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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)!

Be sure to mark your calendars for the following upcoming events:
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[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
diophantine with factorials and exponents
skellyrah   2
N an hour ago by maromex
find all positive integers $a,b,c$ such that $$ a! + 5^b = c^3 $$
2 replies
skellyrah
3 hours ago
maromex
an hour ago
A scalene triangle and nine point circle
ariopro1387   2
N an hour ago by Mysteriouxxx
Source: Iran Team selection test 2025 - P12
In a scalene triangle $ABC$, points $Y$ and $X$ lie on $AC$ and $BC$ respectively such that $BC \perp XY$. Points $Z$ and $T$ are the reflections of $X$ and $Y$ with respect to the midpoints of sides $BC$ and $AC$, respectively. Point $P$ lies on segment $ZT$ such that the circumcenter of triangle $XZP$ coincides with the circumcenter of triangle $ABC$.
Prove that the nine-point circle of triangle $ABC$ passes through the midpoint of segment $XP$.
2 replies
ariopro1387
May 27, 2025
Mysteriouxxx
an hour ago
My journey to IMO
MTA_2024   6
N an hour ago by Fly_into_the_sky
Note to moderators: I had no idea if this is the ideal forum for this or not, feel free to move it wherever you want ;)

Hi everyone,
I am a random 14 years old 9th grader, national olympiad winner, and silver medalist in the francophone olympiad of maths (junior section) Click here to see the test in itself.
While on paper, this might seem like a solid background (and tbh it kinda is); but I only have one problem rn: an extreme lack of preparation (You'll understand very soon just keep reading :D ).
You see, when the francophone olympiad, the national olympiad and the international kangaroo ended (and they where in the span of 4 days!!!) I've told myself :"aight, enough math, take a break till summer" (and btw, summer starts rh in July and ends in October) and from then I didn't seriously study maths.
That was until yesterday, (see, none of our senior's year students could go because the bachelor's degree exam and the IMO's dates coincide). So they replaced them with us, junior students. And suddenly, with no previous warning, I found myself at the very bottom of the IMO list of participants. And it's been months since I last "seriously" studied maths.
I'm really looking forward to this incredible journey, and potentially winning a medal :laugh: . But regardless of my results I know it'll be a fantastic journey with this very large and kind community.
Any advices or help is more than welcome <3 .Thank yall for helping me reach and surpass a ton of my goals.
Sincerely.
6 replies
MTA_2024
5 hours ago
Fly_into_the_sky
an hour ago
Infinite number of sets with an intersection property
Drytime   7
N 2 hours ago by HHGB
Source: Romania TST 2013 Test 2 Problem 4
Let $k$ be a positive integer larger than $1$. Build an infinite set $\mathcal{A}$ of subsets of $\mathbb{N}$ having the following properties:

(a) any $k$ distinct sets of $\mathcal{A}$ have exactly one common element;
(b) any $k+1$ distinct sets of $\mathcal{A}$ have void intersection.
7 replies
Drytime
Apr 26, 2013
HHGB
2 hours ago
m^m+ n^n=k^k
parmenides51   2
N 3 hours ago by Assassino9931
Source: 2021 Ukraine NMO 11.6
Are there natural numbers $(m,n,k)$ that satisfy the equation $m^m+ n^n=k^k$ ?
2 replies
parmenides51
Apr 4, 2021
Assassino9931
3 hours ago
Find the value
sqing   14
N 3 hours ago by Yiyj
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
14 replies
sqing
Jun 22, 2024
Yiyj
3 hours ago
A circle tangent to AB,AC with center J!
Noob_at_math_69_level   6
N 3 hours ago by awesomeming327.
Source: DGO 2023 Team P2
Let $\triangle{ABC}$ be a triangle with a circle $\Omega$ with center $J$ tangent to sides $AC,AB$ at $E,F$ respectively. Suppose the circle with diameter $AJ$ intersects the circumcircle of $\triangle{ABC}$ again at $T.$ $T'$ is the reflection of $T$ over $AJ$. Suppose points $X,Y$ lie on $\Omega$ such that $EX,FY$ are parallel to $BC$. Prove that: The intersection of $BX,CY$ lie on the circumcircle of $\triangle{BT'C}.$

Proposed by Dtong08math & many authors
6 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
3 hours ago
Easy functional equation
fattypiggy123   15
N 5 hours ago by ariopro1387
Source: Singapore Mathematical Olympiad 2014 Problem 2
Find all functions from the reals to the reals satisfying
\[f(xf(y) + x) = xy + f(x)\]
15 replies
fattypiggy123
Jul 5, 2014
ariopro1387
5 hours ago
Iran TST Starter
M11100111001Y1R   5
N 5 hours ago by DeathIsAwe
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[
a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i}
\]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[
a_n < \frac{1}{1404}
\]
5 replies
M11100111001Y1R
May 27, 2025
DeathIsAwe
5 hours ago
Very odd geo
Royal_mhyasd   1
N 5 hours ago by Royal_mhyasd
Source: own (i think)
Let $\triangle ABC$ be an acute triangle with $AC>AB>BC$ and let $H$ be its orthocenter. Let $P$ be a point on the perpendicular bisector of $AH$ such that $\angle APH=2(\angle ABC - \angle ACB)$ and $P$ and $C$ are on different sides of $AB$, $Q$ a point on the perpendicular bisector of $BH$ such that $\angle BQH = 2(\angle ACB-\angle BAC)$ and $R$ a point on the perpendicular bisector of $CH$ such that $\angle CRH=2(\angle ABC - \angle BAC)$ and $Q,R$ lie on the opposite side of $BC$ w.r.t $A$. Prove that $P,Q$ and $R$ are collinear.
1 reply
Royal_mhyasd
5 hours ago
Royal_mhyasd
5 hours ago
Calculating sum of the numbers
Sadigly   5
N 5 hours ago by aokmh3n2i2rt
Source: Azerbaijan Junior MO 2025 P4
A $3\times3$ square is filled with numbers $1;2;3...;9$.The numbers inside four $2\times2$ squares is summed,and arranged in an increasing order. Is it possible to obtain the following sequences as a result of this operation?

$\text{a)}$ $24,24,25,25$

$\text{b)}$ $20,23,26,29$
5 replies
Sadigly
May 9, 2025
aokmh3n2i2rt
5 hours ago
Swap to the symmedian
Noob_at_math_69_level   7
N 5 hours ago by awesomeming327.
Source: DGO 2023 Team P1
Let $\triangle{ABC}$ be a triangle with points $U,V$ lie on the perpendicular bisector of $BC$ such that $B,U,V,C$ lie on a circle. Suppose $UD,UE,UF$ are perpendicular to sides $BC,AC,AB$ at points $D,E,F.$ The tangent lines from points $E,F$ to the circumcircle of $\triangle{DEF}$ intersects at point $S.$ Prove that: $AV,DS$ are parallel.

Proposed by Paramizo Dicrominique
7 replies
Noob_at_math_69_level
Dec 18, 2023
awesomeming327.
5 hours ago
Find (AB * CD) / (AC * BD) & prove orthogonality of circles
Maverick   15
N 6 hours ago by Ilikeminecraft
Source: IMO 1993, Day 1, Problem 2
Let $A$, $B$, $C$, $D$ be four points in the plane, with $C$ and $D$ on the same side of the line $AB$, such that $AC \cdot BD = AD \cdot BC$ and $\angle ADB = 90^{\circ}+\angle ACB$. Find the ratio
\[\frac{AB \cdot CD}{AC \cdot BD}, \]
and prove that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal. (Intersecting circles are said to be orthogonal if at either common point their tangents are perpendicuar. Thus, proving that the circumcircles of the triangles $ACD$ and $BCD$ are orthogonal is equivalent to proving that the tangents to the circumcircles of the triangles $ACD$ and $BCD$ at the point $C$ are perpendicular.)
15 replies
Maverick
Jul 13, 2004
Ilikeminecraft
6 hours ago
f(x+f(x)+f(y))=x+f(x+y)
dangerousliri   10
N Today at 4:54 PM by jasperE3
Source: FEOO, Shortlist A5
Find all functions $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ such that for any positive real numbers $x$ and $y$,
$$f(x+f(x)+f(y))=x+f(x+y)$$Proposed by Athanasios Kontogeorgis, Grecce, and Dorlir Ahmeti, Kosovo
10 replies
dangerousliri
May 31, 2020
jasperE3
Today at 4:54 PM
IMO ShortList 2002, algebra problem 4
orl   62
N Apr 10, 2025 by Ihatecombin
Source: IMO ShortList 2002, algebra problem 4
Find all functions $f$ from the reals to the reals such that \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz)  \] for all real $x,y,z,t$.
62 replies
orl
Sep 28, 2004
Ihatecombin
Apr 10, 2025
IMO ShortList 2002, algebra problem 4
G H J
Source: IMO ShortList 2002, algebra problem 4
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orl
3647 posts
#1 • 12 Y
Y by amatysten, Davi-8191, tenplusten, e_plus_pi, mathematicsy, Adventure10, megarnie, ImSh95, Stuart111, Mango247, Rounak_iitr, cubres
Find all functions $f$ from the reals to the reals such that \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz)  \] for all real $x,y,z,t$.
This post has been edited 4 times. Last edited by orl, Sep 27, 2005, 4:57 PM
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inom
118 posts
#2 • 5 Y
Y by Tawan, Adventure10, ImSh95, Mango247, cubres
Set $x=y=z=t=0$ then we will get $f(0)(2f(0)-1)=0$.
So we will have two cases 1) $f(0)=o$ and 2)$f(0)=\frac{1}{2}$
1) $f(0)=0$
Set $z=t=0$ then we will have $f(x)f(y)=f(xy)$ $(*)$
we know that $(*)$ is Cauchy's equation and its solutions are:
(a) $f(x)=|x|^c$ (where $c$ is constant an we will look for it),
(b) $f(x)=sgnx\times x$ (c) $f(x)=0$
in (a) put instead of each function which is given we will yield:
$|xy|^c+|xt|^c+|zy|^c+|zt|^c=|xy-zt|^c+|xt+yz|^c$
now denote $|xy|=a, |xt|=b, |zy|=d$ and $|zt|=f$
$a^c+b^c+d^c+f^c=(a-f)^c+(d+b)^c$, so we have new function as example $f(a)=a^c$.
Set $b=d=f=a$ then we will have $4a^c=(2a)^c$
take $c$ root from both sides, and have $4^{\frac{1}{c}} a =2a$
cancel $a$ an have $4^{\frac{1}{c}} =2$ from here $c=2$ only solution, so $f(x)=x^2$
(b) cannot be solution because it will give $f(x)=|x|$
(c) is obviously solution
Now we will look in case 2) $f(0)=\frac{1}{2}$
Set $z=x$ and $y=t=0$ then $2f(x)=1\rightarrow f(x)=\frac{1}{2}$
So the solutions are:
$f(x)=x^2$, $f(x)=0$, and $f(x)=\frac{1}{2}$, for all $x$ element of real numbers.
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Albanian Eagle
1693 posts
#3 • 7 Y
Y by vsathiam, Adventure10, ImSh95, Vladimir_Djurica, Mango247, cubres, and 1 other user
cauchy equation is valid only for rationals.
we need an extra condition in this case (we can show that f is monotone) ;)
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wojto111
42 posts
#4 • 4 Y
Y by Adventure10, ImSh95, Mango247, cubres
can you show how to prove that $f(x)$ is monotone. I think that this is only dificulty of this problem.
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olorin
588 posts
#5 • 7 Y
Y by Tawan, xdiegolazarox, Adventure10, ImSh95, Mango247, X.Luser, cubres
$x=z,y=t=0 \Rightarrow 2f(x)\cdot 2f(0)=2f(0)$ for all $x\in\mathbb{R}$ $\Rightarrow f(0)=0$ or $f\equiv{1\over 2}$.
Assume $f(0)=0$.

$z=t=0 \Rightarrow f(xy)=f(x)f(y)$ for all $x,y\in\mathbb{R}$ (*)
$x=y=0 \Rightarrow f(-zt)=f(z)f(t)=f(zt)$ for all $z,t\in\mathbb{R}$
So $f(-x)=f(x)$ for all $x\in\mathbb{R}$ (**)
and $f(x)=f(\sqrt{x})^{2}\geq 0$ for all $x\geq 0$ (***)
and $f(x)=f(x\cdot 1)=f(x)f(1)$ for all $x\in\mathbb{R}$ $\Rightarrow f(1)=1$ or $f\equiv 0$.
Assume $f(1)=1$.

$x=y=z=t=1\Rightarrow f(2)=4$
Then (*) gives $f(2^{r})=4^{r}=(2^{r})^{2}$ for all $r\in\mathbb{Q}$. (****)

And now
$x=t,y=z \Rightarrow (f(x)+f(y))^{2}=f(x^{2}+y^{2})=f(\sqrt{x^{2}+y^{2}})^{2}\\ \hspace*{0.9in}\Rightarrow f(y)\leq f(x)+f(y)=f(\sqrt{x^{2}+y^{2}})$
for all $x,y\geq 0$ using (***).
So $f(x)=f(\sqrt{(\sqrt{x^{2}-y^{2}})^{2}+y^{2}})\geq f(y)$ for all $x\geq y\geq 0$.
This and (****) give $f(x)=x^{2}$ for all $x>0$
and (**) and $f(0)=0$ give
$f(x)=x^{2}$ for all $x\in\mathbb{R}$.
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nealth
303 posts
#6 • 4 Y
Y by ImSh95, Adventure10, Mango247, cubres
I'll skip $ f(0) = 0$.
Set $ x = y, z = t$, we have $ (f(x) + f(z))^2 = f(x^2 - z^2) + f(2xz)$.
Set $ x = t, z = y$, we have $ (f(x) + f(z))^2 = f(x^2 + z^2)$.

Therefore $ f(x^2 + z^2) - f(x^2 - z^2) = f(2xz)$.
\[ \lim_{z\rightarrow 0}\dfrac{f(x^2 + z^2) - f(x^2 - z^2)}{2z^2} = \lim_{z\rightarrow 0}\dfrac{f(2xz)}{2z^2}
\]
Let $ h(x) = \frac {f(x)}{x^2}$, then $ f'(x^2) = \lim_{z\rightarrow 0}h(2xz)\cdot 2x^2$. So $ \lim_{r\rightarrow 0}h(r) = \dfrac{f'(x^2)}{2x^2}$. This means $ \dfrac{f'(x^2)}{2x^2}$ is constant for all $ x$, therefore $ f(x)$ is constant or $ f(x) = kx^2$, and we can easily check that $ k = 1$.
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MellowMelon
5850 posts
#7 • 6 Y
Y by Jc426, ImSh95, Adventure10, Mango247, cubres, and 1 other user
Where did you show the function was differentiable? :(
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nealth
303 posts
#8 • 4 Y
Y by ImSh95, Adventure10, Mango247, cubres
MellowMelon wrote:
Where did you show the function was differentiable? :(

I just found out the same problem. But it can be fixed easily:

$ \lim_{z\rightarrow 0}\frac{f(x^{2}+z^{2})-f(x^{2}-z^{2})}{2z^{2}}=\lim_{z\rightarrow 0}h(2xz)\cdot 2x^{2}$

Since the limit exist (we are fixing $ x$ and let $ z$ go to $ 0$), the funciton must be differentiable. Is that right?
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MellowMelon
5850 posts
#9 • 6 Y
Y by Jc426, ImSh95, Adventure10, Mango247, cubres, and 1 other user
That requires continuity at $ 0$, which you also have the burden of showing. (trying to use derivatives is almost certainly going to fail, by the way)
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cnyd
394 posts
#10 • 6 Y
Y by Tawan, vsathiam, ImSh95, Adventure10, Mango247, cubres
here is my solution;

$ (f(x) + f(z))(f(y) + f(t)) = f(xy - zt) + f(xt + yz)$

if $ x = y = z = t = 0$ $ \implies$ $ 2f(0)2f(0) = 2f(0)$

$ \implies$ $ f(0) = \frac {1}{2}$ or $ f(0) = 0$

$ i)f(0) = \frac {1}{2}$

$ x = z = 0$ $ \implies$ $ 1[f(y) + f(t)] = 1$ $ \implies$ $ f(y) + f(t) = 1$ $ \implies$ $ f(x) = \frac {1}{2}$

$ ii)f(0) = 0$ $ \implies$

$ z = t = 0$ $ \implies$ $ f(x)f(y) = f(xy)$

$ \implies$ $ f(1)(f(1) - 1) = 0$ $ \implies$ $ f(1) = 0$ or $ f(1) = 1$

if $ f(1) = 0$,$ y = t = 1$ $ \implies$ $ f(x - z) + f(x + z) = 0$

if $ x = z$ $ \implies$ $ f(x) = 0$

$ f(1) = 1$ $ \implies$ $ f(x + z) + f(x - z) = 2f(x) + 2f(z)$

$ \implies$ $ f(2x) = 4f(x) = f(2)f(x)$ $ \implies$ $ f(2) = 4$

$ \implies$ $ 2f(a) + 2f(b) = f(a - b) + f(a + b)$

$ f(1) = 1,f(2) = 4$ ,Let $ f(n) = n^{2}$

$ a = nb$ $ \implies$ $ 2f(nb) + 2f(b) = f((n - 1)b) + f((n + 1)b)$

$ \implies$ $ (2n^{2} + 2)f(b) = [(n - 1)^{2} + f(n + 1)]f(b)$

$ \implies$ $ f(n + 1) = (n + 1)^{2}$ $ \implies$ $ f(x) = x^{2}$ $ \forall x\in\mathbb{N}$

$ 2f(a) + 2f(b) = f(a - b) + f(a + b)$ ,if $ a = 0,b = 1$ $ \implies$ $ f(1) = f( - 1) = 1$

$ \implies$ $ f(x) = f( - x)$ $ \implies$ $ f(x) = x^{2}$ $ \forall x\in\mathbb{Z}$

$ f(xy) = f(x)f(y)$ if $ y = \frac {1}{x}(x\in\mathbb{Z})$ $ \implies$ $ f(\frac {1}{x}) = \frac {1}{x^{2}}$

$ \implies$ $ f(\frac {p}{q}) = \frac {p^{2}}{q^{2}}$ $ \implies$ $ f(x) = x^{2}$ $ \forall x\in\mathbb{Q}$

We can analyze this function $ x > 0$

$ f(xy) = f(x)f(y)$, Let $ y = \frac {p}{q}$,$ p > q$

$ \implies$ $ f(x\frac {p}{q}) = f(x)\frac {p^{2}}{q^{2}}$ $ \implies$ $ f(x)$ is increasing function for $ x > 0$

Let $ \exists u\in\mathbb{R^{ + }}$,$ f(u) > u^{2}$ $ \implies$

$ u^{2} < r^{2} = f(r) < f(u),(r\in\mathbb{Q})$ $ \implies$ $ f(u) > f(r),r > u$ Contradiction!

Let $ \exists u\in\mathbb{R^{ + }}$,$ f(u) < u^{2}$ $ \implies$

$ f(u) < f(r) = r^{2} < u^{2}$ $ \implies$ $ f(u) < f(r),u > r$ Contradiction

$ \implies$ $ f(x) = x^{2}$ $ \forall\mathbb{R^{ + }}$

$ f(x) = f( - x)$ $ \implies$ $ f(x) = x^{2}$

$ \implies$ $ \boxed{f(x) = 0},\boxed{f(x) = \frac {1}{2}},\boxed{f(x) = x^{2}}$
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cnyd
394 posts
#11 • 4 Y
Y by ImSh95, Adventure10, Mango247, cubres
my solution is wrongly for increasing

$ x=t,y=z$ $ \implies$

$ [f(a) + f(c)]^{2} = f(a^{2} + c^{2})$

$ \implies$ $ x > 0$ $ \implies$ $ f(x)\geq 0$

$ \implies$ $ f(a^{2} + c^{2})\geq f(a^{2})$ $ \implies$ $ f$ is increasing for $ x > 0$

the rest is easy
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JuanOrtiz
366 posts
#12 • 6 Y
Y by Tawan, Anar24, ImSh95, Adventure10, Mango247, cubres
By setting $x=0=z$ we see that for all $y,t$ we must have $2f(0)(f(y)+f(t))=2f(0)$. Therefore $f=1/2$ or $f(0)=0$. (Note that $f=1/2$ works, so WLOG $f(0)=0$).

By setting $z=t=0$ we see that $f(x)f(y)=f(xy)$ for any $xy$. In particular we get that $f(p) \ge 0$ for any positive $p$, by setting $x=y=\sqrt{p}$.

By setting $x=\frac{zt}{y}$ for any $z,t,y$ and by using multiplicity we get that

$f(\frac{zt^2}{y}+yz)=(f(x)+f(z))(f(y)+f(t))=f(xy)+f(xt)+f(zy)+f(zt)=2f(zt)+f(\frac{zt^2}{y})+f(zy)$

and by multiplting by $f(y)/f(z)$ we get that

$f(t^2+y^2)=(f(t)+f(y))^2$

by using multiplicity. Let $g=\sqrt{f}$ be defined on $\mathbb{R}^{+}$. Then for all $y,t$ positive, we get that

$g(t^2+y^2)=g(t)^2+g(y)^2=g(t^2)+g(y^2)$

and so $g$ is additive. Since it's defined on the positives we get that $g(x)=kx$ for a constant $x$, and also $k$ must be positive (or $0$). So we have that $f(x)=Cx^2$ for positive $x$. Since $f$ is multiplicative we get that $C=1$ or $C=0$, since $C \ge 0$. If $C=0$ then because of multiplicity, $f=0$. Otherwise, $f(x)=x^2$ for positive $x$.

Finally, in the original equation we have, by having $x,y,t$ very big positives and $z$ a small negative number such that $xt+yz > 0$, that

$(x^2+f(z))(y^2+t^2) = (xy-zt)^2 + (xt+yz)^2=(xy)^2+(xt)^2+z^2(y^2+t^2)$

and so $f(z)(y^2+t^2)=z^2(t^2+y^2)$ and so $f(z)=z^2$.

So we have $f=1/2$; $f=0$ or $f(x)=x^2$. Done.
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fclvbfm934
759 posts
#13 • 5 Y
Y by Tawan, ImSh95, Adventure10, Mango247, cubres
The solutions are $f(x) = x^2$, $f(x) = 0$, and $f(x) = \frac{1}{2}$.

Set all variables to $0$ and we get $f(0) = 0$ or $f(0) = \frac{1}{2}$. If we have that $f(0) = \frac{1}{2}$, letting $x = z = 0$ and $t = y$ gives us $2f(y) = 1$, so we see that $\boxed{f(y) = \frac{1}{2}}$ for all $y \in \mathbb{R}$.

So now assume that $f(0) = 0$. Plugging in $y = t = 1, z = 0$, we have
$$2f(1)f(x) = 2f(x).$$If $\boxed{f(x) = 0}$ for all real $x$, we have another solution. Otherwise, assume there exists an $x$ such that $f(x) \neq 0$, so we get $f(1) = 1$.

Now letting $y = t = z = 1$, we get
$$2f(x) + 2 = f(x+1) + f(x-1).$$This is a recurrence relation, and from induction, we can easily show that $f(n) = n^2$ for all $n \in \mathbb{Z}$.

Now we prove multiplicity: set $z = t = 0$ and we get
$$f(x)\cdot f(y) = f(xy).$$Setting $x = q$ and $y = \frac{p}{q}$ for integers $p$ and $q$ gives us
$$f\left( \frac{p}{q} \right) = \frac{f(p)}{f(q)} = \frac{p^2}{q^2},$$so $f(x) = x^2$ for all $x \in \mathbb{Q}$.

To show that $f$ is even: let $x = z$ and $y = 0$ in the original FE, which gives us
$$2f(z)\cdot f(t) = 2f(zt) = f(-zt) + f(zt) \qquad \Rightarrow \qquad f(zt) = f(-zt)$$
Now we will show that $f$ is strictly increasing on the interval $[0, \infty)$. Let $x = y$ and $z = -t$. Then:
\[f(x^2 + t^2)=(f(x) + f(t))^2 = f(x^2) + 2f(xt) + f(t)^2 \ge f(x^2)\]if $x \ge 0$ and $t > 0$. [We have $f(xt) \ge 0$, because $f(xt) = f(\sqrt{xt})f(\sqrt{xt}) \ge 0$]. So now we see that $f$ is strictly increasing.

Because the rationals are dense within the reals, $\boxed{f(x) = x^2}$ for all $x \ge 0$, and because $f$ is even, we're done. So in conclusion, our answers are: $f(x) = x^2$, $f(x) = 0$, and $f(x) = \frac{1}{2}$.

Footnote: In case the density argument wasn't entirely clear, suppose for some positive irrational number $x$, we have $f(x) = x^2 + \varepsilon$ for some $\varepsilon > 0$ (the proof for $\varepsilon < 0$ is analogous). Then, take a rational number $x < r < \sqrt{x^2 + \varepsilon}$, which is possible since between any two real numbers there is a rational number. Since we had shown $f(r) = r^2$, and we had shown $f$ is monotone, we must have
$$f(r) \ge f(x) \qquad \Rightarrow \qquad r^2 \ge x^2 + \varepsilon$$which is a contradiction since $r < \sqrt{x^2 + \varepsilon}$.
This post has been edited 1 time. Last edited by fclvbfm934, Sep 6, 2022, 1:33 AM
Reason: Fixing formatting
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Konigsberg
2238 posts
#14 • 4 Y
Y by ImSh95, Adventure10, Mango247, cubres
Is Cauchy's equation's solution for monotone functions citable?
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mcdonalds106_7
1138 posts
#17 • 3 Y
Y by ImSh95, Adventure10, cubres
Let $P(x,y,z,t)$ be the given assertion. $P(0,0,0,0)$ implies that $f(0)=0$ or $f(0)=\dfrac 12$. If $f(0)=\dfrac 12$, then $P(0,0,0,t)$ gives us the solution $f(t)=\dfrac 12$.

Now, we can assume that $f(0)=0$. $P(x,y,0,0)$ gives that $f(x)f(y)=f(xy)$, and $P(0,0,z,t)$ gives that $f(z)f(t)=f(-zt)$, so $f$ is multiplicative and even. Let $Q$ be the assertion that $f(x)f(y)=f(xy)$. Then, $Q(1,1)$ gives that $f(1)=0$ or $f(1)=1$. If $f(1)=0$, then $Q(1,y)$ gives the solution $f(y)=0$. Otherwise, assume $f(1)=1$. Then $Q\left(x,\dfrac 1x\right)$ gives that $f(x)\neq 0$ for all $x\neq 0$.

Now, $P(x,1,1,1)$ gives that $2=f(x+1)-2f(x)+2f(x-1)$, so using the base cases $f(0)=0$ and $f(1)=1$, we can get that $f(n)=n^2$ for all integers $n$. Then, $Q\left(b, \dfrac ab\right)$ gives that $f(q)=q^2$ for all rationals $q$.

Now, $Q(x,x)$ gives that $f(x^2)=f(x)^2>0$ for all $x\neq 0$. Furthermore, for any positive reals $x<w$, $P\left(x,\sqrt{w^2-x^2},\sqrt{w^2-x^2},x\right)$ gives that $\left[f(x)+f\left(\sqrt{w^2-x^2}\right)\right]^2=f(w)^2$, which implies that $f$ is strictly increasing over the positive reals.

But since $f(q)=q^2$ over the positive rationals and $f$ is increasing over the positive reals, we can use a Cauchy-esque argument to show that $f(x)=x^2$ for all positive reals, and hence over all reals. Hence, the solutions are $f(x)=0, f(x)=\dfrac 12, f(x)=x^2$.
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