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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
Old problem
kwin   1
N 22 minutes ago by ant_
Let $a, b, c \ge 0$ and $ ab+bc+ca>0$. Prove that:
$$ \frac{1}{(a+b)^2} + \frac{1}{(b+c)^2} + \frac{1}{(c+a)^2} + \frac{15}{(a+b+c)^2} \ge \frac{6}{ab+bc+ca}$$Is there any generalizations?
1 reply
kwin
Yesterday at 1:12 PM
ant_
22 minutes ago
IMO Genre Predictions
ohiorizzler1434   35
N 31 minutes ago by ohiorizzler1434
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
35 replies
ohiorizzler1434
May 3, 2025
ohiorizzler1434
31 minutes ago
Inequality
giangtruong13   0
32 minutes ago
Source: My friend
Let $a,b,c>0$ such that $a+b+c=3$. Find the minimum: $$P= \sum_{cyc} \frac{a}{b} + \sum_{cyc} \frac{1}{a^3+b^3+abc}$$
0 replies
1 viewing
giangtruong13
32 minutes ago
0 replies
something interesting...
SunnyEvan   0
an hour ago
Source: old result
Let $x$, $y$, $z$ be non-negative real numbers, no two of which are zero. Such that $ x+y+z=3.$ Prove that :
$$ \sum \frac{16(9-xyz)}{9(z+x)^2(y+3)^2} \geq \frac{2xyz}{\sum x^2} + \frac{\sum (z^2-x^2)(z^2-y^2)(z+x)(z+y)}{\sum(x+y)^3(y+z)^3} $$
0 replies
SunnyEvan
an hour ago
0 replies
Jbmo 2011 Problem 4
Eukleidis   13
N 2 hours ago by Adventure1000
Source: Jbmo 2011
Let $ABCD$ be a convex quadrilateral and points $E$ and $F$ on sides $AB,CD$ such that
\[\tfrac{AB}{AE}=\tfrac{CD}{DF}=n\]

If $S$ is the area of $AEFD$ show that ${S\leq\frac{AB\cdot CD+n(n-1)AD^2+n^2DA\cdot BC}{2n^2}}$
13 replies
Eukleidis
Jun 21, 2011
Adventure1000
2 hours ago
an algebra question
kjhgyuio   1
N 2 hours ago by aidan0626
.........
1 reply
kjhgyuio
2 hours ago
aidan0626
2 hours ago
we can find one pair of a boy and a girl
orl   16
N 2 hours ago by ezpotd
Source: Vietnam TST 2001 for the 42th IMO, problem 3
Some club has 42 members. It’s known that among 31 arbitrary club members, we can find one pair of a boy and a girl that they know each other. Show that from club members we can choose 12 pairs of knowing each other boys and girls.
16 replies
orl
Jun 26, 2005
ezpotd
2 hours ago
teleporting wizard starts on point (2017, 101), 4 moves
parmenides51   1
N 2 hours ago by jasperE3
Source: 2018 USAIMEO #2 p5 (Mock AIME -USAJMO) https://artofproblemsolving.com/community/c594864h1572209p9658908
A teleporting wizard starts on the point $(2017, 101)$ and can teleport to other Cartesian coordinates with only $1$ of $4$ moves: $(x, y) \to (x + y, y)$, $(x, y) \to (x - y, y)$ when $x > y$, $(x, y) \to (x, x + y)$, and $(x, y) \to (x, y - x)$ when $y > x$.

(a) Let $P(x)$ be any polynomial with positive integer coefficients that passes through $(0, 0)$. Show that for all such $P(x)$, there exists a unique point on the curve where the wizard can land on.

(b) For each $P(x)$, let $S$ be this unique point. Find the equation of the graph that contains all potential $S$.
1 reply
parmenides51
Nov 17, 2023
jasperE3
2 hours ago
IMO ShortList 2001, combinatorics problem 1
orl   33
N 2 hours ago by ihategeo_1969
Source: IMO ShortList 2001, combinatorics problem 1
Let $A = (a_1, a_2, \ldots, a_{2001})$ be a sequence of positive integers. Let $m$ be the number of 3-element subsequences $(a_i,a_j,a_k)$ with $1 \leq i < j < k \leq 2001$, such that $a_j = a_i + 1$ and $a_k = a_j + 1$. Considering all such sequences $A$, find the greatest value of $m$.
33 replies
orl
Sep 30, 2004
ihategeo_1969
2 hours ago
CooL geo
Pomegranat   0
3 hours ago
Source: Idk

In triangle \( ABC \), \( D \) is the midpoint of \( BC \). \( E \) is an arbitrary point on \( AC \). Let \( S \) be the intersection of \( AD \) and \( BE \). The line \( CS \) intersects with the circumcircle of \( ACD \), for the second time at \( K \). \( P \) is the circumcenter of triangle \( ABE \). Prove that \( PK \perp CK \).
0 replies
1 viewing
Pomegranat
3 hours ago
0 replies
Coefficient Problem
P162008   2
N 3 hours ago by cazanova19921
Consider the polynomial $g(x) = \prod_{i=1}^{7} \left(1 + x^{i!} + x^{2i!} + x^{3i!} + \cdots + x^{(i-1)i!} + x^{ii!}\right)$
Find the coefficient of $x^{2025}$ in the expansion of $g(x).$
2 replies
P162008
Yesterday at 12:16 PM
cazanova19921
3 hours ago
Shooting An Invisible Tank
Aryan27   0
3 hours ago
Source: 239 MO
An invisible tank is on a $100 \times 100 $ table. A cannon can fire at any $k$ cells of the board after that the tank will move to one of the adjacent cells (by side). Then the process is repeated. Find the smallest value of $k$ such that the cannon can definitely shoot the tank after some time.
0 replies
Aryan27
3 hours ago
0 replies
Showing Tangency
Itoz   1
N 3 hours ago by ja.
Source: Own
The circumcenter of $\triangle ABC$ is $O$. Line $AO$ meets line $BC$ at point $D$, and there is a point $E$ on $\odot(ABC)$ such that $AE \perp BC$. Line $DE$ intersects $\odot(ABC)$ at point $F$. The perpendicular bisector of line segment $BC$ intersects line $AB$ at point $K$, and line $AB$ intersects $\odot(CFK)$ at point $L$.

Prove that $\odot(AFL)$ is tangent to $\odot (OBC)$.
1 reply
Itoz
Yesterday at 1:57 PM
ja.
3 hours ago
2-var inequality
sqing   7
N 3 hours ago by mathuz
Source: Own
Let $ a,b\geq 0    $. Prove that
$$ \frac{a }{a^2+2b^2+1}+ \frac{b }{b^2+2a^2+1}\leq \frac{1}{\sqrt{3}} $$$$   \frac{a }{2a^2+ b^2+2ab+1}+ \frac{b }{2b^2+ a^2+2ab+1}  \leq \frac{1}{\sqrt{5}} $$$$ \frac{a }{2a^2+ b^2+ ab+1}+ \frac{b }{2b^2+ a^2+ ab+1} \leq \frac{1}{2} $$$$\frac{a }{a^2+2b^2+2ab+1}+ \frac{b }{b^2+2a^2+2ab+1}\leq \frac{1}{2} $$
7 replies
sqing
Today at 1:39 AM
mathuz
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
Y by
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
Y by
Solution
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fractals
3028 posts
#4
Y by
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
Y by
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
5604 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
169 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
11291 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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