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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

[list][*]Over 60 summer classes are starting at the Virtual Campus on July 7th - check out the math and language arts options for middle through high school levels.
[*]At AoPS Online, we have accelerated sections where you can complete a course in half the time by meeting twice/week instead of once/week, starting on July 8th:
[list][*]MATHCOUNTS/AMC 8 Basics
[*]MATHCOUNTS/AMC 8 Advanced
[*]AMC Problem Series[/list]
[*]Plus, AoPS Online has a special seminar July 14 - 17 that is outside the standard fare: Paradoxes and Infinity
[*]We are expanding our in-person AoPS Academy locations - are you looking for a strong community of problem solvers, exemplary instruction, and math and language arts options? Look to see if we have a location near you and enroll in summer camps or academic year classes today! New locations include campuses in California, Georgia, New York, Illinois, and Oregon and more coming soon![/list]

MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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Are you tired of the heat and thinking about Fall? You can plan your Fall schedule now with classes at either AoPS Online, AoPS Academy Virtual Campus, or one of our AoPS Academies around the US.

Our full course list for upcoming classes is below:
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0 replies
jwelsh
Jul 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
IMO Genre Predictions
ohiorizzler1434   114
N 10 minutes ago by Shan3t
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
114 replies
ohiorizzler1434
May 3, 2025
Shan3t
10 minutes ago
Nice 60 degree triangle property
Kratsneb   1
N 23 minutes ago by RANDOM__USER
Let $ABC$ be a triangle with $\angle A = 60 ^{\circ}$. $AD$, $BE$ and $CF$ are the altitudes, $M$ is the midpoint of $BC$. $BE \cap (ABC) = P$, $CF \cap (ABC) = Q$. Let $I$ and $J$ be incenters of $\triangle CEP$ and $\triangle BFQ$. Prove that $I$, $J$, $D$, $M$ lie on a circle.
IMAGE
It works for $\angle A = 120 ^{\circ}$ too.
1 reply
Kratsneb
Thursday at 2:13 PM
RANDOM__USER
23 minutes ago
Bashy chessboard problem
BR1F1SZ   1
N 34 minutes ago by Gggvds1
Source: 2024 May Olympiad L2 P1
A $4\times 8$ grid is divided into $32$ unit squares. There are square tiles of sizes $1 \times 1$, $2 \times 2$, $3 \times 3$ and $4 \times 4$. The goal is to completely cover the grid using exactly $n$ of these tiles.
[list=a]
[*]Is it possible to do this if $n = 19$?
[*]Is it possible to do this if $n = 14$?
[*]Is it possible to do this if $n = 7$?
[/list]
Note: The tiles cannot overlap or extend beyond the grid.
1 reply
BR1F1SZ
Jan 25, 2025
Gggvds1
34 minutes ago
AOPS MO problem selection committee selection
MathMaxGreat   7
N an hour ago by Jackson0423
Problems to select the one that will be in the AOPS MO problem selection committee

(All old problems but I will not tell you were)
7 replies
MathMaxGreat
an hour ago
Jackson0423
an hour ago
AOPS MO Introduce
MathMaxGreat   28
N an hour ago by Jackson0423
$AOPS MO$

Problems: post it as a private message to me or @jerryZYang, please post it in $LATEX$ and have answers

6 Problems for two rounds, easier than $IMO$

If you want to do the problems or be interested, reply ’+1’
Want to post a problem reply’+2’ and message me
Want to be in the problem selection committee, reply’+3’
28 replies
MathMaxGreat
Today at 1:04 AM
Jackson0423
an hour ago
Extremum problem
Virgil Nicula   1
N an hour ago by Mathzeus1024
Source: Crux Mathematicorum
Ascertain $\min\left\{x+y+\sqrt{x^2+y^2}\right\}$ , where $a>0$ , $b>0$ are given and

$\left\|\begin{array}{c}
x>a\ \wedge\ y>b\\\\
\frac ax\ +\ \frac by\ =\ 1\end{array}\right\|$ . Present (if is possibly) a geometrical interpretation.
1 reply
Virgil Nicula
Aug 4, 2013
Mathzeus1024
an hour ago
f(xf(y)-yf(x))=f(xy)-xf(y)
ChildFlower   1
N an hour ago by RagvaloD
Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all $x, y \in \mathbb{R}$ we have:
$$f(xf(y)-yf(x))=f(xy)-xf(y)$$
1 reply
ChildFlower
2 hours ago
RagvaloD
an hour ago
Parallel lines (extension of previous problem)
RANDOM__USER   3
N an hour ago by Lil_flip38
Source: Own
Let \(D\) be an arbitrary point on the side \(BC\) in a triangle \(\triangle{ABC}\). Let \(E\) and \(F\) be the intersection of the lines parallel to \(AC\) and \(AB\) through \(D\) with \(AB\) and \(AC\). Let \(G\) be the intersection of \((AFE)\) with \((ABC)\). Let \(M\) be the midpoint of \(BC\) and \(X\) the intersection of \(AM\) with \((ABC)\). Let \(H\) be the intersection of \((XMG)\) with \(BC\). Prove that \(EF\) is parallel to \(AH\).

IMAGE

Note: This is another property of a configuration I posted before where one needed to prove that \(X, D\) and \(G\) are collinear. There are surprisingly many properties in the configuration posted earlier :P
3 replies
RANDOM__USER
5 hours ago
Lil_flip38
an hour ago
inequalities
pennypc123456789   6
N an hour ago by ohiorizzler1434
If $a,b,c$ are positive real numbers, then
$$
\frac{a + b}{a + 7b + c} + \dfrac{b + c}{b + 7c + a}+\dfrac{c + a}{c + 7a + b} \geq \dfrac{2}{3}$$
we can generalize this problem
6 replies
pennypc123456789
Apr 17, 2025
ohiorizzler1434
an hour ago
Constructing a Sequence with Density 1/lnn
EthanWYX2009   0
an hour ago
Source: 2025 May 谜之竞赛-6
Prove that there exists a strictly increasing infinite sequence of positive integers \(\{a_n\}\) satisfying the following two conditions: [list]
[*]For any positive integers \(x \leq y < z\), the following inequality holds:
\[
\sum_{n=x}^{y} a_n \neq a_z;
\][*]There exist a positive constant \(c\) and a positive integer \(M\) such that for any integer \(m > M\), the set \(\{1, 2, \cdots, m\}\) contains at least \(\frac{cm}{\ln m}\) elements appearing in the sequence \(\{a_n\}\).
[/list]
Created by Zheng Yang
0 replies
EthanWYX2009
an hour ago
0 replies
How about an AOPS MO?
MathMaxGreat   37
N an hour ago by benjaminchew13
I am planning to make a $APOS$ $MO$, we can post new and original problems, my idea is to make an competition like $IMO$, 6 problems for 2 rounds
Any idea and plans?
37 replies
MathMaxGreat
Yesterday at 2:37 AM
benjaminchew13
an hour ago
sum(a(b+c)/a^2+2bc) <= 2
Math_addicted_ICGNA   4
N an hour ago by SunnyEvan
Let: $a,b,c >0$
Could you prove this inequality or add a condition to make it true, thank you:
$$\dfrac{a(b+c)}{a^2+2bc} + \dfrac{b(c+a)}{b^2+2ca} + \dfrac{c(a+b)}{c^2+2ab} \le 2$$
4 replies
Math_addicted_ICGNA
Jul 9, 2025
SunnyEvan
an hour ago
Moves on the blackboard
MarkBcc168   2
N an hour ago by math-olympiad-clown
Source: 2019 Baltic Way P5
The $2m$ numbers
$$1\cdot 2, 2\cdot 3, 3\cdot 4,\hdots,2m(2m+1)$$are written on a blackboard, where $m\geq 2$ is an integer. A move consists of choosing three numbers $a, b, c$, erasing them from the board and writing the single number
$$\frac{abc}{ab+bc+ca}$$After $m-1$ such moves, only two numbers will remain on the blackboard. Supposing one of these is $\tfrac{4}{3}$, show that the other is larger than $4$.
2 replies
MarkBcc168
Nov 18, 2019
math-olympiad-clown
an hour ago
number theory or algebra ?
tabel   0
an hour ago
Source: Romanian NMO 2025 9th grade shortlist
Let $n \geq 1$ be an integer and let $x_1, x_2, \ldots, x_n$ be positive integers (i.e., $x_i \in \mathbb{N}^*$ for all $i$), such that
\[
x_k \leq k \quad \text{for all } k = 1, 2, \ldots, n,
\]and
\[
x_1 + x_2 + \cdots + x_n \text{ is an odd integer}.
\]Prove that there exist $\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n \in \{1, -1\}$ such that
\[
\varepsilon_1 x_1 + \varepsilon_2 x_2 + \cdots + \varepsilon_n x_n = 1.
\]
0 replies
tabel
an hour ago
0 replies
continuous function
lolm2k   17
N Jun 6, 2025 by hung9A
Let $f: \mathbb R \rightarrow \mathbb R$ be a continuous function such that $f(f(f(x))) = x^2+1, \forall x \in \mathbb R$ show that $f$ is even.
17 replies
lolm2k
Mar 24, 2018
hung9A
Jun 6, 2025
continuous function
G H J
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lolm2k
158 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $f: \mathbb R \rightarrow \mathbb R$ be a continuous function such that $f(f(f(x))) = x^2+1, \forall x \in \mathbb R$ show that $f$ is even.
This post has been edited 1 time. Last edited by lolm2k, Mar 24, 2018, 5:52 PM
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Lamp909
98 posts
#2 • 4 Y
Y by matinyousefi, lolm2k, Adventure10, Mango247
Substituting $x=f(x)$ we get that $$f(x^{2}+1)=f(x)^{2}+1$$. Now setting $\pm x$ we get that $f(x)^{2}=f(-x)^{2}$. Since $f$ is continious we get that $f$ is either even or odd for all x. Indeed, if we assume that for some x $f(x)=f(-x)$ then for all $y$ very close to $x$ we must have that $f(x)$ is very close to $f(y)$ which means that $f(-y)$ will be very close to $f(-x)=f(x)$ but this will be impossible if $f(-y)=-f(y)$. Analoguously, for $f(x)=-f(-x)$. If $f$ is odd then $f(0)=0$. Setting $x=0$ in the equation we get $0=1$. Whence, a contradiction.
This post has been edited 3 times. Last edited by Lamp909, Mar 24, 2018, 7:04 PM
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Lamp909
98 posts
#4 • 2 Y
Y by Adventure10, Mango247
The above arguments are valid only for $ x \neq 0$ but when we get that the function is odd for nonzero $x$ it is then easy to prove it for $x=0$ by taking $x$ to converge to 0.
This post has been edited 1 time. Last edited by Lamp909, Mar 24, 2018, 6:32 PM
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lolm2k
158 posts
#5 • 2 Y
Y by Adventure10, Mango247
yep, your argument can be made for $x=1$ since $f(1)$ cannot be zero. Then extend for all reals, good solution!
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lolm2k
158 posts
#6 • 1 Y
Y by Adventure10
actually now that I thought about it nothing garantees we can extend our result to all reals, all we proved is that if f(x) = f(-x) for some number there is a ball centered in x such that everyone in there also satisfies it. I would like to see a rigorous extension of this result, which seems to be true.

Please tell me if this is ok:
you proved if $f(x) = f(-x)$ there is this open ball around $x$ such that this equality happens.
Let $A = \{ x \in \mathbb R, f(x) = f(-x) \}$ then $A$ contains and is contained in the union of balls around each element of $A$ such that $f(x) = f(-x)$. So $A$ can be thought as an union of open sets and therefore it is open.
So is $B = \{x \in \mathbb R , f(x) = -f(-x) \}$.
So we would have a contradition of the fact that $\mathbb R$ is conected. Therefore either A or B must be empty.
This post has been edited 2 times. Last edited by lolm2k, Mar 24, 2018, 8:31 PM
Reason: typo
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mkhayech
63 posts
#7 • 4 Y
Y by lolm2k, MNJ2357, Adventure10, Mango247
Lamp909 wrote:
we get that $f(x)^{2}=f(-x)^{2}$. Since $f$ is continious we get that $f$ is either even or odd for all x.
This is false. Consider the function defined $f(x)=1+x$ for $x < 0$
$ = 1 -x$ for $x$ between $0$ and $1$
$= x-1$ for $x>1$
Assume there exists a such as $f(-a)=-f(a)$ different from $0$. Then there exists $b$ such as $f(b)=f(-b)=0$. Now suppose $f(x)=0$ for some $x$ then $x^{2}+1=f(f(0))$ so $x$ is at most $2$ values hence $f(x)=0$ iff $ x=b$ or $x= -b$. This means if $|x| > b$ , $f(x)=-f(-x)$ and $f(x)=f(-x)$ otherwise. Since $f$ is continuous, we can easily prove that it's surjective, but then $f^{3}$ would be surjective as well, contradiction.
This post has been edited 2 times. Last edited by mkhayech, Mar 28, 2018, 6:20 PM
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lolm2k
158 posts
#8 • 2 Y
Y by Adventure10, Mango247
why there exists b such as f(b)=f(-b)=0 ?
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mkhayech
63 posts
#9 • 3 Y
Y by lolm2k, Adventure10, Mango247
Because f(-a)=-f(a) so one of them is negative and the other is postive, f is continous.

I am on mobile so didnt elaborate some parts, but if you try to draw what the graph of f can look like you will get the idea
This post has been edited 1 time. Last edited by mkhayech, Mar 24, 2018, 9:32 PM
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lolm2k
158 posts
#10 • 2 Y
Y by Adventure10, Mango247
i believe in most you sayed, indeed b cannot be zero so there is b and -b. and it feels pretty natural that |x| > b leads to f(x) = -f(-x) and in order to avoid f(0) =0 we must have f(x) = f(-x) inside |x|<b. I'm having trouble seeing f must be surjective tho
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MNJ2357
645 posts
#11 • 2 Y
Y by lolm2k, Adventure10
The fact that $f(a)=f(b)$ then $a^2=b^2$ (...#) finishes the problem.
If $f(0) > 0$, there exists a positive real number $\epsilon$ such that $$|x|<\epsilon \Longrightarrow f(x)>0 \Longrightarrow f(x)=f(-x)$$, then $f(x)$ is even for all real numbers $x$ (by #)
We can use the same arguement for $f(0)<0$.
This post has been edited 1 time. Last edited by MNJ2357, Mar 27, 2018, 12:17 AM
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lolm2k
158 posts
#12 • 2 Y
Y by Adventure10, Mango247
^ you're wrong, read mkhayech first answer
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mkhayech
63 posts
#13 • 3 Y
Y by lolm2k, Adventure10, Mango247
$f$ is surjective because: $f(f(f(x)))= x^{2} + 1 $ so every value bigger than $1$ is in the range.$ f(b)=0 $ and $f$ iscontinous so every positive number is in the range. Now if $c < 0$ then we can find $x< -b $such that $f(x)=-f(-x)=-(-c)=c$ (since $-c$is positive so it's in the range).
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MNJ2357
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#14 • 2 Y
Y by Adventure10, Mango247
lolm2k wrote:
^ you're wrong, read mkhayech first answer
That is why I mentioned $$f(a)=f(b) \text{  then  } a^2=b^2 $$If this doesn't answer your question, could you kindly tell me which part is wrong?
This post has been edited 1 time. Last edited by MNJ2357, Mar 25, 2018, 7:28 AM
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lolm2k
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#15 • 2 Y
Y by Adventure10, Mango247
the big mistake is that even if there is an interval (which it does exist) in which $f(x)=f(-x)$ this does NOT implies in a obvious manner that the function is even for all reals.

if $f(0)=0$ then the functio does not exist so i'm not sure why you considered this case, not that this is too relevant it is just weird you tried to prove f is even when this happens
This post has been edited 2 times. Last edited by lolm2k, Mar 27, 2018, 12:18 AM
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achen29
561 posts
#16 • 1 Y
Y by Adventure10
When substituting by x=f(x); aren't you assuming that the function has a fixed point? (I might be mistaken tho)
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lolm2k
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#17 • 2 Y
Y by Adventure10, Mango247
never did that particular substitution, only when i assumed $f(0) = 0$ to prove this leads to contradiction
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MNJ2357
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#18 • 3 Y
Y by lolm2k, Adventure10, Mango247
lolm2k wrote:
the big mistake is that even if there is an interval (which it does exist) in which $f(x)=f(-x)$ this does not implies in a obvious manner that the function is even for all reals.

if $f(0)=0$ then the function does not exist so i'm not sure why you considered this case, not that this is too relevant it is just weird you tried to prove f is even when this happens

I didn't realize that there are no functions $f$ such that $f(0)=0$. Fixed!! :D
And if $f$ increases for $x \in (0,\epsilon)$, $f$ increases for all $x>0$, or else because of continuity, there will be two positive reals $a,b$ such that $f(a)=f(b)$. Do the same thing for $x<0$, and we're done.
If my post isn't clear, here is another solution:

Since there are no solutions for $f(x) \in \{x,-x\}$
(if then $x^2+1=f(f(f(x))) \in \{x,-x\}$)
If $f(0)>0 \Longrightarrow f(x)>|x| \forall x\in \mathbb{R}$ so $f$ is even.
same for $f(0)<0$, so we are done!!
This post has been edited 3 times. Last edited by MNJ2357, Mar 27, 2018, 2:16 AM
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hung9A
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#19
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Note that $f(x^2 + 1) = f(f(f(f(x)))) = f(x)^2 + 1, \forall x \in \mathbb{R}$, so $f(x)^2 = f(-x)^2, \forall x \in \mathbb{R}$. In addition, if $f(a) = f(b)$ for some $a, b \in \mathbb{R}$ then $a^2 + 1 = f(f(f(a))) = f(f(f(b))) = b^2 + 1$, so $a^2 = b^2$.

Consider the two sets
$$A = \{x \in \mathbb{R}: f(x) \neq f(-x)\}, B = \{x \in \mathbb{R}: f(x) \neq -f(-x)\}.$$Assume that both sets are non-empty. Then there exists $a, b \geq 0$ such that $a \in A, b \in B$. Let $I$ be the closed segment with endpoints $a, b$, and $J$ be the closed segment with endpoints $-a, -b$. Then $f$ is injective on $I$ and injective on $J$, so $f$ is monotonic on $I$ and on $J$.
Now if $f(a)f(b) > 0$ then $f(-a)f(-b) < 0$. This means that $f$ has a zero on $J$ but not on $I$, but this cannot happen since $f(x)^2 = f(-x)^2, \forall x \in \mathbb{R}$. We get a similar contradiction if $f(a)f(b) < 0$.

Therefore either $A$ or $B$ is non-empty, so $f$ is either odd or even. Since $f(f(f(x))) = x^2 + 1$ is an even function, $f$ is also an even function.
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