-f(x)=f(-x)$. This implies that if $x\geq 0, y\leq 0$ or $x\leq 0, y\geq 0$, we have $f(x)+f(y)=f(x+y)$.

However, we can also get $f(x^2-y^2)+f(y^2)=f(x^2)$, and setting $x>y$ gives us that $f(x)+f(y)=f(x+y)$ also holds true when both $x,y\geq 0$. Finally, by negating everything and using $-f(x)=f(-x)$, this also holds true when both $x,y\leq 0$. Hence, $f$ satisfies Cauchy.

Then, we plug in $x+1$ into $f(x^2)=xf(x)$ and use the fact that $f$ satisfies Cauchy to get \[f((x+1)^2)=f(x^2)+f(x)+f(x+1)=(x+1)f(x+1)=xf(x)+xf(1)+f(x)+f(1),\]and by using $f(x^2)=xf(x)$, we can cancel things to get $f(x+1)=f(1)(x+1)$, giving $f(x)=cx$ where $c=f(1)$.
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Markas
150 posts
#82
Y by
We will show that the answer is $f(x) = cx$. These work since $(x^2 - y^2).c = x.cx - y.cy$. Now we need to show these are the only solutions. Let x = 0, we get $f(-y^2) = -yf(y)$. Let y = 0, we get $f(x^2) = xf(x)$. By these two we get that f is odd and also that $f(0) = 0$. So now using what we got we can write the starting equation as $f(x^2 - y^2) + f(y^2) = f(x^2)$. Using this and the fact that f is odd we get that f is additive. Now plugging in x = x + 1 in $f(x^2) = xf(x)$ and using that f is additive we get $f((x+1)^2 ) = (x+1)f(x+1)$ $\Rightarrow$ $f(x^2 + 2x + 1) = f(x^2) + 2f(x) + f(1) = (x+1)f(x) + (x+1)f(1)$ which after clearing things up gives us that $f(x) = f(1)x = cx$ $\Rightarrow$ there are no other solutions and we are ready.
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fearsum_fyz
56 posts
#83 • 2 Y
Y by alexanderhamilton124, poirasss
We claim that the only solution is $\boxed{f(x) = cx}$ for some constant $c$. It is easy to verify that this satisfies the given condition. Now we will prove that it is the only solution.

$\underline{P}(x, x) \implies \boxed{f(0) = 0}$
$\underline{P}(x, y) - \underline{P}(y, x) \implies f(y^2 - x^2) = - f(x^2 - y^2) \implies \boxed{f \text{ is odd}}$.

$\underline{P}(x, 0) \implies \boxed{f(x^2) = x f(x)}$
$\implies f(x^2 - y^2) = f(x^2) - f(y^2)$
$\implies \boxed{f(x - y) = f(x) - f(y) \text{ for positive } x, y} \ldots \textcircled{1}$.

Now choose some positive $x$.
$\underline{P}(x - 1, x) \implies f(1 - 2x) = (x - 1)f(x - 1) - xf(x)$
$\overset{\textcircled{1}}{\implies} f(1) - f(2x) = (x - 1)(f(x) - f(1)) - xf(x)$
$\implies f(2x) - f(x) = xf(1)$
$\overset{\textcircled{1}}{\implies} f(x) = xf(1)$

So $f(x) = cx$ for positive $x$. Since $f$ is odd, this is also true for negative $x$.
We are done.
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alexanderhamilton124
401 posts
#85
Y by
From $f(x^2) = xf(x)$, and $f$ is odd, we can directly use Cauchy to finish :wink:
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Siddharthmaybe
119 posts
#86
Y by
alexanderhamilton124 wrote:
From $f(x^2) = xf(x)$, and $f$ is odd, we can directly use Cauchy to finish :wink:

wait what?
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alexanderhamilton124
401 posts
#87 • 1 Y
Y by S_14159
I don't know if it's well known (somebody told me it was in BJV (?)), but look at the last property here
This post has been edited 1 time. Last edited by alexanderhamilton124, Dec 23, 2024, 7:22 PM
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ItsBesi
147 posts
#89 • 2 Y
Y by alexanderhamilton124, S_14159
alexanderhamilton124 wrote:
I don't know if it's well known (somebody told me it was in BJV (?)), but look at the last property here

Here is another problem you can use this trick. The problem is from 2022 Kosovo TST P1

https://artofproblemsolving.com/community/c6h2797116p24625420
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mathwiz_1207
105 posts
#90
Y by
We claim the solutions are either $\boxed{f \equiv 0}$ and $\boxed{f = cx}$. It is easy to verify these both work. Now, we will show that these are the only functions satisfying the given conditions. Plugging in $x = 0$,
\[f(x^2) = xf(x)\]Plugging in $y = 0$,
\[f(-y^2) = -yf(y) = -f(y^2)\]So, $f(a) = -f(-a)$ for all $a$, thus it is odd. Now, we prove the following claim:

$f$ is additive.
We can rewrite the condition as
\[f(x^2 - y^2) = f(x^2) - f(y^2) \implies f(a - b) = f(a) - f(b)\]for all nonnegative reals $a, b$. Now, set $a = x + y, b = y$, for nonnegative $x, y$. This gives
\[f(y) = f(x+y) - f(x) \implies f(x) + f(y) = f(x + y)\]for all $x, y \geq 0$. Since $f$ is odd, we have
\[-f(-x-y) = f(x + y) = f(x) + f(y) = -f(-x) - f(-y) \implies f(-x-y) = f(-x) + f(-y)\]so $f$ is also additive over all the nonpositive reals. Now, rewrite
\[f(x) - f(-y) = f(x + y) \implies f(x) = f(x + y) + f(-y)\]We can set $x = -y + a$ for any $a \geq y \geq 0$, therefore the above equation rewrites as
\[f(a -y) = f(a) + f(-y)\]So, $f(x) + f(y) = f(x + y)$ is also true for all pairs $(a, b)$ with $a \geq 0 \geq b$.


In conclusion, $f$ is additive over all the real numbers. Thus, we have
\[f((x + 1)^2) = f(x^2) + f(2x) + f(1) = xf(x) + 2f(x) + f(1)\]\[f((x + 1)^2) = (x + 1)f(x + 1) = (x + 1)f(x) + xf(1) + f(1)\]Setting the two equations equal, we get
\[(x + 2)f(x) = (x + 1)f(x) + xf(1) \implies f(x) = xf(1) \implies f(x) = cx\]for some $c \in \mathbb{R}$, so we are done.
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eg4334
636 posts
#91
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First, $f(0)=0$ from $P(0, 0)$. Also $y=0$ gives $f(x^2)=xf(x)$ so we have $f(x^2-y^2)=f(x^2)-f(y^2)$. This also tells us that $f$ is odd. In other words $f(a+b)=f(a)+f(b)$ for one of $a, b$ being positive. We can naturally extend this to the other sign cases. Now we consider $f((x+1)^2) = (x+1)(f(x)+f(1)) = f(x^2+2x+1)=xf(x)+2f(x)+f(1)$. This tells us $f(x)=xf(1)$, or $f(x) = kx$.
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ray66
48 posts
#92
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Let $P(x,y)$ be the assertion that the functional equation holds for $(x,y)$. Plugging $(0,0)$ gives $f(0)=0$ and plugging $P(x,-y)$ gives $f$ is odd. Plugging $P(x,0)$ gives $f(x^2)=xf(x)$ for all real $x$. So for nonnegative real $x$ $P(\sqrt{x},0)$ gives $f(x)=\sqrt{x} f(\sqrt{x}) = x^{\frac{3}{4}}f(x^{\frac{1}{4}}) = \ldots = xf(1)$, and we can extend this to all negatives because $f$ is odd. So $f(x)=cx$ for all real $x$. Plugging this into the original equation gives $c(x^2-y^2)=cx^2-cy^2$, so $f(x)=cx$ is the only solution.
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Marcus_Zhang
980 posts
#93
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Cool ig
This post has been edited 1 time. Last edited by Marcus_Zhang, Mar 14, 2025, 1:46 AM
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Maximilian113
575 posts
#94
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Let $P(x, y)$ denote the assertion. Then $P(x, 0) \implies f(x^2)=xf(x) \implies f(x)=-f(-x).$ Therefore $$P(x) \iff P(x-y)=P(x)-P(y)$$for all nonnegative $x, y.$ Therefore, for positive $x, y$ we have $$f(y)=f(x+y-x)=f(x+y)-f(x) \implies f(x+y)=f(x)+f(y).$$As $f(x)$ is odd, it follows that $f(x)$ is additive. Hence $$(x+1)(f(x)+f(1))=(x+1)f(x+1)=f((x+1)^2) = f(x^2)+2f(x)+f(1) \implies f(x)=xf(1).$$This solution clearly works.
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blueprimes
363 posts
#95
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We claim the answer is $f(x) \equiv cx$ for any constant $c$ which clearly works. Now we prove they are the only ones.

Note that the assertions $(x, 0)$ and $(-x, 0)$ for nonzero $x$ implies $f$ is odd. Now $y = 0$ gives $f(x^2) = x f(x)$ so in fact $f(x^2 - y^2) + f(y^2) = f(x^2)$. Imposing $x^2 > y^2$, replacing $x^2 - y^2 \mapsto x, y^2 \mapsto y$ yields $f(x) + f(y) = f(x + y)$ for all $x, y \ge 0$.

We use this fact to our advantage to decompose terms: Consider $(x, y) \mapsto (x + 1, x)$ for $x \ge 0$, we have
\begin{align*}
2 f(x) + f(1) &= f(2x + 1) \\
&= (x + 1) f(x + 1) - x f(x) \\
&= (x + 1) [f(x) + f(1)] - x f(x) \\
&= f(x) + x f(1) + f(1) \\
\end{align*}so $f(x) \equiv x f(1)$ for $x \ge 0$. But $f$ is odd, so $f(x) \equiv cx$ for some constant $x$ as needed.
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Ilikeminecraft
676 posts
#96
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I claim that $f(x) = ax$.

By taking $x = 0, y = 0$ as seperate cases, we see that $f(x^2) = xf(x), f(-y^2)= - yf(y).$ Thus, we also see that $f(x) = -f(-x).$

Thus, we can rewrite our equation as $f(x^2 - y^2) = f(x^2) - f(y^2),$ or $f(x^2 - y^2) + f(y^2) = f(x^2).$ For $x \geq y,$ we see that the domain is all positive, and thus, $(x + 1.434)^2 + (f(x))^2 > 1.$ Now we substitute $x = u + 1$ to get that $f(x) = xf(1). $ Thus, we are done.
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