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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)!

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[*]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]

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0 replies
jwelsh
Jul 1, 2025
0 replies
AOPS MO Introduce
MathMaxGreat   3
N a minute ago by BR1F1SZ
$AOPS MO$

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’
3 replies
MathMaxGreat
2 hours ago
BR1F1SZ
a minute ago
How about an AOPS MO?
MathMaxGreat   33
N 7 minutes ago by Bread10
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?
33 replies
MathMaxGreat
Yesterday at 2:37 AM
Bread10
7 minutes ago
Interesting Spiral
VitaPretor   2
N 9 minutes ago by Bread10
a) We start at $(0,0)$ and walk $400$ feet north and turn $90$ degrees to the right.
We then walk 75% of $400$ or $300$ feet east and turn $90$ degrees to the right.
We next walk 75% of $300$ or $225$ feet south and turn $90$ degrees to the right.
We repeat the process indefinitely of walking 75% of the distance that we last walked and turning $90$ degrees to the right forming a spiral. What are the exact coordinates we approach after repeating the process indefinitely?

b) We start at $(0,0)$ and walk $x$ feet north and turn $90$ degrees to the right.
We then walk $x * y$ feet to the east and turn $90$ degrees to the right.
We next walk $x * y^2$ feet south and turn $90$ degrees to the right.
We repeat the process indefinitely of walking $y$ times the distance that we last walked and turning $90$ degrees to the right forming a spiral.
Assume $x$ > $0$ and $0$ < $y$ < $1$. If the point we eventually approach is $(50,60)$ find the ordered pair $(x,y)$.
2 replies
VitaPretor
Yesterday at 12:45 AM
Bread10
9 minutes ago
Easy with 3 var and parameter
mihaig   1
N an hour ago by pooh123
Source: Own
Find the smallest real constant $K$ such that
$$18+3abc\geq7\left(ab+bc+ca\right)$$for all $a,b,c\geq K$ satisfying $a+b+c=3.$
1 reply
mihaig
Yesterday at 6:23 AM
pooh123
an hour ago
No more topics!
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
158 posts
#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
645 posts
#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
50 posts
#19
Y by
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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