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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Easy P4 combi game with nt flavour
Maths_VC   1
N 3 hours ago by p.lazarov06
Source: Serbia JBMO TST 2025, Problem 4
Two players, Alice and Bob, play the following game, taking turns. In the beginning, the number $1$ is written on the board. A move consists of adding either $1$, $2$ or $3$ to the number written on the board, but only if the chosen number is coprime with the current number (for example, if the current number is $10$, then in a move a player can't choose the number $2$, but he can choose either $1$ or $3$). The player who first writes a perfect square on the board loses. Prove that one of the players has a winning strategy and determine who wins in the game.
1 reply
Maths_VC
May 27, 2025
p.lazarov06
3 hours ago
Central sequences
EeEeRUT   14
N 3 hours ago by HamstPan38825
Source: EGMO 2025 P2
An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$.

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots, $ there are infinitely many positive integers $n$ with $a_n = b_n$.
14 replies
EeEeRUT
Apr 16, 2025
HamstPan38825
3 hours ago
Elementary Problems Compilation
Saucepan_man02   32
N 4 hours ago by atdaotlohbh
Could anyone send some elementary problems, which have tricky and short elegant methods to solve?

For example like this one:
Solve over reals: $$a^2 + b^2 + c^2 + d^2  -ab-bc-cd-d +2/5=0$$
32 replies
Saucepan_man02
May 26, 2025
atdaotlohbh
4 hours ago
Random Points = Problem
kingu   5
N 4 hours ago by happypi31415
Source: Chinese Geometry Handout
Let $ABC$ be a triangle. Let $\omega$ be a circle passing through $B$ intersecting $AB$ at $D$ and $BC$ at $F$. Let $G$ be the intersection of $AF$ and $\omega$. Further, let $M$ and $N$ be the intersections of $FD$ and $DG$ with the tangent to $(ABC)$ at $A$. Now, let $L$ be the second intersection of $MC$ and $(ABC)$. Then, prove that $M$ , $L$ , $D$ , $E$ and $N$ are concyclic.
5 replies
+1 w
kingu
Apr 27, 2024
happypi31415
4 hours ago
No more topics!
Unexpected FE
Taco12   18
N May 7, 2025 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
May 7, 2025
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
872 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
1926 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
872 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
1557 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
4191 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
6882 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
799 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
Y by
Solution
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eibc
600 posts
#12
Y by
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
167 posts
#13
Y by
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
1417 posts
#14
Y by
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
Y by
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
636 posts
#16
Y by
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
Y by
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
674 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
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#21
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Solution
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