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EeEeRUT   4
N 15 minutes ago by lksb
Source: Thailand MO 2025 P6
Find all function $f: \mathbb{R}^+ \rightarrow \mathbb{R}$,such that the inequality $$f(x) + f\left(\frac{y}{x}\right) \leqslant \frac{x^3}{y^2} + \frac{y}{x^3}$$holds for all positive reals $x,y$ and for every positive real $x$, there exist positive reals $y$, such that the equality holds.
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EeEeRUT
Wednesday at 6:45 AM
lksb
15 minutes ago
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problem about equation
jred   2
N an hour ago by Truly_for_maths
Source: China south east mathematical Olympiad 2006 problem4
Given any positive integer $n$, let $a_n$ be the real root of equation $x^3+\dfrac{x}{n}=1$. Prove that
(1) $a_{n+1}>a_n$;
(2) $\sum_{i=1}^{n}\frac{1}{(i+1)^2a_i} <a_n$.
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jred
Jul 4, 2013
Truly_for_maths
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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
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Miquel-point
Apr 6, 2025
lksb
May 13, 2025
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Polynomial divisible by x^2+1
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Source: Romanian IMO TST 1981, P1 Day 1
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Miquel-point
491 posts
Miquel-point
#1 h  Apr 6, 2025, 6:13 PM • 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
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luutrongphuc
#2 h  May 12, 2025, 4:03 PM
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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
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lksb
#3 h  May 13, 2025, 2:40 AM
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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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