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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.

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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
Find all functions $f$ is strictly increasing : \(\mathbb{R^+}\) \(\rightarrow\)
guramuta   4
N a few seconds ago by guramuta
Find all functions $f$ is strictly increasing : \(\mathbb{R^+}\) \(\rightarrow\) \(\mathbb{R^+}\) such that:
i) $f(2x)$ \(\geq\) $2f(x)$
ii) $f(f(x)f(y)+x) = f(xf(y)) + f(x) $
4 replies
guramuta
Yesterday at 1:45 PM
guramuta
a few seconds ago
Is this FE is solvable?
ItzsleepyXD   1
N 3 minutes ago by jasperE3
Source: Own , If not appear somewhere before
Find all function $f : \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in  \mathbb{R}$ . $$f(x+f(y))+f(x+y)=2x+f(y)+f(f(y))$$. Original
1 reply
ItzsleepyXD
3 hours ago
jasperE3
3 minutes ago
Hard combi
EeEApO   2
N 14 minutes ago by EeEApO
In a quiz competition, there are a total of $100 $questions, each with $4$ answer choices. A participant who answers all questions correctly will receive a gift. To ensure that at least one member of my family answers all questions correctly, how many family members need to take the quiz?

Now, suppose my spouse and I move into a new home. Every year, we have twins. Starting at the age of $16$, each of our twin children also begins to have twins every year. If this pattern continues, how many years will it take for my family to grow large enough to have the required number of members to guarantee winning the quiz gift?
2 replies
EeEApO
Yesterday at 6:08 PM
EeEApO
14 minutes ago
Equilateral triangle formed by circle and Fermat point
Mimii08   1
N 18 minutes ago by srirampanchapakesan
Source: Heard from a friend
Hi! I found this interesting geometry problem and I would really appreciate help with the proof.

Let ABC be an acute triangle, and let T be the Fermat (Torricelli) point of triangle ABC. Let A1, B1, and C1 be the feet of the perpendiculars from T to the sides BC, AC, and AB, respectively. Let ω be the circle passing through points A1, B1, and C1. Let A2, B2, and C2 be the second points where ω intersects the sides BC, AC, and AB, respectively (different from A1, B1, C1).

Prove that triangle A2B2C2 is equilateral.

1 reply
Mimii08
6 hours ago
srirampanchapakesan
18 minutes ago
Geometry Parallel Proof Problem
CatalanThinker   1
N 29 minutes ago by ItzsleepyXD
Source: No source found, just yet, please share if you find it though :)
Let M be the midpoint of the side BC of triangle ABC. The bisector of the exterior angle of point A intersects the side BC in D. Let the circumcircle of triangle ADM intersect the lines AB and AC in E and F respectively. If the midpoint of EF is N, prove that MN || AD.
I have done some constructions, but still did not quite get to the answer, see diagram attached below
1 reply
CatalanThinker
an hour ago
ItzsleepyXD
29 minutes ago
Inspired by Kosovo 2010
sqing   0
36 minutes ago
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
0 replies
sqing
36 minutes ago
0 replies
Kosovo MO 2010 Problem 5
Com10atorics   20
N an hour ago by sqing
Source: Kosovo MO 2010 Problem 5
Let $x,y$ be positive real numbers such that $x+y=1$. Prove that
$\left(1+\frac {1}{x}\right)\left(1+\frac {1}{y}\right)\geq 9$.
20 replies
Com10atorics
Jun 7, 2021
sqing
an hour ago
Fourth powers and square roots
willwin4sure   39
N an hour ago by awesomeming327.
Source: USA TSTST 2020 Problem 4, by Yang Liu
Find all pairs of positive integers $(a,b)$ satisfying the following conditions:
[list]
[*] $a$ divides $b^4+1$,
[*] $b$ divides $a^4+1$,
[*] $\lfloor\sqrt{a}\rfloor=\lfloor \sqrt{b}\rfloor$.
[/list]

Yang Liu
39 replies
willwin4sure
Dec 14, 2020
awesomeming327.
an hour ago
Interesting inequalities
sqing   1
N an hour ago by sqing
Source: Own
Let $ a,b >0 $ and $ a^2-ab+b^2\leq 1 $ . Prove that
$$a^4 +b^4+\frac{a }{b +1}+ \frac{b }{a +1} \leq 3$$$$a^3 +b^3+\frac{a^2}{b^2+1}+ \frac{b^2}{a^2+1} \leq 3$$$$a^4 +b^4-\frac{a}{b+1}-\frac{b}{a+1} \leq 1$$$$a^4+b^4 -\frac{a^2}{b^2+1}- \frac{b^2}{a^2+1}\leq 1$$$$a^3+b^3 -\frac{a^3}{b^3+1}- \frac{b^3}{a^3+1}\leq 1$$
1 reply
sqing
2 hours ago
sqing
an hour ago
Sum of 1/(a^5(b+2c))^2 at least 1/3 [USA TST 2010 2]
MellowMelon   42
N 2 hours ago by Adywastaken
Let $a, b, c$ be positive reals such that $abc=1$. Show that \[\frac{1}{a^5(b+2c)^2} + \frac{1}{b^5(c+2a)^2} + \frac{1}{c^5(a+2b)^2} \ge \frac{1}{3}.\]
42 replies
MellowMelon
Jul 26, 2010
Adywastaken
2 hours ago
Weird function?
ItzsleepyXD   2
N 2 hours ago by ItzsleepyXD
Source: Own
Find all functions \( f: \mathbb{R} \rightarrow \mathbb{R} \) such that for all \( x, y \in \mathbb{R} \),
\[
f(x + f(2y)) + f(x^2 - y) = f(f(x)) f(x + 1) + 2y - f(y).
\]
2 replies
ItzsleepyXD
Apr 11, 2025
ItzsleepyXD
2 hours ago
Almost similar one but more answer lol
ItzsleepyXD   0
2 hours ago
Source: Own , Modified
Find all non decreasing functions or non increasing function $f \colon \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in \mathbb{R}$

$$ f(x+f(y))=f(x)+f(y) \text{ or } f(f(f(x)))+y$$.
0 replies
ItzsleepyXD
2 hours ago
0 replies
A lot of unexpected answer from non decreasing function
ItzsleepyXD   0
2 hours ago
Source: Own
Find all non decreasing function $f : \mathbb{R} \to \mathbb{R}$ such that for all $x,y \in  \mathbb{R}$ and $m,n \in \mathbb{N}_0$ such that $m+n \neq 0$ there exist $m',n' \in \mathbb{N}_0$ such that $m'+n'=m+n+1$ and $$f(f^m(x)+f^n(y))=f^{m'}(x)+f^{n'}(y)$$. Note : $f^0(x)=x$ and $f^{n}(x)=f(f^{n-1}(x))$ for all $n \in \mathbb{N}$ . original
0 replies
ItzsleepyXD
2 hours ago
0 replies
Cute Inequality
EthanWYX2009   0
3 hours ago
Let $a_1,\ldots ,a_n\in\mathbb R\backslash\{0\},$ determine the minimum and maximum value of
\[\frac{\sum_{i,j=1}^n|a_i+a_j|}{\sum_{i=1}^n|a_i|}.\]
0 replies
EthanWYX2009
3 hours ago
0 replies
Unexpected FE
Taco12   18
N Wednesday at 6:49 PM by lpieleanu
Source: 2023 Fall TJ Proof TST, Problem 3
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $x$ and $y$, \[ f(2x+f(y))+f(f(2x))=y. \]
Calvin Wang and Zani Xu
18 replies
Taco12
Oct 6, 2023
lpieleanu
Wednesday at 6:49 PM
Unexpected FE
G H J
Source: 2023 Fall TJ Proof TST, Problem 3
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Taco12
1757 posts
#1 • 2 Y
Y by ItsBesi, rightways
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $x$ and $y$, \[ f(2x+f(y))+f(f(2x))=y. \]
Calvin Wang and Zani Xu
This post has been edited 2 times. Last edited by Taco12, Oct 6, 2023, 1:06 AM
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EthanWYX2009
862 posts
#2 • 1 Y
Y by ys-lg
$P(0,0)\Rightarrow f(f(0))=0.$
$P(0,y)\Rightarrow y=f(f(y))+f(f(0))=f(f(y)).$
$\Rightarrow f(2x+f(y))=y-2x.$
$f(2x+f(2y+f(z)))=f(2x+z-2y)=2y+f(z)-2x.$
$t=2x-2y\Rightarrow f(t+z)=f(z)-t.$
$z=0\Rightarrow f(t)=f(0)-t=c_1-t.$$\Rightarrow \forall 2\mid x,f(x)=c_1-x.$
$z=1\Rightarrow f(t+1)=f(1)-t.\Rightarrow \forall 2\nmid x,f(x)=c_2-x$
$\Rightarrow f(x)=c-x.\blacksquare$
This post has been edited 2 times. Last edited by EthanWYX2009, Oct 6, 2023, 1:31 AM
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cj13609517288
1915 posts
#3
Y by
@above no ! so close tho

First three steps are same as above. Since $f$ is an involution, $f(y)$ can be $0$ or $1$.
Varying $x$ yields that $f(2x)=C_1-2x$ for some $C_1$ and $f(2x+1)=C_2-(2x+1)$ for some $C_2$.

Finally note that if $C_1$ is even then $C_2$ has to be even, and vice versa. If $C_1$ is odd then $C_2=C_1$. So our solution is $C_1,C_2$ even and $C_1=C_2$ odd. This works upon checking.

Remark
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EthanWYX2009
862 posts
#4 • 4 Y
Y by cj13609517288, BigJoJo, LLL2019, ys-lg
I didn't check the final result....... I'm so fool
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MathLuis
1524 posts
#5
Y by
The Speedrun begins!.
Let $P(x,y)$ the assertion of the following F.E., i claim that $f(2n)=c-2n, f(2n+1)=c_1-2n-1$ works for all integers $n$ where either $c=c_1$ are odd or $c,c_1$ are both even.
$P(0,x)$
$$f(f(x))+f(f(0))=x \implies f \; \text{bijective}$$Also $P(0,0)$ gives $f(f(0))=0$ so infact $f$ is an involution (i.e. $f(f(x))=x$).
Hence our functional equation became $f(2x+f(y))=y-2x$, now set $f(c)=0$ then by $P(x,c)$ we get that $f(x)=c-x$ for all even $x$.
Also by $P(x,f(1))$ we get $f(2x+1)=(f(1)+1)-2x-1$ for all integers $x$ so $f(x)=c_1-x$ for all odd $x$, now if $c_1$ is odd then notice that $x=f(f(x))=f(c_1-x)=c-c_1+x$ for all odd $x$. so $c=c_1$ odd, if $c_1$ is even, but $c$ odd then $x=f(f(x))=f(c-x)=c_1-c+x$ for all $x$ even so $c=c_1$ which cant happen so $c$ is also even in this case. Hence our claim is true and we are done :D.
This post has been edited 1 time. Last edited by MathLuis, Oct 6, 2023, 1:49 AM
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tadpoleloop
311 posts
#6
Y by
Taco12 wrote:
Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that for all integers $x$ and $y$, \[ f(2x+f(y))+f(f(2x))=y. \]
Calvin Wang and Zani Xu

What an interesting answer.

Claim:
$$\boxed{f(x) = \left\{ \begin{matrix}a-x & x \text{ even}\\b-x & x \text{ odd}\end{matrix}\right. \quad\text{where }a=b\text{ or a,b both even}}$$
Proof:
$P(0,0) \implies f(f(0)) = 0$
$P(0,x)\implies f(f(x)) = x$
$P(x,f(y))\implies f(2x+y) = f(y) - 2x$
In particular $f(2x) = f(0) - 2x$ and $f(2x+1) = (f(1)+1)-(2x+1)$

So $a=f(0)$ and $b = f(1) + 1$ as per our claim.

Plugging into our involution requirement we see that if either $a$ or $b$ are odd that $a=b$ $\square$
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john0512
4187 posts
#7
Y by
The answer is $f(x)=c-x$ for even $x$ and $f(x)=d-x$ for odd $x$ where either $c,d$ are both even or $c=d.$ These solutions clearly work.

Letting $x=y=0$, we get that $f(f(0))=0$. Letting only $x=0$, we have $$f(f(y))+f(f(0))=y$$so $$y=f(f(y)).$$Thus, $f$ is an involution.

Let $f(0)=c$ (and $f(c)=0$).

With this, the original functional equation implies that $$f(2x+f(y))+2x=y.$$Letting $y=2x$ gives $$f(2x+f(2x))=0.$$Since involutions are both injective and surjective, we can de-nest this into $$2x+f(2x)=c$$$$f(2x)=c-2x.$$Thus, for all even $n$, we have $$f(n)=c-n.$$
Plugging in $x=1$, we have that $$f(2+f(y))=y-2.$$Applying $f$ to both sides and using the fact that $f$ is an involution, $$2+f(y)=f(y-2)$$$$f(y)-f(y-2)=-2.$$Thus, by induction, if $f(1)=d-1$, then $$f(odd)=d-odd$$for all odd integers $odd$.

However, note that $$f(f(2x))=f(c-2x)=2x.$$If $c$ is even, then this equation is clearly true since $c-2x$ is even. Then, if $c$ is even, then we have that all even inputs lead to even outputs. Hence, if $d$ is odd, then odd also maps to even, contradicting surjectivity. Thus, if $c$ is even, then $d$ must also be even. Otherwise, if $c$ is odd, then we must have $$d-(c-2x)=2x$$$$d=c.$$Thus we are done.
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v_Enhance
6877 posts
#8 • 2 Y
Y by ESAOPS, Marcus_Zhang
Solution from Twitch Solves ISL:

The answers are \[ f(x) = \begin{cases} a - x & x \equiv 0 \pmod 2 \\ b - x & x \equiv 1 \pmod 2 \end{cases} \]where $a$ and $b$ are either both even, or $a = b$. It can be checked that all of these work, so we prove they're the only solutions.
Let $P(x,y)$ be the given assertion.
  • $P(0,0) \implies f(f(0)) = 0$.
  • $P(0,t) \implies f(f(t)) = t$.
  • $P(1,f(z)) \implies f(z+2)=z-2$.
The last equation $f(z+2) = z-2$ implies $f$ takes the above form for some $a$ and $b$, so we'd be done if we could show the parity condition. If $a$ is odd, then plug in $x=0$ to deduce $a=b$; if $b$ is odd, plug in $x=1$ to deduce $b=a$. This finishes the problem.
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shendrew7
795 posts
#9 • 1 Y
Y by rstenetbg
It's easy to get $f(f(0))=0$ from $(0,0)$ and $f(f(n))=n$ from $(0,n)$. Then our condition can be rewritten as
\[y=f(2x+f(y))+2x.\]
We then substitute values to determine the value of $f$ based on parity:
\begin{align*}
(-t,f(2t)):& \quad f(2t)=-2t+f(0) \implies f(x)=-x+c \text{ for even } x \\
(-t, f(2t+1)):& \quad f(2t+1)=-2t+f(1) \implies f(x)=-x+d \text{ for odd } x
\end{align*}
We finish by using casework on the parities of $c$ and $d$ with our involution $f(f(n))$. We get the following solution, which can be easily tested:
\[\boxed{f(x) = \begin{cases} -x+c & \text{for even } x \\ -x+d & \text{for odd } x \end{cases} \quad \text{where }a,b \text{ both even or } a=b.}\]
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vsamc
3789 posts
#11
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Solution
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eibc
600 posts
#12
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All solutions are of the form $f(x) = -x + c$ for odd $x$ and $f(x) = -x + d$ for even $x$ where $c, d$ are integers such that either $c = d$ or $c \equiv d \equiv 0 \pmod 2$. With some effort, we can verify that these all work.

Denote the original assertion as $P(x, y)$. From $P(0, 0)$ we have $f(f(0)) = 0$. Then from $P(0, y)$, we have $f(f(y)) = y$, which implies that $f$ is bijective. This also lets us rewrite the original equation as $f(2x + f(y)) + 2x = y$.

Then, from $P(1, f(x))$, we have $f(x + 2) + 2 = f(x)$, which implies that there exists integers $c, d$ such that $f(x) = -x + c$ when $x$ is odd and $f(x) = -x + d$ when $x$ is even. From $f(f(0)) = 0$, we see that:
  • If $d$ is even, then $c$ must also be even, as if $c$ is odd then $1 = f(f(1)) = f(-1 + c) = 1 - c + d \equiv 0 \pmod 2$, a contradiction. Any pair $(c, d)$ with $c$ and $d$ both even will work, as mentioned above.
  • If $d$ is odd, then $0 = f(f(0)) = f(d) = -d + c$, so $c = d$, which also works.
Having exhausted all cases, we are done.
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Martin2001
152 posts
#13
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Plug in $P(0,0)$ then $P(0,y)$ to get $f(f(y))=y.$ Thus, $f$ is an involution. The original equation is now
$$y=f(f(y)+2x)+2x.$$Then we can separate into cases even and odd like this :
\begin{align*}
P(-t, f(2t))=f(2t)=-2t+f(0) &\rightarrow f(x)=-x+c \\
p(-t, f(2t+1))=f(2t+1)=-2t+f(1) &\rightarrow f(x)=-x+d.
\end{align*}If $d$ is even, $c$ must also be even, as if $c$ is odd then $1=f(f(1))=f(-1+c)=1-c+d \equiv 0 \pmod 2,$ contradiction.
\newline If $d$ is odd, then $c=d,$ because $0=f(f(0))=f(d)=-d+c,$ which works. Thus, our answer is
$$\boxed{f(x) = \begin{cases} -x+c & \text{for even } x \\ -x+d & \text{for odd } x \end{cases} \quad \text{where }c,d \text{ both even or } c=d.}$$
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Jndd
1416 posts
#14
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We claim that the solution is $f(x)=-x+c$ for even $x$ and $f(x)=-x+d$ for odd $x$ where $c=d$ or $c$ and $d$ are both even. It is easy enough to do cases and check that these solutions work.

Let $P(x,y)$ denote the assertion. $P(0,0)$ gives us $f(f(0))=0$, so by $P(0,y)$, we get \[f(f(y))+f(f(0))=f(f(y))=y,\]meaning $f$ is an involution, and is therefore bijective. Now, We have \[f(2x+f(y))+f(f(2x))=f(2x+f(y))+2x=y,\]giving $f(2x+f(0))=-2x$. Thus, for even $x$, we have $f(x)=-x+f(0)$. We also have $f(2x+f(1))=1-2x$, giving $f(x)=-x+(f(1)+1)$ for odd $x$. Let $c=f(0)$ and $d=f(1)+1$, so we can write $f(x)=-x+c$ for even $x$ and $f(x)=-x+d$ for odd $x$.

Now, suppose $y$ is even in our original equation. Then, we have \[f(2x+f(y))+f(f(2x))=f(2x-y+c)+2x=y,\]so if $c$ is even, then this works. Otherwise, if $c$ is odd, then we must have $c=d$.

Now, suppose $y$ is odd in our original case. Then, we have \[f(2x+f(y))+f(f(2x))=f(2x-y+d)+2x=y,\]so if $d$ is even then this works. Otherwise, we must have $c=d$.

This means that either $c=d$, or $c$ and $d$ are both even, as desired.
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Eka01
204 posts
#15
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I might be missing something cuz I'm sleepy but here goes :-
Put $x=0$ to get $f(f(y))+f(f(0))=y$. Putting $y=0$ gives $f(f(y))=y$. This implies $f$ is bijective.
Now let $a$ be such that $f(a)=0$. Putting $y=a$ gives us that $f(2x)+2x=a$ since $f(f(2x)=2x$ giving us that $f(2x)=a-2x$.
Now let $b$ be such that $f(b)=1$. Putting $y=b$ gives us that $f(2x+1)=b-2x$.

Now using these expressions, we see that either $a$ is even and $b$ is odd or $a+1=b$ in order to satisfy the given equation.
Hence our solution is $\boxed{f(2x)=a-2x, f(2x+1)=b-2x}$ where $a,b$ satisfy the above conditions.
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eg4334
637 posts
#16
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if $P$ is the assertion then $P(0, 0) \implies f(f(0))=0$. Then $P(0, y) \implies f(f(y))=y$. We also have $2x+f(y)=f(y-2x)$ which when $x=1$ gives us $f(y)=f(y-2)-2$. Thus our solutions are determined by $f(0)$ and $f(1)$, say write it as $f(x) = a-x$ when $x$ is odd and $f(x)=b-x$ when $x$ is even. Now $f(f(0)) = 0$ tells us that $f(b) = 0$. Therefore, $b$ can be even or $a=b$. A similar analysis on $f(f(1))=1$ tells us that $a$ is even or $a=b$. Thus, we have $f(x)=a-x$ when $x$ is odd and $f(x)=b-x$ when $x$ is even for some $a, b$ such that they are both even or equal.
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Marcus_Zhang
980 posts
#17
Y by
Storage
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Maximilian113
575 posts
#18
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Let $P(x, y)$ denote the assertion. Then $P(0,0) \implies f(f(0))=0.$ Hence $$P(0, x) \implies f(f(x))=x,$$so $f(x)$ is bijective. Thus the assertion becomes $$P(x, y) \iff f(2x+f(y))=y-2x.$$Thus $$P(x, f(0)) \implies f(2x)=f(0)-2x, P(x, f(1)) \implies f(2x+1)=f(1)-2x.$$If $f(0)$ is odd, we have that $$f(f(2x)) = f(f(0)-2x)=f(1)-f(0)+2x \implies f(1)=f(0).$$This holds also when $f(1)$ is odd. Hence, the solutions are $$f(2x)=a-x, f(2x+1)=b-x$$where either $a, b$ are both even or $a=b.$
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Ilikeminecraft
623 posts
#19
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I claim the answer is $f(x) = -x + c$ if $c$ is some odd integer, or $f(x) = -x + c_1$ if $x$ is even and $f(x) = -x + c_2$ if $x$ is odd, and $c_1, c_2$ are even integers.

Let $P(x, y)$ denote our assertion. From $P(0, 0),$ we see that $f(f(0)) = 0.$ From $P(0, x),$ we get that $f(f(x)) = x,$ using the fact that $f(f(0)) = 0.$ By rearranging our given equation, and taking $f$ on both sides, we see that $2x + f(y) = f(y - 2x).$ Let $f(0) = c_1, f(1) = c_2.$ Note that by plugging in $y = 0, 1$, $f(2x) = c_1 - 2x, f(2x + 1) = -2x + c_2.$ Now we consider two seperate cases.
\begin{enumerate}
\item If $c_1\equiv1\pmod2,$ we plug it back into our original equation to see that $-2x = f(2x + f(0)) = f(2x + c_1),$ and thus we have that $-2x - c_1 + 1 + c_2 = -2x.$ Thus, $1 + c_2 = c_1,$ and hence $f(0) = f(1) + 1.$ Thus, we can write $f(x) = -x + c.$
\item If $c_1 \equiv 0\pmod 2,$ clearly we can't say anything, and both solutions are valid.
\end{enumerate}
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lpieleanu
3001 posts
#21
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Solution
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