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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.

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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
Geometry
gggzul   0
31 minutes ago
In trapezoid $ABCD$ segments $AB$ and $CD$ are parallel. Angle bisectors of $\angle A$ and $\angle C$ meet at $P$. Angle bisectors of $\angle B$ and $\angle D$ meet at $Q$. Prove that $ABPQ$ is cyclic
0 replies
gggzul
31 minutes ago
0 replies
hard problem
Cobedangiu   5
N 39 minutes ago by KhuongTrang
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
5 replies
Cobedangiu
Yesterday at 4:24 PM
KhuongTrang
39 minutes ago
Nordic 2025 P3
anirbanbz   9
N 42 minutes ago by Tsikaloudakis
Source: Nordic 2025
Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. Let $E$ and $F$ be points on the line segments $AC$ and $AB$ respectively such that $AEHF$ is a parallelogram. Prove that $\vert OE \vert = \vert OF \vert$.
9 replies
anirbanbz
Mar 25, 2025
Tsikaloudakis
42 minutes ago
Aime type Geo
ehuseyinyigit   1
N an hour ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
an hour ago
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N an hour ago by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
an hour ago
Which numbers are almost prime?
AshAuktober   5
N 2 hours ago by Jupiterballs
Source: 2024 Swiss MO/1
If $a$ and $b$ are positive integers, we say that $a$ almost divides $b$ if $a$ divides at least one of $b - 1$ and $b + 1$. We call a positive integer $n$ almost prime if the following holds: for any positive integers $a, b$ such that $n$ almost divides $ab$, we have that $n$ almost divides at least one of $a$ and $b$. Determine all almost prime numbers.
original link
5 replies
AshAuktober
Dec 16, 2024
Jupiterballs
2 hours ago
Inequality involving square root cube root and 8th root
bamboozled   1
N 2 hours ago by arqady
If $a,b,c,d,e,f,g,h,k\in R^+$ and $a+b+c=d+e+f=g+h+k=8$, then find the minimum value of $\sqrt{ad^3 g^4} +\sqrt[3]{be^3 h^4} + \sqrt[8]{cf^3 k^4}$
1 reply
bamboozled
4 hours ago
arqady
2 hours ago
If $b^n|a^n-1$ then $a^b >\frac {3^n}{n}$ (China TST 2009)
Fang-jh   16
N 2 hours ago by Aiden-1089
Source: Chinese TST 2009 6th P1
Let $ a > b > 1, b$ is an odd number, let $ n$ be a positive integer. If $ b^n|a^n-1,$ then $ a^b > \frac {3^n}{n}.$
16 replies
Fang-jh
Apr 4, 2009
Aiden-1089
2 hours ago
Confusing inequality
giangtruong13   1
N 2 hours ago by Natrium
Source: An user
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum: $$P= \sum_{cyc} \frac{a}{b} + \sum_{cyc} \frac{1}{a^3+b^3+abc}$$
1 reply
giangtruong13
Yesterday at 8:04 AM
Natrium
2 hours ago
Two equal angles
jayme   3
N 2 hours ago by jayme
Dear Mathlinkers,

1. ABCD a square
2. I the midpoint of AB
3. 1 the circle center at A passing through B
4. Q the point of intersection of 1 with the segment IC
5. X the foot of the perpendicular to BC from Q
6. Y the point of intersection of 1 with the segment AX
7. M the point of intersection of CY and AB.

Prove : <ACI = <IYM.

Sincerely
Jean-Louis
3 replies
jayme
May 2, 2025
jayme
2 hours ago
50 points in plane
pohoatza   13
N 2 hours ago by cursed_tangent1434
Source: JBMO 2007, Bulgaria, problem 3
Given are $50$ points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least $130$ scalene triangles with vertices of that color.
13 replies
pohoatza
Jun 28, 2007
cursed_tangent1434
2 hours ago
D1024 : Can you do that?
Dattier   5
N 2 hours ago by SimplisticFormulas
Source: les dattes à Dattier
Let $x_{n+1}=x_n^2+1$ and $x_0=1$.

Can you calculate $\sum\limits_{i=1}^{2^{2025}} x_i \mod 10^{30}$?
5 replies
Dattier
Apr 29, 2025
SimplisticFormulas
2 hours ago
Hardest N7 in history
OronSH   25
N 2 hours ago by sansgankrsngupta
Source: ISL 2023 N7
Let $a,b,c,d$ be positive integers satisfying \[\frac{ab}{a+b}+\frac{cd}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d}.\]Determine all possible values of $a+b+c+d$.
25 replies
OronSH
Jul 17, 2024
sansgankrsngupta
2 hours ago
Friends Status are changing
lminsl   64
N 3 hours ago by SteppenWolfMath
Source: IMO 2019 Problem 3
A social network has $2019$ users, some pairs of whom are friends. Whenever user $A$ is friends with user $B$, user $B$ is also friends with user $A$. Events of the following kind may happen repeatedly, one at a time:
[list]
[*] Three users $A$, $B$, and $C$ such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends, change their friendship statuses such that $B$ and $C$ are now friends, but $A$ is no longer friends with $B$, and no longer friends with $C$. All other friendship statuses are unchanged.
[/list]
Initially, $1010$ users have $1009$ friends each, and $1009$ users have $1010$ friends each. Prove that there exists a sequence of such events after which each user is friends with at most one other user.

Proposed by Adrian Beker, Croatia
64 replies
lminsl
Jul 16, 2019
SteppenWolfMath
3 hours ago
FE over R
IAmTheHazard   19
N Apr 1, 2025 by Bardia7003
Source: ELMO Shortlist 2024/A3
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu
19 replies
IAmTheHazard
Jun 22, 2024
Bardia7003
Apr 1, 2025
FE over R
G H J
G H BBookmark kLocked kLocked NReply
Source: ELMO Shortlist 2024/A3
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IAmTheHazard
5001 posts
#1 • 3 Y
Y by ItsBesi, ihatemath123, pho1234
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu
This post has been edited 1 time. Last edited by IAmTheHazard, Jun 22, 2024, 3:41 PM
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MarkBcc168
1595 posts
#2
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Solution
This post has been edited 1 time. Last edited by MarkBcc168, Jun 22, 2024, 3:42 PM
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VicKmath7
1389 posts
#3
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Solution
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fractals
3028 posts
#4
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Let $P(x,y)$ denote the relation $f(x+f(y))+xy=f(x)f(y)+f(x)+y$.
Solution
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CyclicISLscelesTrapezoid
372 posts
#5 • 1 Y
Y by pho1234
The solutions are $f(x) \equiv x$ and $f(x) \equiv -x$, which work.

Let $P(x,y)$ denote the assertion that
\[f(x+f(y))+xy=f(x)f(y)+f(x)+y.\]$P(0,x)$ gives $f(f(x))=f(0)f(x)+f(0)+x$. Notice that $P(f(x),y)$ gives
\[f(f(x)+f(y))+yf(x)=f(f(x))f(y)+f(f(x))+y,\]and subtracting this by its symmetric variant with $x$ and $y$ swapped gives
\[yf(x)-xf(y)=f(f(x))f(y)+f(f(x))+y-f(f(y))f(x)-f(f(y))-x.\]Using $P(0,x)$ to rewrite all the nested $f$'s, the RHS simplifies to $xf(y)-yf(x)$, so we have $xf(y)=yf(x)$. Plugging in $y=1$, we get $f(x)=xf(1)$. If $f(1)=c$, then the functional equation simplifies to $c(x+cy)+xy=c^2xy+cx+y$, which is only true for $c=1,-1$, as desired. $\square$
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KevinYang2.71
421 posts
#6
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Solved with Orthogonal..

We claim the only functions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$. It is easy to check that these work.

Let $P(x,y)$ denote the given assertion. $P(0,x)$ gives $f(f(x))=f(0)f(x)+f(0)+x$. Comparing $P(f(x),1)$ and $P(f(1),x)$ gives $f(x)\equiv f(1)x$. We can show that $f(1)=\pm 1$, as desired. $\square$
This post has been edited 2 times. Last edited by KevinYang2.71, Jun 23, 2024, 2:26 AM
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megarnie
5606 posts
#7 • 1 Y
Y by KevinYang2.71
The only solutions are $f(x) = x$ and $f(x) = -x$, which work. Let $P(x,y)$ denote the given assertion.

Case 1: $f(0) = 0$
Then $P(0,x): f(f(x)) = x$, so $f$ is an involution.

$P(1, f(x)): f(x + 1)= x f(1) + f(1)$, so $f$ is linear, meaning $f(x) = cx$ for some constant $c$. Since $f$ is an involution, either $c = 1$ or $c = -1$.

Case 2: $f(0) \ne 0$.
Claim: $f$ is injective.
Proof: If $f(a)= f(b)$, then $P(0,a)$ compared with $P(0,b)$ gives $a = b$. $\square$

Claim: $x + f(x)$ is injective.
Proof: Suppose $a + f(a) = b + f(b)$. Then $P(x,x): f(x + f(x)) - (x + f(x)) = (f(x) - x)(x + f(x))$, so comparing $x = a$ and $x = b$ gives that $(a + f(a)) (f(a) - a)) = (a + f(a)) (f(b) - b)$. If $a + f(a) = 0$, then $P(a,a): f(0) = 0$, absurd. Hence $a + f(a) \ne 0$, so $f(a) - a = f(b)  - b$. Hence \[(f(a) - a) + (f(a) + a) = (f(b) - b) + (f(b) + b) \implies f(a) = f(b) \implies a = b\]$\square$

Now, the equation gives \[ f(x + f(y)) - f(x) - y = f(x) f(y) - xy,\]so swapping $x,y$ here gives $f(x + f(y)) - f(x) - y = f(y + f(x)) - f(y) - x$, so $f(x + f(y))  + (x + f(y)) = f(y + f(x)) + (y + f(x))$, which by our earlier claim implies $x + f(y) = y + f(x)\implies f(x) - x = f(y) - y$, so $f(x) = x + c$ for some constant $c$.

$P(0,0): f(f(0)) = f(0)^2 + f(0)$, so $2c = c^2 + c\implies c\in \{0,1\}$. Checking, we see that $c = 1$ fails, so $c = 0$, but this is absurd since $f(0) \ne 0$.
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P2nisic
406 posts
#8
Y by
IAmTheHazard wrote:
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu

$f(x)f(y)f(z)=f(x+f(y))f(z)+xyf(z)-f(z)f(x)-yf(z)=f(x+f(y)+f(z))+(x+f(y))f(z)-f(x+f(y))-z+xyf(z)-f(x+f(z))-xz+f(x)+z-yf(z)$

Now by the symmetry of $y,z$ we get that:

$f(x+f(y)+f(z))+(x+f(y))f(z)-f(x+f(y))-z+xyf(z)-f(x+f(z))-xz+f(x)+z-yf(z)=$
$f(x+f(z)+f(y))+(x+f(z))f(y)-f(x+f(z))-y+xzf(y)-f(x+f(y))-xy+f(x)+y-zf(y)$
$\Rightarrow xf(z)+xyf(z)-xz-yf(z)=xf(y)+xzf(y)-xy-zf(y)$

In this for $x=0$ we get that:
$-yf(z)=-zf(y)\Rightarrow \frac{f(z)}{z}=\frac{f(y)}{y}\Rightarrow f(x)=cx$

Now in the start we have that:
$cx+c^2y+xy=c^2xy+cx+y\Rightarrow c^2y=y\Rightarrow c=+-1$
So we get that: $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$.



In the same way we can solve the problem if $f : \mathbb{R^+}\to\mathbb{R^+}$
This post has been edited 1 time. Last edited by P2nisic, Jun 23, 2024, 11:20 AM
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X.Allaberdiyev
103 posts
#9
Y by
Cute one.
$P(0,x)$ -> $f(f(x))=f(x)f(0)+f(0)+x$ $(1)$
By looking at $P(f(x),y)$ and $P(f(y),x)$ we have $f(f(x)+f(y))=f(f(x))f(y)+f(f(x))+y-f(x)y=f(f(y))f(x)+f(f(y))+x-f(y)x$, and by using $(1)$ we observe that $xf(y)=yf(x)$, which means that $f(x)=cx$. And by plugging it into equation one can prove that only solutions are $\boxed{f(x)\equiv x}$ and $\boxed{f(x)\equiv -x}$.
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ItsBesi
146 posts
#13
Y by
I am not sure about the surjectivety part
$\textbf{Answer:}$ $f(x)=\pm x \forall x \in \mathbb{R}$

$\textbf{Solution:}$

Let $P(x,y)$-denote the given assertion.

$\textbf{Claim:}$ $f$-bijective

$\textbf{Proof:}$

$P(0,x) \implies f(f(x))=f(x)f(0)+f(0)+x \implies f$-is surjective

f-injective

Since $f$ is both injective and surjective we get that $f$-is bijective. $\square$


$\textbf{Claim:}$ $f(0)=0$
$\textbf{Proof:}$

Since $f$-is surjective there exists an $\alpha$ such that $f(\alpha)=0 \iff \exists \alpha \in \mathbb{R} : f(\alpha)=0$

$P(\alpha,0) \implies f(\alpha+f(0))=0=f(\alpha) \implies f(\alpha+f(0))=f(\alpha) \stackrel{f-injective}{\implies} \alpha+f(0)=\alpha \implies f(0)=0.$ $\square$

$\textbf{Claim:}$ $f(x)= \pm x \forall \in \mathbb{R}$

$\textbf{Proof:}$
$P(0,x) \implies f(f(x))=x$ $...(*)$

$P(x,x) \implies f(x+f(x))+x^2=f(x)^2+f(x)+x \implies f(x+f(x))-x-f(x)=f(x)^2-x^2$
$...(3)$

$P(f(x),f(x)) \stackrel{(*)}{\implies} f(x+f(x))+f(x)^2=x^2+x+f(x) \implies f(x+f(x))-x-f(x)=x^2-f(x)^2$ $...(4)$

$(3)-(4) \implies f(x)^2-x^2=x^2-f(x)^2 \implies f(x)^2=x^2 \implies f(x)= \pm x \forall \in \mathbb{R}$ $\blacksquare$

$\textbf{POINTWISE TRAP!}$
This post has been edited 1 time. Last edited by ItsBesi, Jul 3, 2024, 5:37 PM
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omar1tun
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#14
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This solution might be one of the lamest ever :
$P(x,y):f(x+f(y))+xy=f(x)f(y)+f(x)+y$
We will just simplify $P(f(y),1)$ to find the answer :

$ \boxed{f(f(y)+f(1)) +f(y)-1= f(f(y))(f(1)+1)}$

Now : $P(0,y)$ , gives $f(f(y))=f(0)(f(y)+1)+y$
So for the right side we get that : $\boxed{f(f(y))(f(1)+1)=f(y).f(0)(f(1)+1)+f(0)(f(1)+1)+y(f(1)+1)}$
For the left side using $P(f(1),y)$ , we get that :
$f(f(y)+f(1))=f(f(1))(f(y)+1)+y(1-f(1))=[f(0)(f(1)+1)+1](f(y)+1)+y(1-f(1))$

So :$ \boxed{f(f(y)+f(1))+f(y)-1=f(0)(f(1)+1).f(y)+f(0)(f(1)+1)+f(y)+1+f(y)-1+y(1-f(1))}$
Identifying both sides will be left with :
$2f(y)+y(1-f(1)) = y(f(1)+1)$ , thus $f(y)=yf(1)$ , the rest is obvious .
This post has been edited 2 times. Last edited by omar1tun, Aug 1, 2024, 5:23 PM
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omar1tun
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#15
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For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x)
This post has been edited 1 time. Last edited by omar1tun, Aug 1, 2024, 5:28 PM
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omar1tun
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#16
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For above that equation doesn't imply that f is surjective as changing x to some value will also change f(x) an example is $f(x)=(x+1)^2$ ,will get on the right : $ (x+1)^2+(x+1)$ , which doesn't take all values in $R$ such as $-4$
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ihatemath123
3446 posts
#17
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The solutions are $f(x) = x$ and $f(x)=-x$, which are both easy to verify.

Claim: $f(0) = 0$.
Proof: Assume that $f(0) \neq 0$ FTSOC. Taking $P(x,0)$ gives us
\[f(x+f(0)) = f(x) \cdot (1 + f(0)),\]so incrementing $x$ by $f(0)$ gives some geometric sequence. But the original equation can be rewritten as
\[f(x+f(y)) - f(x)(1+f(y)) = y(1-x),\]and if we increment $x$ by $f(0)$ here, the LHS will multiply by $1+f(0)$. However, the RHS cannot grow exponentially since it is negative for all $x>1$ and positive for all $x < 1$, something no exponential function does. This gives us a contradiction.

Now, taking $P(0,x)$ gives us $f(f(x)) = x$, implying surjectivity, and $P(1,x)$ gives us
\[f(1+f(x)) = f(1)(1+f(x)),\]which is enough to imply that $f$ is linear. Plugging this into our original equation gives the claimed solutions.
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bo18
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#18
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f(x+f(y))+x.y=f(x).f(y)+f(x)+y
Let' s call this P(x,y)
P(0,0) gives us
f(f(0))=f(0)^2+f(0) this call *
P(x, 0) gives us
f(x+f(0))=f(x).f(0)+f(x)
x->-f(0) gives us
f(0)=f(-f(0)).f(0)+f(-f(0))
from this, we can get that:
f(0)/f(f(0))=f(0)+1
from this, * gives us
f(0)^2=f(-f(0)).f(f(0))=A
P(-f(0), f(0))
f(-f(0)+f(f(0)))-f(0)^2=A+f(-f(0))+f(0)
f(-f(0)+f(0)^2+f(0))=2.f(0)^2+f(0)+f(-f(0))
this give
f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0))
P(f(0), f(0)^2)
f(f(0)+f(f(0)^2))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2
f(f(0)+f(0)^2+f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2
f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2
Now, P(2f(f(0)), -f(0)), which give us
f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0)
Now P(x, f(0)) W
f(x+f(f(0)))+f(0).x=f(f(0)).f(x)+f(x)+f(0)
x->f(f(0)), I don' t know why I didn't put this here W
f(2f(f(0)))+f(0).f(f(0))=f(f(0)).f(f(f(0)))+f(f(f(0)))+f(0)
Now I want to find f(f(f(0)))
if we take this P(0, y) we will get this
f(f(y))=f(0).f(y)+f(0)+y
y->f(0), f(f(f(0)))=f(f(0)).f(0)+2.f(0)
if we call f(0) to be (a) and f(f(0)) to be (b), where b=a^2+a( from the beginning)
Then, f(2f(f(0)))+f(0).f(f(0))=f(f(0)).(f(f(0)).f(0)+2.f(0))+f(f(0)).f(0)+2.f(0)+f(0)
we will get this
f(2f(f(0)))+a.b=b.(b.a+2.a)+b.a+2.a+a, which is equivalent to f(2f(f(0)))=b(b.a+2.a)+3.a
f(2f(f(0))+f(-f(0)))-2.f(f(0)).f(0)=f(2f(f(0))).f(-f(0))+f(2f(f(0)))-f(0), Now, from this, we have:
f(2f(f(0))+f(-f(0)))-2.a.b=(b.(b.a+2.a)+3.a).f(-f(0))+b.(b.a+2.a)+3.a-a, call this T
From this, f(2f(f(0))+f(-f(0)))+f(0)^3=f(f(0)).f(f(0)^2)+f(f(0))+f(0)^2, we have that the left side is equal to, also I want to say that f(f(0)^2)=f(0)^2+f(f(0))+f(-f(0))=a^2+b+f(-f(0))
Now, nightmare is coming
The left side=b(a^2+b+f(-f(0)))+b+a^2-a^3
But we have that: f(0)^2=f(f(0)).f(-f(0)), which is equal to a^2=b.f(-f(0)), i.e the left side=a^2.b+b^2+a^2+b+a^2-a^3=a^2.b+b^2+2.a^2+b-a^3, call D
Now-if we multiply whole equation with b, we will get this
(The left side).b=2.b^2.a+(b^2.a+2.a.b+3.a).a^2+b^3.a+2.a.b^2+2.a.b
From D multiply with (b), we get this
a^2.b^2+b^3+2.a^2.b+b^2-a^3.b to be equal to 2.b^2.a+b^2.a^3+2.a^3.b+3.a^3+b^3.a+2.a.b^2+2.a.b
Now, if this above wasn' t nightmare, after we will replaceable b with a^2+a- this will be a real bad nightmare, let's start (teeth)
a^2.(a^2+a)^2+(a^2+a)^3+2.a^2.(a^2+a)+(a^2+a)^2-a^3.(a^2+a)=2.(a^2+a)^2.a+(a^2+a)^2.a^3+2.a^3.(a^2+a)+3.a^3+(a^2+a)^3.a+2.a.(a^2+a)^2+2.a.(a^2+a)
After we use Wolfram Alpha, oops, just a joke, I figured it all myself (sad) , we will get this
+2.a^7+3.a^6+6.a^5+5.a^4+4.a^3+a^2=0, the solution of this two in Real numbers, the one of them is approximately to some digit-negative and stranger, the other solution is 0
Then we have a=0=f(0)
Then, if we put this in this:
f(f(y)=f(0).f(y)+f(0)+y=y, i.e the function is involution
Now, let P(f(x), f(y)) gives us
f(f(x)+y)+f(x).f(y)=x.y+f(y)+x
But P(y,x) gives us
f(f(x)+y)+x.y=f(x).f(y)+f(y)+x, call G
Let sum this two, then this gives us
f(f(x)+y)=f(y)+x
Put this in G gives
x.y=f(x).f(y), y->x gives f(x)^2=x^2
f(x)=+x, f(x)=-x
This post has been edited 1 time. Last edited by bo18, Sep 18, 2024, 5:58 PM
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Davud29_09
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#19
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P(f(x),y) ,change x,y after this and y=0 implies that f(f(x))×(f(0)+1)=f(x)×f(f(0))+f(f(0))+x-x×f(0).
Plug x=0 we get f(f(y))=f(0)×f(y)+f(0)+y we get f is injective. Also we use this equation above and we get f(x)×(f(0)²+f(0)-f(f(0))) =-2x×f(0)-(f(0)²+f(0)-f((0))) if f(0)²+f(0)-f(f(0)) isn't equal 0 then f(x)=ax+b we check and we get f(x)=x or f(x)=-x. Other case implies that f(0)=0.In common equation we plug x=0 and we get f(f(x))=x.In next steps P(x,f(y)) and we change x,y we get x×f(y)=y×f(x) y=1 and we get f(x)=cx.It is easy to check and find f(x)=x and f(x)=-x answers.We are done.
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lksb
171 posts
#20
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$P(0,x)\implies f(f(x))=f(0)(f(x)+1)$
$P(f(x), y)\implies f(f(x)+f(y))+yf(x)=f(0)(f(x)+1)(f(y)+1)+xf(y)+x+y$
$P(f(x),y)-P(x, f(y))\implies  xf(y)=yf(x)\implies \boxed{f(x)=\pm x}$
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awesomeming327.
1712 posts
#21
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The answer is $f(x)=x$ or $f(x)=-x$, which clearly work. Now we prove that this is the only solution. Let $P(x,y)$ denote the assertion, and let $f(0)=c$.

We have
\begin{align*}
P(0,0) &\implies f(c) = c^2+c \\
P(0,x) &\implies f(f(x)) = cf(x) + c + x \\
P(x,0) &\implies f(x+c) = (c+1)f(x)
\end{align*}
Claim 1: $f(0)=0$.
Take $P(c,1)$ which gives us on one hand,
\[f(c+f(1))+c=(c^2+c)(f(1)+1)+1\]But on the other hand, we have
\begin{align*}
f(f(1)+c)+c &= (c+1)f(f(1)) + c \\
&= (c+1)(cf(1) + c + 1) + c \\
&=(c^2+c)(f(1)+1)+2c+1
\end{align*}which proves our claim.

Now, $P(0,x)$ gives $f(f(x))=x$. Thus when we use $P(f(x),y)$ we get
\[f(f(x)+f(y))+f(x)y=xy+x+y\]Since $xy+x+y-f(f(x)+f(y))$ is symmetric, we must have $f(x)y=f(y)x$, implying that $f(x)/x$ is a constant. Since $f$ is an involution, this constant is $\pm 1$.
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jasperE3
11293 posts
#22
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IAmTheHazard wrote:
Find all functions $f : \mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$,
$$f(x+f(y))+xy=f(x)f(y)+f(x)+y.$$
Andrew Carratu

Let $P(x,y)$ be the assertion $f(x+f(y))+xy=f(x)f(y)+f(x)+y$.
$P(0,x)\Rightarrow f(f(x))=f(0)(f(x)+1)+x$
$P(f(x),y)\Rightarrow f(f(x)+f(y))-f(0)(f(x)+1)(f(y)+1)=xf(y)-yf(x)+x+y$
Swapping $x,y$ gives $yf(x)-xf(y)=xf(y)-yf(x)$, then setting $y=1$, we have $f(x)=xf(1)$. Testing, only $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ work.
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Bardia7003
20 posts
#23
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Probably similar solutions are already mentioned. Anyway, I'll go on with mine.
Let $P(x, y)$ denote the given assertion.
$P(0, x): \underline{f(f(x)) = f(0)(f(x) + 1) + x}$
$P(f(x), y): f(f(x) + f(y)) + f(x)y = f(f(x))(f(y) + 1) + y = (f(0)(f(x) + 1) + x)(f(y) + 1) + y = f(0)f(x)f(y) + f(0)f(y) + xf(y) + f(0)f(x) + f(0) + x + y$
$\rightarrow f(f(x) + f(y)) = (x + y) + f(0)f(x)f(y) + f(0)(f(x)f(y) + f(y) + f(x) + 1) + xf(y) - f(x)y$
Now, if we swap $x,y$, the left side stays the same, and all the parts in the right side stay the same, except the $xf(y) - f(x)y$, so:
$$xf(y) - f(x)y = yf(x) - f(y)x \rightarrow 2xf(y) = 2yf(x) \rightarrow \frac{f(x)}{x} = \frac{f(y)}{y} \rightarrow \underline{f(x) = kx}$$And by checking this format in the equation, we find that only $k = 1, -1$ are the solutions, therefore the only answers are $\boxed{f(x) = x \quad \forall x \in \mathbb{R}}$ and $\boxed{f(x) = -x \quad \forall x \in \mathbb{R}}$. $\blacksquare$ :)
This post has been edited 1 time. Last edited by Bardia7003, Apr 1, 2025, 4:34 PM
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