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Find the value
sqing   18
N 14 minutes ago by Yiyj
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
18 replies
sqing
Jun 22, 2024
Yiyj
14 minutes ago
find question
mathematical-forest   5
N an hour ago by Jupiterballs
Are there any contest questions that seem simple but are actually difficult? :-D
5 replies
mathematical-forest
Thursday at 10:19 AM
Jupiterballs
an hour ago
Own made functional equation
Primeniyazidayi   10
N an hour ago by Phat_23000245
Source: own(probably)
Find all functions $f:R \rightarrow R$ such that $xf(x^2+2f(y)-yf(x))=f(x)^3-f(y)(f(x^2)-2f(x))$ for all $x,y \in \mathbb{R}$
10 replies
+1 w
Primeniyazidayi
May 26, 2025
Phat_23000245
an hour ago
Tough inequality
TUAN2k8   4
N an hour ago by Phat_23000245
Source: Own
Let $n \ge 2$ be an even integer and let $x_1,x_2,...,x_n$ be real numbers satisfying $x_1^2+x_2^2+...+x_n^2=n$.
Prove that
$\sum_{1 \le i < j \le n} \frac{x_ix_j}{x_i^2+x_j^2+1} \ge \frac{-n}{6}$
4 replies
TUAN2k8
May 28, 2025
Phat_23000245
an hour ago
Guess period of function
a1267ab   9
N 2 hours ago by HamstPan38825
Source: USA TST 2025
Let $n$ be a positive integer. Ana and Banana play a game. Banana thinks of a function $f\colon\mathbb{Z}\to\mathbb{Z}$ and a prime number $p$. He tells Ana that $f$ is nonconstant, $p<100$, and $f(x+p)=f(x)$ for all integers $x$. Ana's goal is to determine the value of $p$. She writes down $n$ integers $x_1,\dots,x_n$. After seeing this list, Banana writes down $f(x_1),\dots,f(x_n)$ in order. Ana wins if she can determine the value of $p$ from this information. Find the smallest value of $n$ for which Ana has a winning strategy.

Anthony Wang
9 replies
1 viewing
a1267ab
Dec 14, 2024
HamstPan38825
2 hours ago
Inequality with abc=1
tenplusten   11
N 2 hours ago by sqing
Source: JBMO 2011 Shortlist A7
$\boxed{\text{A7}}$ Let $a,b,c$ be positive reals such that $abc=1$.Prove the inequality $\sum\frac{2a^2+\frac{1}{a}}{b+\frac{1}{a}+1}\geq 3$
11 replies
tenplusten
May 15, 2016
sqing
2 hours ago
Central sequences
EeEeRUT   13
N 3 hours ago by v_Enhance
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 $a_n = b_n$.
13 replies
EeEeRUT
Apr 16, 2025
v_Enhance
3 hours ago
Interesting inequality
sqing   0
3 hours ago
Source: Own
Let $ a,b,c\geq  0 , a^2+b^2+c^2 =3.$ Prove that
$$ a^4+ b^4+c^4+6abc\leq9$$$$ a^3+ b^3+  c^3+3( \sqrt{3}-1)abc\leq 3\sqrt 3$$
0 replies
sqing
3 hours ago
0 replies
IMO Shortlist 2014 C7
hajimbrak   19
N 3 hours ago by quantam13
Let $M$ be a set of $n \ge 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $AB$ and $CD$ sharing a common interior point and replace them by the segments $AC$ and $BD$ if none of them is present at this moment. Prove that it is impossible to perform $n^3 /4$ or more such moves.

Proposed by Vladislav Volkov, Russia
19 replies
hajimbrak
Jul 11, 2015
quantam13
3 hours ago
<BAC = 2 <ABC wanted, AC + AI = BC given , incenter I
parmenides51   3
N 4 hours ago by LeYohan
Source: 2020 Dutch IMO TST 1.1
In acute-angled triangle $ABC, I$ is the center of the inscribed circle and holds $| AC | + | AI | = | BC |$. Prove that $\angle BAC = 2 \angle ABC$.
3 replies
parmenides51
Nov 21, 2020
LeYohan
4 hours ago
China South East Mathematical Olympiad 2014 Q3B
sqing   5
N 4 hours ago by MathLuis
Source: China Zhejiang Fuyang , 27 Jul 2014
Let $p$ be a primes ,$x,y,z $ be positive integers such that $x<y<z<p$ and $\{\frac{x^3}{p}\}=\{\frac{y^3}{p}\}=\{\frac{z^3}{p}\}$.
Prove that $(x+y+z)|(x^5+y^5+z^5).$
5 replies
sqing
Aug 17, 2014
MathLuis
4 hours ago
Polynomial divisible by x^2+1
Miquel-point   2
N May 13, 2025 by lksb
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
2 replies
Miquel-point
Apr 6, 2025
lksb
May 13, 2025
Polynomial divisible by x^2+1
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G H BBookmark kLocked kLocked NReply
Source: Romanian IMO TST 1981, P1 Day 1
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Miquel-point
499 posts
#1 • 1 Y
Y by PikaPika999
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
This post has been edited 1 time. Last edited by Miquel-point, Apr 6, 2025, 6:17 PM
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luutrongphuc
58 posts
#2
Y by
Let $Q(x)=-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)$
It is suffices to prove that $Q(i)=0$ and $Q(-i)=0$
$\textbf{Claim 1:}     p \equiv 1 \pmod{4}$
We have: $P\left(x^{p^k}\right)= \frac{x^{p^{k+1}}-1}{x^{p^k}-1}$
Then $P\left((-i)^{p^k}\right)=P\left(i^{p^k}\right) =\frac{i^{p^{k+1}}-1}{i^{p^k}-1}=1$
So that $Q(i)=Q(-i)=0$. $\blacksquare$
$\textbf{Claim 1:}     p \equiv 3 \pmod{4}$
Then $\begin{cases}
3^{k+1} \equiv 1 \pmod{4}, & \text{if } 2 \mid k \\
3^{k} \equiv 3 \pmod{4}, & \text{if } 2 \nmid k
\end{cases}$
So, for $k$ is even, we have: $P\left(i^{p^k}\right)=P(i)=\frac{i^{p}-1}{i-1}=- \frac{i+1}{i-1}=i$
for $k$ is odd, we have: $P\left(i^{p^k}\right)=P(-i)=-i$
So : $Q(i) =-1 +(-i)^{\frac{n}{2}}=-1+1=0$
Similary, $Q(-i)=0$. $\blacksquare$
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lksb
183 posts
#3
Y by
The product telescopes to $\frac{x^{p^n}-1}{x-1}$, therefore, all left to do is prove that $x^{p^n}-1\equiv x-1\pmod{x^2+1}$, which is trivial by analizing $x^k-1\pmod{x^2+1}$ for $k$ modulo $4$
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