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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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0 replies
jwelsh
Jul 1, 2025
0 replies
Walk on coordinate plane
Aiden-1089   2
N 16 minutes ago by YaoAOPS
Source: APMO 2025 Problem 2
Let $\alpha$ and $\beta$ be positive real numbers. Emerald makes a trip in the coordinate plane, starting off from the origin $(0,0)$. Each minute she moves one unit up or one unit to the right, restricting herself to the region $| x-y | <2025$, in the coordinate plane. By the time she visits a point $(x,y)$ she writes down the integer $\lfloor x\alpha + y\beta \rfloor$ on it. It turns out that Emerald wrote each non-negative integer exactly once. Find all the possible pairs $(\alpha, \beta)$ for which such a trip would be possible.
2 replies
Aiden-1089
Today at 3:37 PM
YaoAOPS
16 minutes ago
Polynomial NT with analytic flavour?
Aiden-1089   10
N 21 minutes ago by YaoAOPS
Source: APMO 2025 Problem 3
Let $P(x)$ be a non-constant polynomial with integer coefficients such that $P(0) \neq 0$. Let $a_1, a_2, a_3, \dots$ be an infinite sequence of integers such that $P(i - j)$ divides $a_i-a_j$ for all distinct positive integers $i,j$. Prove that the sequence $a_1, a_2, a_3, \dots$ must be constant, that is, $a_n$ equals a constant $c$ for all positive integers $n$.
10 replies
Aiden-1089
Today at 3:40 PM
YaoAOPS
21 minutes ago
Finite Process with Pairwise Non-Coprime Increments
jungle_wang   1
N 35 minutes ago by megarnie
Source: 2025 IMOC
Let $n$ be a positive integer with $n>1$. Initially, there are $n$ positive integers. In each move, Chien-Chin may choose a number from the $n$ numbers that is not coprime with every other number, and increase it by $1$.
Prove that Chien-Chin can make only finitely many moves.

1 reply
jungle_wang
Today at 1:47 PM
megarnie
35 minutes ago
Number theory-Very hard conjucture
slimshady360   4
N an hour ago by slimshady360
I'm making this conjucture on a problem i'm solving but i can't find a pattern ...
4 replies
slimshady360
an hour ago
slimshady360
an hour ago
Shaded Squares in k Rows Occupy k Columns
Viliciri   0
3 hours ago
A rectangular grid has $m$ rows and $n$ columns, with $1 < m \leq n$. At least $m$ squares of this grid are shaded such that for all $1 \leq k \leq m$, the collection of shaded squares in any $k$ rows of this grid together occupy at least $k$ columns. For example, the following grid is not a valid shading because the shaded cells in three rows -- the $1$st, $3$rd, and $4$th rows -- only occupy two columns, namely columns $2$ and $3$.
IMAGE
Prove that it is possible to choose $1$ shaded square from each of the $m$ rows in this grid such that no two chosen squares lie in the same column.
0 replies
Viliciri
3 hours ago
0 replies
10 Problems
Sedro   60
N 4 hours ago by Sedro
Title says most of it. I've been meaning to post a problem set on HSM since at least a few months ago, but since I proposed the most recent problems I made to the 2025 SSMO, I had to wait for that happen. (Hence, most of these problems will probably be familiar if you participated in that contest, though numbers and wording may be changed.) The problems are very roughly arranged by difficulty. Enjoy!

Problem 1: An sequence of positive integers $u_1, u_2, \dots, u_8$ has the property for every positive integer $n\le 8$, its $n^\text{th}$ term is greater than the mean of the first $n-1$ terms, and the sum of its first $n$ terms is a multiple of $n$. Let $S$ be the number of such sequences satisfying $u_1+u_2+\cdots + u_8 = 144$. Compute the remainder when $S$ is divided by $1000$.

Problem 2 (solved by fruitmonster97): Rhombus $PQRS$ has side length $3$. Point $X$ lies on segment $PR$ such that line $QX$ is perpendicular to line $PS$. Given that $QX=2$, the area of $PQRS$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 3 (solved by Math-lover1): Positive integers $a$ and $b$ satisfy $a\mid b^2$, $b\mid a^3$, and $a^3b^2 \mid 2025^{36}$. If the number of possible ordered pairs $(a,b)$ is equal to $N$, compute the remainder when $N$ is divided by $1000$.

Problem 4 (solved by CubeAlgo15): Let $ABC$ be a triangle. Point $P$ lies on side $BC$, point $Q$ lies on side $AB$, and point $R$ lies on side $AC$ such that $PQ=BQ$, $CR=PR$, and $\angle APB<90^\circ$. Let $H$ be the foot of the altitude from $A$ to $BC$. Given that $BP=3$, $CP=5$, and $[AQPR] = \tfrac{3}{7} \cdot [ABC]$, the value of $BH\cdot CH$ can be expressed as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Compute $m+n$.

Problem 5 (solved by maromex): Anna has a three-term arithmetic sequence of integers. She divides each term of her sequence by a positive integer $n>1$ and tells Bob that the three resulting remainders are $20$, $52$, and $R$, in some order. For how many values of $R$ is it possible for Bob to uniquely determine $n$?

Problem 6 (solved by Mathsll-enjoy): There is a unique ordered triple of positive reals $(x,y,z)$ satisfying the system of equations \begin{align*} x^2 + 9 &= (y-\sqrt{192})^2 + 4 \\ y^2 + 4 &= (z-\sqrt{192})^2 + 49 \\ z^2 + 49 &= (x-\sqrt{192})^2 + 9. \end{align*}The value of $100x+10y+z$ can be expressed as $p\sqrt{q}$, where $p$ and $q$ are positive integers such that $q$ is square-free. Compute $p+q$.

Problem 7 (solved by sami1618): Let $S$ be the set of all monotonically increasing six-term sequences whose terms are all integers between $0$ and $6$ inclusive. We say a sequence $s=n_1, n_2, \dots, n_6$ in $S$ is symmetric if for every integer $1\le i \le 6$, the number of terms of $s$ that are at least $i$ is $n_{7-i}$. The probability that a randomly chosen element of $S$ is symmetric is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Compute $p+q$.

Problem 8: For a positive integer $n$, let $r(n)$ denote the value of the binary number obtained by reading the binary representation of $n$ from right to left. Find the smallest positive integer $k$ such that the equation $n+r(n)=2k$ has at least ten positive integer solutions $n$.

Problem 9 (solved by Math-lover1, sami1618): Let $p$ be a quadratic polynomial with a positive leading coefficient. There exists a positive real number $r$ such that $r < 1 < \tfrac{5}{2r} < 5$ and $p(p(x)) = x$ for $x \in \{ r,1,  \tfrac{5}{2r} , 5\}$. Compute $p(20)$.

Problem 10 (solved by aaravdodhia, sami1618): Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=995$ and $ab+bc+ca$ is a multiple of $995$.
60 replies
Sedro
Jul 10, 2025
Sedro
4 hours ago
f^(f(n))(n)=n over [10], # of solutions (ARML 2021 I-8)
jasperE3   4
N 4 hours ago by torch
For a given function $f$, define $f^1(x)=f(x)$, and for $k\ge2$, define
$$f^k(x)=f(f^{k-1}(x)).$$For example, $f^3(x)=f(f(f(x)))$. Of the $10!$ functions $g$ whose domain and range are the set $S=\{1,2,3,\ldots,10\}$, compute the number of functions such that $g^{g(n)}(n)=n$ for all $n\in S$.
4 replies
jasperE3
Jun 7, 2021
torch
4 hours ago
Beautiful geo
justalonelyguy   5
N 4 hours ago by sami1618
Let \( \triangle ABC \) be an acute triangle with \( AB < AC \). The internal angle bisectors of angles \( \angle ABC \) and \( \angle ACB \) intersect at point \( I \). Let \( D \) be the feet of the perpendicular from \( I \) to \( BC \). Let points \( K \) and \( L \) lie on segments \( IB \) and \( IC \), respectively, such that \( \angle KDL = 90^\circ \). Prove that \[
    \angle KAL = \frac{1}{2} \angle BAC.
    \]
5 replies
justalonelyguy
Jul 28, 2025
sami1618
4 hours ago
Geometry Problem
Rice_Farmer   6
N 4 hours ago by Rice_Farmer
Let $w_1$ ad $w_2$ be two circles intersecting at $P$ and $Q.$ The tangent like closer to $Q$ touches $w_1$ and $w_2$ at $M$ and $N$ respectively. If $PQ=3,NQ=2,$ and $MN=PN,$ find $QM.$

hint
6 replies
Rice_Farmer
Today at 3:09 AM
Rice_Farmer
4 hours ago
Challenge: Make as many positive integers from 2 zeros
Biglion   48
N 5 hours ago by huajun78
How many positive integers can you make from at most 2 zeros, any math operation and cocatination?
New Rule: The successor function can only be used at most 3 times per number
Starting from 0, 0=0
48 replies
Biglion
Jul 2, 2025
huajun78
5 hours ago
If OAB and OAC share equal angles and sides, why aren't they congruent?
Merkane   1
N 6 hours ago by nudinhtien

Problem 1.39 (CGMO 2012/5). Let ABC be a triangle. The incircle of ABC is tangent
to AB and AC at D and E respectively. Let O denote the circumcenter of BCI .
Prove that ∠ODB = ∠OEC. Hints: 643 89 Sol: p.241

While I have solved the problem, I encountered a step that seems logically sound but leads to a contradiction, and I would like help identifying the flaw.

Here is the reasoning I followed:

The quadrilateral ABOC is cyclic.

OB = OC.

∠OAB = ∠OCB.
Similarly, ∠OAC = ∠OBC.

From symmetry and the above, it seems that ∠OAB = ∠OAC.

Since OA is a shared side, I concluded that triangle OAB ≅ triangle OAC.


But clearly, OAB and OAC are not congruent.
Where exactly is the logical error in this argument?
1 reply
Merkane
Today at 4:33 AM
nudinhtien
6 hours ago
Limit of a sequence defined by recurrence and exponential factor
JackMinhHieu   2
N Today at 3:02 PM by JackMinhHieu
Hi everyone,

I came across this interesting sequence defined recursively and wanted to explore its behavior.

Let the sequence (x_n) be defined by:
    x_1 = sqrt(2)
    x_{n+1} = sqrt(2*x_n / (1 + x_n)) for all n >= 1

Define another sequence:
    y_n = 4^n * (1 - 1 / x_n^2)

Question:
Does the sequence (y_n) converge? If so, what is its limit?

Any ideas, observations, or rigorous arguments would be very welcome. Thanks in advance!
2 replies
JackMinhHieu
Today at 9:31 AM
JackMinhHieu
Today at 3:02 PM
Crazy Dice
radioactiverascal90210   2
N Today at 2:59 PM by radioactiverascal90210
A pair of crazy dice are two cubes labeled with integers such that they are not labeled with the same numbers as an ordinary pair of dice , but the probability of rolling any number with the pair of crazy dice is the same as
rolling it with ordinary dice. Find a pair of crazy dice.
2 replies
radioactiverascal90210
Today at 8:16 AM
radioactiverascal90210
Today at 2:59 PM
S(n)=24,S(n^2)=9
whwlqkd   2
N Today at 2:54 PM by whwlqkd
Does there exist positive integer $n$ such that $S(n)=24$ and $S(n^2)=9$ where $S(n)$ is the digit sum of $n$?. I dont even know whether it is unsolved problem or not. And what about $S(n)=3k$,$S(n^2)=9$?
2 replies
whwlqkd
Today at 5:41 AM
whwlqkd
Today at 2:54 PM
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
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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
645 posts
#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
Reason: NOT
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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
158 posts
#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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