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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
Number theory
MathsII-enjoy   0
7 minutes ago
Prove that when $x^p+y^p$ | $(p^2-1)^n$ with $x,y$ are positive integers and $p$ is prime ($p>3$), we get: $x=y$
0 replies
MathsII-enjoy
7 minutes ago
0 replies
F.E....can you solve it?
Jackson0423   1
N 11 minutes ago by MathsII-enjoy
Find all functions \( f : \mathbb{R} \to \mathbb{R} \) such that
\[
f\left(\frac{x^2 - f(x)}{f(x) - 1}\right) = x
\]for all real numbers \( x \) satisfying \( f(x) \neq 1 \).
1 reply
Jackson0423
2 hours ago
MathsII-enjoy
11 minutes ago
IMO Genre Predictions
ohiorizzler1434   41
N 11 minutes ago by DottedCaculator
Everybody, with IMO upcoming, what are you predictions for the problem genres?


Personally I predict: predict
41 replies
ohiorizzler1434
May 3, 2025
DottedCaculator
11 minutes ago
CooL geo
Pomegranat   1
N 15 minutes ago by Pomegranat
Source: Idk

In triangle \( ABC \), \( D \) is the midpoint of \( BC \). \( E \) is an arbitrary point on \( AC \). Let \( S \) be the intersection of \( AD \) and \( BE \). The line \( CS \) intersects with the circumcircle of \( ACD \), for the second time at \( K \). \( P \) is the circumcenter of triangle \( ABE \). Prove that \( PK \perp CK \).
1 reply
Pomegranat
Today at 5:57 AM
Pomegranat
15 minutes ago
Convergent series with weight becomes divergent
P_Fazioli   3
N 2 hours ago by solyaris
Initially, my problem was : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ?

Thinking about the continuous case : if $g:\mathbb{R}_+\longrightarrow\mathbb{R}$ is continuous, positive with $g(x)\underset{{x}\longrightarrow{+\infty}}\longrightarrow +\infty$, does $f$ continuous and positive exist on $\mathbb{R}_+$ such that $\displaystyle\int_0^{+\infty}f(x)\text{d}x$ converges and $\displaystyle\int_0^{+\infty}f(x)g(x)\text{d}x$ diverges ?

To the last question, the answer seems to be yes if $g$ is in the $\mathcal{C}^1$ class, increasing : I chose $f=\dfrac{g'}{g^2}$. With this idea, I had the idea to define $a_n=\dfrac{b_{n+1}-b_n}{b_n^2}$ but it is not clear that it is ok, even if $(b_n)$ is increasing.

Now I have some questions !

1) The main problem : is it true that if we fix $(b_n)$ positive such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$, then there exists $(a_n)$ positive such that $\displaystyle\sum_{n\geq 0}a_n$ converges and $\displaystyle\sum_{n\geq 0}a_nb_n$ diverges ? And if $(b_n)$ is increasing ?
2) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\left(\frac{b_{n+1}}{b_n}-1\right)$ diverges ?
3) is it true that if we fix $(b_n)$ positive increasing such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}\underset{{n}\longrightarrow{+\infty}}\longrightarrow 1$, then $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ converges ?
4) if $(b_n)$ is positive increasing and such that $b_n\underset{{n}\longrightarrow{+\infty}}\longrightarrow +\infty$
and $\frac{b_{n+1}}{b_n}$ does not converge to $1$, can $\displaystyle\sum_{n\geq 0}\frac{b_{n+1}-b_n}{b_n^2}$ diverge ?
5) for the continuous case, is it true if we suppose $g$ only to be continuous ?

3 replies
P_Fazioli
Today at 5:37 AM
solyaris
2 hours ago
Putnam 1954 B1
sqrtX   6
N 3 hours ago by AshAuktober
Source: Putnam 1954
Show that the equation $x^2 -y^2 =a^3$ has always integral solutions for $x$ and $y$ whenever $a$ is a positive integer.
6 replies
sqrtX
Jul 17, 2022
AshAuktober
3 hours ago
Cube Colouring Problems
Saucepan_man02   1
N 5 hours ago by removablesingularity
Could anyone kindly post some problems (and hopefully along the solution thread/final answer) related to combinatorial colouring of cube?
1 reply
Saucepan_man02
May 3, 2025
removablesingularity
5 hours ago
D1026 : An equivalent
Dattier   4
N Today at 4:56 AM by 3ch03s
Source: les dattes à Dattier
Let $u_0=1$ and $\forall n \in \mathbb N, u_{2n+1}=\ln(1+u_{2n}), u_{2n+2}=\sin(u_{2n+1})$.

Find an equivalent of $u_n$.
4 replies
Dattier
May 3, 2025
3ch03s
Today at 4:56 AM
Putnam 1947 B1
sqrtX   3
N Yesterday at 5:38 PM by Levieee
Source: Putnam 1947
Let $f(x)$ be a function such that $f(1)=1$ and for $x \geq 1$
$$f'(x)= \frac{1}{x^2 +f(x)^{2}}.$$Prove that
$$\lim_{x\to \infty} f(x)$$exists and is less than $1+ \frac{\pi}{4}.$
3 replies
sqrtX
Apr 3, 2022
Levieee
Yesterday at 5:38 PM
Double Sum
P162008   2
N Yesterday at 3:01 PM by Etkan
Evaluate $\sum_{a=1}^{\infty} \sum_{b=1}^{a + 2} \frac{1}{ab(a + 2)}.$
2 replies
P162008
Yesterday at 12:19 PM
Etkan
Yesterday at 3:01 PM
Summation
Saucepan_man02   4
N Yesterday at 2:47 PM by Etkan
If $P = \sum_{r=1}^{50} \sum_{k=1}^{r} (-1)^{r-1} \frac{\binom{50}{r}}{k}$, then find the value of $P$.

Ans
4 replies
Saucepan_man02
May 3, 2025
Etkan
Yesterday at 2:47 PM
Trigo + Series
P162008   0
Yesterday at 11:59 AM
$F(r,x) = \tan\left(\frac{\pi}{3} + 3^r x\right)$ and $G(r,x) = \cot\left(\frac{\pi}{6} + 3^r x\right)$

Let $A = \sum_{r=0}^{24} \frac{d(F(r,x))}{dx} \left[\frac{1}{F(r,x) + \frac{1}{F(r,x)}}\right]$

$B = \sum_{r=0}^{24} \frac{-d(G(r,x))}{dx} \left[\frac{1}{G(r,x) + \frac{1}{G(r,x)}}\right]$

$P = \prod_{r=0}^{24} F(r,x), Q = \prod_{r=0}^{24} G(r,x)$

and, $R = \frac{A - B}{\tan x\left[PQ - \frac{1}{3^{25}}\right]}$ where $x \in R$

Find the remainder when $R^{2025}$ is divided by $11.$
0 replies
P162008
Yesterday at 11:59 AM
0 replies
Integration Bee in Czechia
Assassino9931   2
N Yesterday at 9:35 AM by pi_quadrat_sechstel
Source: Vojtech Jarnik IMC 2025, Category II, P3
Evaluate the integral $\int_0^{\infty} \frac{\log(x+2)}{x^2+3x+2}\mathrm{d}x$.
2 replies
Assassino9931
May 2, 2025
pi_quadrat_sechstel
Yesterday at 9:35 AM
Find all continuous functions
bakkune   3
N Yesterday at 3:58 AM by bakkune
Source: Own
Find all continuous function $f, g\colon\mathbb{R}\to\mathbb{R}$ satisfied
$$
(x - k)f(x) = \int_k^x g(y)\mathrm{d}y 
$$for all $x\in\mathbb{R}$ and all $k\in\mathbb{Z}$.
3 replies
bakkune
May 3, 2025
bakkune
Yesterday at 3:58 AM
Integer polynomials
GeoKing   1
N Apr 10, 2025 by eulerleonhardfan
Let f(x) be a monic polynomial of degree n with integer coefficients, and let d1, · · · , dn be pairwise distinct integers. Suppose that for infinitely many prime numbers p there exists an integer kp for which f(kp + d1) ≡ f(kp + d2) ≡ · · · f(kp + dn) ≡ 0 (mod p). Prove that there exists an integer k0 such that f(k0 + d1) = f(k0 + d2) =· · · = f(k0 + dn) = 0
1 reply
GeoKing
Aug 2, 2021
eulerleonhardfan
Apr 10, 2025
Integer polynomials
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GeoKing
518 posts
#1
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Let f(x) be a monic polynomial of degree n with integer coefficients, and let d1, · · · , dn be pairwise distinct integers. Suppose that for infinitely many prime numbers p there exists an integer kp for which f(kp + d1) ≡ f(kp + d2) ≡ · · · f(kp + dn) ≡ 0 (mod p). Prove that there exists an integer k0 such that f(k0 + d1) = f(k0 + d2) =· · · = f(k0 + dn) = 0
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eulerleonhardfan
53 posts
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Credits to gghx for this solution.

Let $f(x) = x^n + a_{n-1}x^{n-1} + \ldots + a_0$.
For a prime $p$ where $p \not \vert n$, if $k_p$ exists, then $$f(x) \equiv (x-k_p-d_1) \ldots (x-k_p-d_n) \pmod p$$and by Vieta's formula, we have $$a_{n-1} \equiv nk_p + \sum_{i=1}^nd_i \pmod p$$$$\Rightarrow k_p \equiv \frac{a_{n-1}-\sum_{i=1}^nd_i}{n} \pmod p$$so taking $k_0 = \frac{a_{n-1}-\sum_{i=1}^nd_i}{n}$, we have $k_p \equiv k_n \pmod p$, so $f(k_0 + d_i) \equiv f(k_p + d_i) \equiv 0 \pmod p$ for infinitely many $p$, i.e. $f(k_0+d_i) = 0$.
By the integral root theorem, all the rational roots of $f$ are integers. Since $k_0+d_i \in \mathbb{Q}$ are the roots of $f$, it follows that $k_0$ is an integer, as desired.
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