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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.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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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
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Hard geometry
Lukariman   1
N 3 minutes ago by Lukariman
Given triangle ABC, a line d intersects the sides AB, AC and the line BC at D, E, F respectively.

(a) Prove that the circles circumscribing triangles ADE, BDF and CEF pass through a point P and P belongs to the circumcircle of triangle ABC.

(b) Prove that the centers of the circles circumscribing triangles ADE, BDF, CEF and ABC are all on the circle.

(c) Let $O_a$,$ O_b$, $O_c$ be the centers of the circles circumscribing triangles ADE, BDF, CEF. Prove that the orthocenter of triangle $O_a$$O_b$$O_c$ belongs to d.

(d) Prove that the orthocenters of triangles ADE, ABC, BDF, CEF are collinear.
1 reply
Lukariman
Yesterday at 12:53 PM
Lukariman
3 minutes ago
Anything real in this system must be integer
Assassino9931   1
N 5 minutes ago by lksb
Source: Al-Khwarizmi International Junior Olympiad 2025 P1
Determine the largest integer $c$ for which the following statement holds: there exists at least one triple $(x,y,z)$ of integers such that
\begin{align*} x^2 + 4(y + z) = y^2 + 4(z + x) = z^2 + 4(x + y) = c \end{align*}and all triples $(x,y,z)$ of real numbers, satisfying the equations, are such that $x,y,z$ are integers.

Marek Maruin, Slovakia
1 reply
Assassino9931
Friday at 9:26 AM
lksb
5 minutes ago
geometry problem
kjhgyuio   0
20 minutes ago
........
0 replies
kjhgyuio
20 minutes ago
0 replies
Concurrency of two lines and a circumcircle
BR1F1SZ   1
N 26 minutes ago by MathLuis
Source: 2025 Francophone MO Juniors P3
Let $\triangle{ABC}$ be a triangle, $\omega$ its circumcircle and $O$ the center of $\omega$. Let $P$ be a point on the segment $BC$. We denote by $Q$ the second intersection point of the circumcircles of triangles $\triangle{AOB}$ and $\triangle{APC}$. Prove that the line $PQ$ and the tangent to $\omega$ at point $A$ intersect on the circumcircle of triangle $\triangle AOB$.
1 reply
BR1F1SZ
2 hours ago
MathLuis
26 minutes ago
IMO Shortlist 2009 - Problem A2
April   93
N 32 minutes ago by ezpotd
Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that:
\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]
Proposed by Juhan Aru, Estonia
93 replies
April
Jul 5, 2010
ezpotd
32 minutes ago
Product of consecutive terms divisible by a prime number
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Seniors P4
Determine all sequences of strictly positive integers $a_1, a_2, a_3, \ldots$ satisfying the following two conditions:
[list]
[*]There exists an integer $M > 0$ such that, for all indices $n \geqslant 1$, $0 < a_n \leqslant M$.
[*]For any prime number $p$ and for any index $n \geqslant 1$, the number
\[
a_n a_{n+1} \cdots a_{n+p-1} - a_{n+p}
\]is a multiple of $p$.
[/list]


0 replies
BR1F1SZ
an hour ago
0 replies
Fixed and variable points
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Seniors P3
Let $\omega$ be a circle with center $O$. Let $B$ and $C$ be two fixed points on the circle $\omega$ and let $A$ be a variable point on $\omega$. We denote by $X$ the intersection point of lines $OB$ and $AC$, assuming $X \neq O$. Let $\gamma$ be the circumcircle of triangle $\triangle AOX$. Let $Y$ be the second intersection point of $\gamma$ with $\omega$. The tangent to $\gamma$ at $Y$ intersects $\omega$ at $I$. The line $OI$ intersects $\omega$ at $J$. The perpendicular bisector of segment $OY$ intersects line $YI$ at $T$, and line $AJ$ intersects $\gamma$ at $P$. We denote by $Z$ the second intersection point of the circumcircle of triangle $\triangle PYT$ with $\omega$. Prove that, as point $A$ varies, points $Y$ and $Z$ remain fixed.
0 replies
BR1F1SZ
an hour ago
0 replies
Use 3d paper
YaoAOPS   7
N an hour ago by EGMO
Source: 2025 CTST p4
Recall that a plane divides $\mathbb{R}^3$ into two regions, two parallel planes divide it into three regions, and two intersecting planes divide space into four regions. Consider the six planes which the faces of the cube $ABCD-A_1B_1C_1D_1$ lie on, and the four planes that the tetrahedron $ACB_1D_1$ lie on. How many regions do these ten planes split the space into?
7 replies
YaoAOPS
Mar 6, 2025
EGMO
an hour ago
Cyclic ine
m4thbl3nd3r   2
N an hour ago by m4thbl3nd3r
Let $a,b,c>0$ such that $a^2+b^2+c^2=3$. Prove that $$\sum \frac{a^2}{b}+abc \ge 4$$
2 replies
m4thbl3nd3r
Yesterday at 3:34 PM
m4thbl3nd3r
an hour ago
Sequence inequality
BR1F1SZ   0
an hour ago
Source: 2025 Francophone MO Seniors P1
Let $a_1, a_2, a_3, \ldots$ be a sequence of positive integers satisfying the following property: for all positive integers $k < \ell$, for all distinct integers $m_1, m_2, \ldots, m_k$ and for all distinct integers $n_1, n_2, \ldots, n_\ell$,
\[
a_{m_1} + a_{m_2} + \cdots + a_{m_k} \leqslant a_{n_1} + a_{n_2} + \cdots + a_{n_\ell}.
\]Prove that there exist two integers $N$ and $b$ such that $a_n = b$ for all $n \geqslant N$.
0 replies
BR1F1SZ
an hour ago
0 replies
GCD and LCM operations
BR1F1SZ   0
2 hours ago
Source: 2025 Francophone MO Juniors P4
Charlotte writes the integers $1,2,3,\ldots,2025$ on the board. Charlotte has two operations available: the GCD operation and the LCM operation.
[list]
[*]The GCD operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{gcd}(a, b)$.
[*]The LCM operation consists of choosing two integers $a$ and $b$ written on the board, erasing them, and writing the integer $\operatorname{lcm}(a, b)$.
[/list]
An integer $N$ is called a winning number if there exists a sequence of operations such that, at the end, the only integer left on the board is $N$. Find all winning integers among $\{1,2,3,\ldots,2025\}$ and, for each of them, determine the minimum number of GCD operations Charlotte must use.

Note: The number $\operatorname{gcd}(a, b)$ denotes the greatest common divisor of $a$ and $b$, while the number $\operatorname{lcm}(a, b)$ denotes the least common multiple of $a$ and $b$.
0 replies
BR1F1SZ
2 hours ago
0 replies
Balanced grids
BR1F1SZ   0
2 hours ago
Source: 2025 Francophone MO Juniors/Seniors P2
Let $n \geqslant 2$ be an integer. We consider a square grid of size $2n \times 2n$ divided into $4n^2$ unit squares. The grid is called balanced if:
[list]
[*]Each cell contains a number equal to $-1$, $0$ or $1$.
[*]The absolute value of the sum of the numbers in the grid does not exceed $4n$.
[/list]
Determine, as a function of $n$, the smallest integer $k \geqslant 1$ such that any balanced grid always contains an $n \times n$ square whose absolute sum of the $n^2$ cells is less than or equal to $k$.
0 replies
BR1F1SZ
2 hours ago
0 replies
Radiant sets
BR1F1SZ   0
2 hours ago
Source: 2025 Francophone MO Juniors P1
A finite set $\mathcal S$ of distinct positive real numbers is called radiant if it satisfies the following property: if $a$ and $b$ are two distinct elements of $\mathcal S$, then $a^2 + b^2$ is also an element of $\mathcal S$.
[list=a]
[*]Does there exist a radiant set with a size greater than or equal to $4$?
[*]Determine all radiant sets of size $2$ or $3$.
[/list]
0 replies
1 viewing
BR1F1SZ
2 hours ago
0 replies
Classic Diophantine
Adywastaken   4
N 2 hours ago by mrtheory
Source: NMTC 2024/6
Find all natural number solutions to $3^x-5^y=z^2$.
4 replies
Adywastaken
Yesterday at 3:39 PM
mrtheory
2 hours ago
IMO 2008, Question 3
delegat   79
N Apr 15, 2025 by shanelin-sigma
Source: IMO Shortlist 2008, N6
Prove that there are infinitely many positive integers $ n$ such that $ n^{2} + 1$ has a prime divisor greater than $ 2n + \sqrt {2n}$.

Author: Kestutis Cesnavicius, Lithuania
79 replies
delegat
Jul 16, 2008
shanelin-sigma
Apr 15, 2025
IMO 2008, Question 3
G H J
Source: IMO Shortlist 2008, N6
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delegat
652 posts
#1 • 14 Y
Y by Davi-8191, OlympusHero, jhu08, megarnie, HWenslawski, Adventure10, Mango247, and 7 other users
Prove that there are infinitely many positive integers $ n$ such that $ n^{2} + 1$ has a prime divisor greater than $ 2n + \sqrt {2n}$.

Author: Kestutis Cesnavicius, Lithuania
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harazi
5526 posts
#2 • 17 Y
Y by imowinner, Adventure10, jhu08, megarnie, HWenslawski, LLL2019, sabkx, Mango247, and 9 other users
http://www.mathlinks.ro/viewtopic.php?search_id=550514793&t=126781 This says much more and the solution presented there gives infinitely many such numbers, not just one, as stated in that link. However, I guess the official solution runs along the ideas used in a problem given in a recent USAMO (proposed by Titu Andreescu and me). In any case, I doubt this is a good problem for IMO.
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harazi
5526 posts
#3 • 33 Y
Y by numbertheorist17, Mediocrity, shinichiman, rafayaashary1, imowinner, jt314, MathbugAOPS, Polynom_Efendi, Illuzion, OlympusHero, mathleticguyyy, Wizard_32, TETris5, jhu08, MatBoy-123, myh2910, SADAT, sabkx, Adventure10, MarioLuigi8972, Jhy2027, yobu, and 11 other users
Sorry for double posting, but it's just as I thought: take $ p-1$ multiple of $ 4$ very large. Then we know there is $ n$ such that $ p|n^2+1$ and of course we may assume that $ n<p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p-n)^2+1$. Now, write $ p-2n=k>0$ and observe that $ p$ divides $ 4n^2+4=(p-k)^2+4$, thus $ p$ divides $ k^2+4$ and so $ k$ is at least $ \sqrt{p-4}$. This ismmediately implies $ p>2n+\sqrt{2n}$ if $ n$ is large enough, that is $ p$ is large enough.
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TomciO
552 posts
#4 • 8 Y
Y by MintTea, jhu08, GuvercinciHoca, Adventure10, Mango247, and 3 other users
Fix a prime $ p$ of the form $ 20m+1$. There are two solutions to the congruence $ n^2 \equiv -1 \pmod{p}$, one of them is less then $ \frac{p-1}{2}$ and one is greater. Let $ \frac{p-1}{2}-k$ be the smaller one ($ k>0$). Now we want find a lower bound for the $ k$, let's see what do we need. Take for simplicity $ \frac{p-1}{2} = a$. We want to have:
$ 2(a-k) + \sqrt{2(a-k)} \leq p-1$
or
$ a-k + \sqrt{\frac{a-k}{2}} \leq a$
which is equivalent to
$ a \leq 2k^2 + k$
and finally
$ p \leq 4k^2 + 2k + 1$.

Ok, now we see that $ (\frac{p-1}{2} - k)^2 \equiv -1 \pmod{p}$ is equivalent to $ (2k+1)^2 \equiv -4 \pmod{p}$. But $ p-4 \equiv 3 \pmod{5}$, so it's not a square. Of course $ (2k+1)^2 \not = 2p-4$ since the parity is different. Therefore $ (2k+1)^2 \geq 3p-4$, i.e. $ 4k^2+4k + 5 \geq 3p$ and it's clear that the desired inequality is satisfied.
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tmbtw
122 posts
#5 • 8 Y
Y by jhu08, Adventure10, Mango247, Mango247, and 4 other users
harazi wrote:
Sorry for double posting, but it's just as I thought: take $ p - 1$ multiple of $ 4$ very large. Then we know there is $ n$ such that $ p|n^2 + 1$ and of course we may assume that $ n < p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p - n)^2 + 1$. Now, write $ p - 2n = k > 0$ and observe that $ p$ divides $ 4n^2 + 4 = (p - k)^2 + 4$, thus $ p$ divides $ k^2 + 4$ and so $ k$ is at least $ \sqrt {p - 4}$. This ismmediately implies $ p > 2n + \sqrt {2n}$ if $ n$ is large enough, that is $ p$ is large enough.

Very nice !
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ywl
117 posts
#6 • 6 Y
Y by Adventure10, Adventure10, jhu08, sabkx, and 2 other users
Seems like people who have seen the problem
Prove that there exists infinite $ n$ such that $ n^4+1$ has a prime divisor larger than $ 2n$

would have some advantages
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fedja
6920 posts
#7 • 11 Y
Y by OlympusHero, math_comb01, Adventure10, jhu08, sabkx, Mango247, and 5 other users
Just a side remark. Even those who did not know that any $ p=4k+1$ divides some $ x^2+1$ could solve the problem like harazi did because all one needs is to have infinitely many primes that divide $ x^2+1$ for some $ x$ and the classical Euclid's proof of the infinitude of primes can be trivially ajusted to give this statement. The opening sentence in TomciO's solution, on the other hand, would tempt me to reduce a few points immediately if I were one of the graders because it is quite doubtful that the contestant could prove it if asked.

Anyway, the first day problems were sort of disappointing :(. Let's see what the second day will bring :)
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jmerry
12096 posts
#8 • 4 Y
Y by Adventure10, jhu08, and 2 other users
Every prime that is congruent to $ 1$ mod $ 4$ appears exactly once as the divisor here, except for $ 5$ and $ 13$. Harazi's "large enough" threshold is $ p = 29$, and $ p = 17$ works easily because $ 17 - 4$ is not a perfect square.
If $ 2n + \sqrt {2n}$ were changed to $ 2n + \sqrt {2n} + 1$, we would need to exclude all primes of the form $ m^2 + 4$ as well. This still leaves infinitely many, but proving that is hard compared to the Euclid-style argument that there are infinitely many primes $ \equiv 1\mod 4$.

[Edit- fixed silly mistake]
This post has been edited 1 time. Last edited by jmerry, Jul 17, 2008, 6:53 AM
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Albanian Eagle
1693 posts
#9 • 4 Y
Y by Adventure10, jhu08, Mango247, and 1 other user
Shouldn't 3rd problems be a little difficult in the sense that the solution should be a little long or at least involve some clever argument or trick?
I mean look at Harazi's post (#3)...
And then a gain P1 asking if you know Power Of a Point, and P2 being cross multiplication+multiplication+addition+completing the square...
Very disappointed.
Again the question is:
Poor choice of leaders or bad ISL?
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dblues
203 posts
#10 • 6 Y
Y by jhu08, Adventure10, Mango247, and 3 other users
Note that for any prime $ p \equiv 1 \pmod{4}$, we can always choose some $ \alpha \in \{ 0, \ldots, \frac{p-3}{2} \}$ such that $ \left(\frac{p-1}{2} - \alpha \right)^2 \equiv -1 \pmod{p}$. Denote $ m = \frac{p-1}{2} - \alpha$. This problem then becomes equivalent to choosing a suitable $ p$, with corresponding $ \alpha$, such that $ p > 2m + \sqrt{2m}$. We check that \[ p > p-1 - 2\alpha + \sqrt{p-1-2\alpha} \Leftrightarrow 2\alpha + 1 > \sqrt{p-1-2\alpha} \Leftrightarrow 4\alpha^2 + 6\alpha + 2-p >0.\] Solving this quadratic inequality, since we assumed $ \alpha \geq 0$, this is equivalent to \[ \alpha > \dfrac{-6+\sqrt{36+16(p-2)}}{8} = \dfrac{-3+\sqrt{4p+1}}{4}.\]

Hence, if $ \alpha > \dfrac{-3+\sqrt{4p+1}}{4}$, then we are done. Suppose not, then we have $ 0 \leq \alpha \leq \dfrac{-3+\sqrt{4p+1}}{4}$. Note that $ 4m^2 = (p-1-2\alpha)^2 \equiv (2\alpha+1)^2 \pmod{p}$. By the bounds on $ \alpha$, we get $ 0 \leq (2\alpha+1)^2 \leq \left( \dfrac{\sqrt{4p+1} - 1}{2} \right)^2$. Now, by assumption, $ 4m^2 \equiv -4 \pmod{p}$, so if $ \left( \dfrac{\sqrt{4p+1} - 1}{2} \right)^2 < p-4$, we get a contradiction. Solving this inequality, this is equilavent to $ p>20$.

Therefore, if $ p$ is sufficiently large (i.e. $ p>20$) and $ p\equiv 1 \pmod{p}$, we can always find some $ m$ such that $ p|(m^2+1)$ and $ p > 2m + \sqrt{2m}$.
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greentreeroad
484 posts
#11 • 3 Y
Y by jhu08, Adventure10, Mango247
harazi wrote:
Sorry for double posting, but it's just as I thought: take $ p - 1$ multiple of $ 4$ very large. Then we know there is $ n$ such that $ p|n^2 + 1$ and of course we may assume that $ n < p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p - n)^2 + 1$. Now, write $ p - 2n = k > 0$ and observe that $ p$ divides $ 4n^2 + 4 = (p - k)^2 + 4$, thus $ p$ divides $ k^2 + 4$ and so $ k$ is at least $ \sqrt {p - 4}$. This ismmediately implies $ p > 2n + \sqrt {2n}$ if $ n$ is large enough, that is $ p$ is large enough.

Similarly I got to the step that $ p$ divides $ k^2 + 4$, obviously if $ \frac{k^2+4}{p}>1$ we will done. So p=$ k^2$+4. So this will lead to that any prime congruent to 1 mod 4 need also be congruent to 5 mod 8.(otherwise p-4 congruent to 5 mod 8 which cannot be $ k^2$) This is clearly a contradiction by Dirichlet's Theorem!

P.S Is there an elementary way to show that there are infinitely many prime of form 8K+1?
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Albanian Eagle
1693 posts
#12 • 8 Y
Y by mathcool2009, Adventure10, jhu08, sabkx, Mango247, and 3 other users
yes. one way for example is considering divisors of $ 2^{2^m} + 1$ :wink:
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orl
3647 posts
#13 • 3 Y
Y by Adventure10, jhu08, Mango247
Author of this problem is Kęstutis Česnavičius, Lithuania., from Jacobs University Bremen which is the host of the 50th International Mathematical Olympiad 2009 in Bremen Germany. It will also be the 20th anniversary after the last IMO in Braunschweig, Germany in 1989. Formerly the German Democratic Republic (GDR) was supposed to host the IMO 1992 but due to the unification with the Federal Republic of Germany (FRG) the event was cancelled and organized by Moscow, Russia, as a substitute host in 1992.
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pavel kozlov
616 posts
#14 • 5 Y
Y by Adventure10, jhu08, Mango247, and 2 other users
If solution of the problem consists only of two simple steps – it isn’t third.
My solution. Consider the set of numbers $ S=${$ p \in \mathbb{P} | p \equiv 1 \mod 4, p \not = 5,13$}.
For every $ p \in S$ let us define set of positive integers $ N_p$ such that $ n \in N_p \leftrightarrow p|n^2+1$ and $ n_p$ - the minimal of them.
It’s easily seen that $ n_p<p$ and also $ n_p<\frac{p}{2}$ else we can take $ p-n_p$ instead of $ n_p$.
Let $ k_p$ be $ p-2n_p$. Then $ k_p^2+4=(p-2n_p)^2+4 \equiv 4(n_p^2+1) \equiv 0 \mod p$, hence $ k^2+4>=p$.
Let us notice that $ k_p>4$. Really, if this isn’t true then there are only two variants: $ k_p=1$ and $ k_p=3$ since $ k_p$ is odd. If $ k_p=1$ then $ p|1^2+4$ and $ p=5$ - contr; if $ k_p=3$ then $ p|3^2+4$ and $ p=13$ -contr.
So we have $ k_p^2+k_p>k_p+4 \ge p$, therefore $ p-k_p+\sqrt{p-k_p}<p$ or $ 2n_p+\sqrt{2n_p}<p$. Since $ |S|=\infty$ then there are infinitely many different integers $ n_p$ such that $ p \in S$, so we are done.
Evidently it’s possible to estimate this value better. For example if we take the set $ T=${$ p \in \mathbb{P} | p \equiv 1 \mod 4, p \equiv 1 \mod 5$} then for every $ p \in T$ $ p-4$ isn’t square of integer and hence $ k_p^2+4 \ge 5p$.
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conan_naruto236
96 posts
#15 • 3 Y
Y by jhu08, Adventure10, Mango247
harazi wrote:
\we may assume that $ n < p$ and even $ 2n\leq p$, simply because $ p$ also divides $ (p - n)^2 + 1$. .
vhy??? :?: :maybe:
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