Stay ahead of learning milestones! Enroll in a class over the summer!

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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
For positive integers \( a, b, c \), find all possible positive integer values o
Jackson0423   12
N 4 minutes ago by Jackson0423
For positive integers \( a, b, c \), find all possible positive integer values of
\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.
\]
12 replies
Jackson0423
Apr 13, 2025
Jackson0423
4 minutes ago
10abc/Π(a+b) inequality
lbh_qys   1
N 17 minutes ago by lbh_qys
Source: Inspired by JK1603JK
Let $a,b,c > 0$, prove that

$$\frac{2a}{2a+b+c} + \frac{2b}{2b+c+a}+\frac{2c}{2c+a+b} \geq \sqrt{1 + \frac{10abc}{(a+b)(b+c)(c+a)}}.$$
1 reply
+1 w
lbh_qys
29 minutes ago
lbh_qys
17 minutes ago
GMO P3 2024
Z4ADies   5
N an hour ago by ihategeo_1969
Source: Geometry Mains Olympiad (GMO) 2024 P3
Let $ABC$ be a scalene triangle. $D$ is the foot of the altitude from $A$ to $BC$. $H$ is the orthocenter, $M$ is the midpoint of $BC$, and $N$ is the midpoint of $AH$. Tangents from $B$ and $C$ to the circumcircle of $ABC$ intersect at $T$. Prove that line $TH$ passes through one of the intersections of circles $ADT$ with $MHN$.

Author:Haris Shaqiri (Kosovo)
5 replies
Z4ADies
Oct 20, 2024
ihategeo_1969
an hour ago
One inequality 2
prof.   0
an hour ago
If $a,b,c$ are positiv real number, such that $abc=1$, prove inequality $$2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
0 replies
prof.
an hour ago
0 replies
No more topics!
Lifting the Exponent Lemma (LTE)
nsato   34
N Dec 13, 2017 by Amir Hossein
As our initial entry of a series of featured articles, Amir Hossein Parvardi (Amir Hossein) has written "Lifting the Exponent Lemma (LTE)," which is a useful result for solving exponential Diophantine equations.
34 replies
nsato
Apr 11, 2011
Amir Hossein
Dec 13, 2017
Lifting the Exponent Lemma (LTE)
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nsato
15653 posts
#1 • 60 Y
Y by Amir Hossein, astamir, bluecarneal, nsun48, rdj5933mile5, shinichiman, avatarofakato, Debdut, bryanbr2, airplanes1, dantx5, silly_mouse, Binomial-theorem, knittingfrenzy18, AdvitiyaBrijesh, ahaanomegas, forthegreatergood, bcp123, minimario, Mathematicalx, Nashimo, mikechen, Devesh14, trumpeter, ais3000, Magikarp1, Wave-Particle, eshan, m1234567, zed1969, JasperL, Wavefunction, rodamaral, mathleticguyyy, Toinfinity, Chipaoo, rg_ryse, Adventure10, Mango247, and 21 other users
As our initial entry of a series of featured articles, Amir Hossein Parvardi (Amir Hossein) has written "Lifting the Exponent Lemma (LTE)," which is a useful result for solving exponential Diophantine equations.
Attachments:
Lifting The Exponent - Version 6.pdf (215kb)
This post has been edited 2 times. Last edited by Amir Hossein, Oct 22, 2016, 6:45 AM
Reason: Edited the link (and my username).
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Amir Hossein
5452 posts
#2 • 36 Y
Y by nsun48, rdj5933mile5, Binomial-theorem, knittingfrenzy18, AdvitiyaBrijesh, ahaanomegas, forthegreatergood, MiMi1376, minimario, ais3000, Magikarp1, Wavefunction, Toinfinity, rg_ryse, MathQurious, Adventure10, and 20 other users
Thank you very much, dear Naoki! :)
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mathmdmb
1547 posts
#3 • 6 Y
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nsato wrote:
As our initial entry of a series of featured articles, Amir Parvardi (amparvardi) has written "Lifting the Exponent Lemma (LTE)," which is a useful result for solving exponential Diophantine equations.
I think it is useful not only for dipohantine equations,also for so many divisibility related problems like IMO-1990,3 :D
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ridgers
713 posts
#4 • 6 Y
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well done Amir!
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FBI__
29 posts
#5 • 5 Y
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Excellent!
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zZzZzZzZz
20 posts
#6 • 5 Y
Y by Adventure10, Mango247, and 3 other users
Can you write a file "Full solution for 33 problem" . Thank you :)
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powerofzeta
55 posts
#7 • 3 Y
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thank you,it is soo interesting .
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mathmdmb
1547 posts
#8 • 5 Y
Y by Amir Hossein, Adventure10, Mango247, and 2 other users
zZzZzZzZz wrote:
Can you write a file "Full solution for 33 problem" . Thank you :)
Here it is:
Attachments:
Solution to 33.pdf (98kb)
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QuantumTiger
609 posts
#9 • 5 Y
Y by Adventure10, Mango247, and 3 other users
Thank you; this is actually quite useful. Thank you, Amir!
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Abdur-Rahman
1 post
#10 • 4 Y
Y by Adventure10, Mango247, and 2 other users
Thank you very much
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hyperspace.rulz
287 posts
#11 • 7 Y
Y by Adventure10, Mango247, and 5 other users
Hi all,

How would we quote this in our solutions?

Cheers,
Hyperspace Rulz!
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Amir Hossein
5452 posts
#12 • 6 Y
Y by Adventure10, Mango247, and 4 other users
hyperspace.rulz wrote:
How would we quote this in our solutions?

Sorry, what do you mean?
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hyperspace.rulz
287 posts
#13 • 4 Y
Y by Adventure10, Mango247, and 2 other users
My apologies amparvardi. What I mean is, suppose we found we could use this lemma in a contest. Would we have to quote and prove it as a lemma or could we just say "By the Lifting The Exponent Lemma"...?

Sorry again,
Hyperspace Rulz!
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Amir Hossein
5452 posts
#14 • 5 Y
Y by Adventure10, Mango247, and 3 other users
Surely depend on the competition. I think it's allowed to use this lemma on IMO without its proof, but I'm not sure about other competitions. So you may try to learn the proof of this theorem and write it on your olympiad papers (however, the whole idea of LTE is its proof, not only the theorem).
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ShahinBJK
113 posts
#15 • 3 Y
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Can you send other versions?
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Amir Hossein
5452 posts
#16 • 6 Y
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Other (and older) versions have been posted and discussed on this topic, and this is the last version of the article.
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NaeunLove
248 posts
#17 • 4 Y
Y by Adventure10, Mango247, and 2 other users
$T$hank $y$ou $s$o $m$uch^^ :D
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mathskid
13 posts
#18 • 3 Y
Y by Adventure10 and 2 other users
Thank you very much amparvardi
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goodar2006
1347 posts
#19 • 4 Y
Y by Amir Hossein, Adventure10, Mango247, and 1 other user
Thanks again Amir. they gave it to us as a reference in the summer camp. I'd really admire your effort.
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Amir Hossein
5452 posts
#20 • 5 Y
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Thanks, Goodarz.

@Goodarz: aghaye Sefidgaran behem goftan. :)
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shinichiman
3212 posts
#21 • 2 Y
Y by Adventure10, Mango247
Thanks a lot!
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Amir Hossein
5452 posts
#22 • 4 Y
Y by javlik, bearytasty, Adventure10, Mango247
I just found a nice article in Yufei Zhao's site (and here are the solutions of the problems).
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astamir
18 posts
#23 • 2 Y
Y by Adventure10, Mango247
Very useful to me. Thank you very much!
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kakashi94
6 posts
#24 • 2 Y
Y by Adventure10, Mango247
Could you give me the full solution for Problem 15? I can't check the condition $x-y \vdots p$.
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Amir Hossein
5452 posts
#25 • 3 Y
Y by kakashi94, Binomial-theorem, Adventure10
Fedja wrote in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=393335]this[/url] topic
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kakashi94
6 posts
#26 • 2 Y
Y by Adventure10, Mango247
amparvardi wrote:
Fedja wrote in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=393335]this[/url] topic
But we don't have $n=mu$ is the number which satisfies $p^n|a^n-b^n$, So why can we get $n=v_p(p^n)\le v_p(a^n-b^n)\le v_p(m)+v_p(a^u-b^u)+v_p(a^u+b^u)$
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shinichiman
3212 posts
#27 • 2 Y
Y by Adventure10, Mango247
Thanks a lot! :D
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triskaidecahedron
70 posts
#28 • 1 Y
Y by Adventure10
For the problem in the section Challenge Problems, can you not just solve Bulgaria 1997 with strong induction?
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bavaria-1
27 posts
#29 • 2 Y
Y by Adventure10, Mango247
Sorry,who can give me solutions of problems?Or where can i take it?
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eshan
471 posts
#30 • 3 Y
Y by Amir Hossein, Ani10, Adventure10
Hello friends.
Sorry for revive, but I have recently proven $LTE$ by myself. And I am posting my proof.
The proof is in the attachment.
Hope you like it. :)

I posted the proof in India forum too because I wanted to be sure that my proof is correct.
Sorry for that also. ;)

Typo fixed.
Attachments:
My Proof for First Form of LTE.pdf (133kb)
This post has been edited 3 times. Last edited by eshan, Nov 17, 2015, 7:48 AM
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trumpeter
3332 posts
#31 • 6 Y
Y by Eugenis, rkm0959, eshan, Amir Hossein, Adventure10, Mango247
I don't think the proof is compltely correct.

(so that I don't forget later, I think you also typoed in the first line - it should be $x^n-y^n=\left(x-y\right)\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)$, but this doesn't affect the lter part of the proof)

In your conversion of $v_p\left(\sum_{k=0}^{n-1}x^ky^{n-k-1}\right)$ to $v_p\left(n\right)$, you assume that if $a\equiv b\pmod p$, then $v_p\left(a\right)=v_p\left(b\right)$. This is not true - take any two numbers $a$ and $b$ such that $v_p\left(a\right)>v_p\left(b\right)\geq1$. Then, $a\equiv b\pmod p$ but $v_p\left(a\right)\neq v_p\left(b\right)$.

Good try, though!
This post has been edited 5 times. Last edited by trumpeter, Nov 17, 2015, 4:48 AM
Reason: credits to CantonMathGuy for the noice pun
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gauss01
58 posts
#32 • 1 Y
Y by Adventure10
When was the lemma first published or invented ?
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Goobersmith
289 posts
#33 • 2 Y
Y by Adventure10, Mango247
Hello nsato, could we include these articles on the aops article website? Should we ask the authors individually?
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gauss01
58 posts
#34 • 2 Y
Y by Adventure10, Mango247
gauss01 wrote:
When was the lemma first published or invented ?

i mean, from the 'mathematical reflection' ?
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Amir Hossein
5452 posts
#35 • 2 Y
Y by Adventure10, Mango247
I have added the solutions to the challenge problems (which were solved by mythical Fedja) here: https://parvardi.com/lte/
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