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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.

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[*]April 22nd (Webinar), 4:00pm PT/7:00pm ET, Competitive Programming at AoPS (USACO).[/list]
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0 replies
jlacosta
Apr 2, 2025
0 replies
Congruence related perimeter
egxa   4
N 14 minutes ago by Geometrineq
Source: All Russian 2025 9.8 and 10.8
On the sides of triangle \( ABC \), points \( D_1, D_2, E_1, E_2, F_1, F_2 \) are chosen such that when going around the triangle, the points occur in the order \( A, F_1, F_2, B, D_1, D_2, C, E_1, E_2 \). It is given that
\[
AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2.
\]Prove that the perimeters of the triangles formed by the lines \( AD_1, BE_1, CF_1 \) and \( AD_2, BE_2, CF_2 \) are equal.
4 replies
egxa
Apr 18, 2025
Geometrineq
14 minutes ago
Killer NT that nobody solved (also my hardest NT ever created)
mshtand1   8
N 14 minutes ago by mshtand1
Source: Ukraine IMO 2025 TST P8
A positive integer number \( a \) is chosen. Prove that there exists a prime number that divides infinitely many terms of the sequence \( \{b_k\}_{k=1}^{\infty} \), where
\[
b_k = a^{k^k} \cdot 2^{2^k - k} + 1.
\]
Proposed by Arsenii Nikolaev and Mykhailo Shtandenko
8 replies
mshtand1
Apr 19, 2025
mshtand1
14 minutes ago
Value of the sum
fermion13pi   1
N 30 minutes ago by RagvaloD
Source: Australia
Calculate the value of the sum

\sum_{k=1}^{9999999} \frac{1}{(k+1)^{3/2} + (k^2-1)^{1/3} + (k-1)^{2/3}}.
1 reply
fermion13pi
4 hours ago
RagvaloD
30 minutes ago
Quad formed by orthocenters has same area (all 7's!)
v_Enhance   37
N 2 hours ago by lpieleanu
Source: USA January TST for the 55th IMO 2014
Let $ABCD$ be a cyclic quadrilateral, and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. Let $W$, $X$, $Y$ and $Z$ be the orthocenters of triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively. Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.
37 replies
v_Enhance
Apr 28, 2014
lpieleanu
2 hours ago
No more topics!
Hard number theory
Hip1zzzil   13
N Mar 30, 2025 by Hip1zzzil
Source: FKMO 2025 P6
Positive integers $a, b$ satisfy both of the following conditions.
For a positive integer $m$, if $m^2 \mid ab$, then $m = 1$.
There exist integers $x, y, z, w$ that satisfies the equation $ax^2 + by^2 = z^2 + w^2$ and $z^2 + w^2 > 0$.
Prove that there exist integers $x, y, z, w$ that satisfies the equation $ax^2 + by^2 + n = z^2 + w^2$, for each integer $n$.
13 replies
Hip1zzzil
Mar 30, 2025
Hip1zzzil
Mar 30, 2025
Hard number theory
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G H BBookmark kLocked kLocked NReply
Source: FKMO 2025 P6
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Hip1zzzil
14 posts
#1
Y by
Positive integers $a, b$ satisfy both of the following conditions.
For a positive integer $m$, if $m^2 \mid ab$, then $m = 1$.
There exist integers $x, y, z, w$ that satisfies the equation $ax^2 + by^2 = z^2 + w^2$ and $z^2 + w^2 > 0$.
Prove that there exist integers $x, y, z, w$ that satisfies the equation $ax^2 + by^2 + n = z^2 + w^2$, for each integer $n$.
This post has been edited 4 times. Last edited by Hip1zzzil, Mar 30, 2025, 1:07 PM
Reason: Better
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Acorn-SJ
59 posts
#2 • 1 Y
Y by seoneo
Please copy the original wording, it’s mildly annoying to watch some phrases missing.
In particular, $m$ is supposed to be a positive integer.

EDIT: $n$ is an integer not a positive integer
This post has been edited 1 time. Last edited by Acorn-SJ, Mar 30, 2025, 11:25 AM
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pokmui9909
185 posts
#3 • 1 Y
Y by seoneo
This would be better:
Quote:
Positive integers $a, b$ satisfy both of the following conditions.
  • For a positive integer $m$, if $m^2 \mid ab$, then $m = 1$.
  • There exist integers $x, y, z, w$ that satisfies the equation $ax^2 + by^2 = z^2 + w^2$ and $z^2 + w^2 > 0$.
Prove that there exist integers $x, y, z, w$ that satisfies the equation $ax^2 + by^2 + n = z^2 + w^2$, for each integer $n$.
This post has been edited 2 times. Last edited by pokmui9909, Mar 30, 2025, 7:10 AM
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whwlqkd
97 posts
#4
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Anyone solved it?
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Acorn-SJ
59 posts
#5
Y by
a failed attempt
This post has been edited 1 time. Last edited by Acorn-SJ, Mar 30, 2025, 6:20 AM
Reason: Typo
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Lufin
10 posts
#6 • 2 Y
Y by Acorn-SJ, jjkim0336
In the original paper, n is an integer, and does not have to be a positive integer.
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seoneo
339 posts
#7
Y by
(Sketch)

From the condition, $a$ and $b$ are square free and coprime.
By the infinite descent method, we have relatively prime $a \alpha, b \beta, \zeta, \omega$ so that $a \alpha^2 +b \beta^2 = \zeta^2 + \omega^2$.

Now consider the expression
\[ (\zeta t + p)^2 + (\omega t +q)^2 -a(\alpha t +r)^2 -b(\beta t +s)^2 = 2(\zeta p + \omega q -a\alpha r -b \beta s)t + p^2 +q^2 -ar^2 -bs^2 \]From Bezout, we have $p,q,r,s$ such that
\[ \zeta p + \omega q -a\alpha r -b \beta s =1 \]so we have all ever, or all odd $n$.

By changing parity of $p^2 +q^2 -ar^2 -bs^2 $, we have all integers.

PS. I thought switching the parity would be easy, but it's actually more subtle than I first thought.
This post has been edited 1 time. Last edited by seoneo, Mar 30, 2025, 8:17 AM
Reason: To add PS
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seoneo
339 posts
#8
Y by
This FKMO problem was not accurately translated at first. If the original wording is not preserved in the current post, it may be better to create a properly translated version and redirect others to that post.

P.S. It looks like it's been fixed now.
This post has been edited 1 time. Last edited by seoneo, Mar 31, 2025, 6:31 AM
Reason: To fix grammer.
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segment
24 posts
#10 • 1 Y
Y by MihaiT
seoneo wrote:
(Sketch)

From the condition, $a$ and $b$ are square free and coprime.
By the infinite descent method, we have relatively prime $a \alpha, b \beta, \zeta, \omega$ so that $a \alpha^2 +b \beta^2 = \zeta^2 + \omega^2$.

Now consider the expression
\[ (\zeta t + p)^2 + (\omega t +q)^2 -a(\alpha t +r)^2 -b(\beta t +s)^2 = 2(\zeta p + \omega q -a\alpha r -b \beta s)t + p^2 +q^2 -ar^2 -bs^2 \]From Bezout, we have $p,q,r,s$ such that
\[ \zeta p + \omega q -a\alpha r -b \beta s =1 \]so we have all ever, or all odd $n$.

By changing parity of $p^2 +q^2 -ar^2 -bs^2 $, we have all integers.

Solved same, the motivation was experimenting $a=b=1$, realizing that we can make a linear function of one variable.
I thought \[ \zeta p + \omega q -a\alpha r -b \beta s =1 \]is not enough for the case when $a,b,\alpha,\beta,\gamma,\delta$ are all odd, so for that case I used \[ \zeta p + \omega q -a\alpha r -b \beta s =2 \]and eventually I got all odds, $4k$s, $4k+2$s.
This post has been edited 2 times. Last edited by segment, Mar 30, 2025, 8:06 AM
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MihaiT
748 posts
#11
Y by
very nice!
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GreekIdiot
179 posts
#12
Y by
Acorn-SJ wrote:
Please copy the original wording, it’s mildly annoying to watch some phrases missing.
In particular, $m$ is supposed to be a positive integer.

Doesnt change a lot if $m$ is negative now, does it?
Also in case $ab$ is not squarefree then $m=1$ is a positive integer...
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Acorn-SJ
59 posts
#13
Y by
GreekIdiot wrote:
Acorn-SJ wrote:
Please copy the original wording, it’s mildly annoying to watch some phrases missing.
In particular, $m$ is supposed to be a positive integer.

Doesnt change a lot if $m$ is negative now, does it?
Also in case $ab$ is not squarefree then $m=1$ is a positive integer...


Yes, you could say that, but better preserve the wording for the sake of archiving. Also, a much more important phrase was omitted, so I added that in.
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GreekIdiot
179 posts
#14
Y by
yeah, just saw the edit
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Hip1zzzil
14 posts
#15
Y by
Thank you for all the replies, Sorry for my bad English. I'm working on it though :)
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