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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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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
Three operations make any number
awesomeming327.   2
N 17 minutes ago by happymoose666
Source: own
The number $3$ is written on the board. Anna, Boris, and Charlie can do the following actions: Anna can replace the number with its floor. Boris can replace any integer number with its factorial. Charlie can replace any nonnegative number with its square root. Prove that the three can work together to make any positive integer in finitely many steps.
2 replies
awesomeming327.
3 hours ago
happymoose666
17 minutes ago
Inspired by RMO 2006
sqing   3
N 36 minutes ago by Marrelia
Source: Own
Let $ a,b >0  . $ Prove that
$$  \frac {a^{2}+1}{b+k}+\frac { b^{2}+1}{ka+1}+\frac {2}{a+kb}  \geq \frac {6}{k+1}  $$Where $k\geq 0.03 $
$$  \frac {a^{2}+1}{b+1}+\frac { b^{2}+1}{a+1}+\frac {2}{a+b}  \geq 3  $$
3 replies
sqing
Saturday at 3:24 PM
Marrelia
36 minutes ago
Inspired by 2025 Beijing
sqing   10
N 2 hours ago by ytChen
Source: Own
Let $ a,b,c,d >0  $ and $ (a^2+b^2+c^2)(b^2+c^2+d^2)=36. $ Prove that
$$ab^2c^2d \leq 8$$$$a^2bcd^2 \leq 16$$$$ ab^3c^3d \leq \frac{2187}{128}$$$$ a^3bcd^3 \leq \frac{2187}{32}$$
10 replies
1 viewing
sqing
Saturday at 4:56 PM
ytChen
2 hours ago
IMO 2017 Problem 4
Amir Hossein   116
N 3 hours ago by cj13609517288
Source: IMO 2017, Day 2, P4
Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$.

Proposed by Charles Leytem, Luxembourg
116 replies
Amir Hossein
Jul 19, 2017
cj13609517288
3 hours ago
A sharp one with 3 var
mihaig   10
N 3 hours ago by mihaig
Source: Own
Let $a,b,c\geq0$ satisfying
$$\left(a+b+c-2\right)^2+8\leq3\left(ab+bc+ca\right).$$Prove
$$ab+bc+ca+abc\geq4.$$
10 replies
mihaig
May 13, 2025
mihaig
3 hours ago
Another right angled triangle
ariopro1387   1
N 3 hours ago by lolsamo
Source: Iran Team selection test 2025 - P7
Let $ABC$ be a right angled triangle with $\angle A=90$. Point $M$ is the midpoint of side $BC$ And $P$ be an arbitrary point on $AM$. The reflection of $BP$ over $AB$ intersects lines $AC$ and $AM$ at $T$ and $Q$, respectively. The circumcircles of $BPQ$ and $ABC$ intersect again at $F$. Prove that the center of the circumcircle of $CFT$ lies on $BQ$.
1 reply
ariopro1387
Yesterday at 4:13 PM
lolsamo
3 hours ago
four points lie on a circle
pohoatza   78
N 4 hours ago by ezpotd
Source: IMO Shortlist 2006, Geometry 2, AIMO 2007, TST 1, P2
Let $ ABCD$ be a trapezoid with parallel sides $ AB > CD$. Points $ K$ and $ L$ lie on the line segments $ AB$ and $ CD$, respectively, so that $AK/KB=DL/LC$. Suppose that there are points $ P$ and $ Q$ on the line segment $ KL$ satisfying \[\angle{APB} = \angle{BCD}\qquad\text{and}\qquad \angle{CQD} = \angle{ABC}.\]Prove that the points $ P$, $ Q$, $ B$ and $ C$ are concyclic.

Proposed by Vyacheslev Yasinskiy, Ukraine
78 replies
pohoatza
Jun 28, 2007
ezpotd
4 hours ago
JBMO TST Bosnia and Herzegovina 2023 P4
FishkoBiH   2
N 4 hours ago by Stear14
Source: JBMO TST Bosnia and Herzegovina 2023 P4
Let $n$ be a positive integer. A board with a format $n*n$ is divided in $n*n$ equal squares.Determine all integers $n$3 such that the board can be covered in $2*1$ (or $1*2$) pieces so that there is exactly one empty square in each row and each column.
2 replies
FishkoBiH
Yesterday at 1:38 PM
Stear14
4 hours ago
Does there exist 2011 numbers?
cyshine   8
N 4 hours ago by TheBaiano
Source: Brazil MO, Problem 4
Do there exist $2011$ positive integers $a_1 < a_2 < \ldots < a_{2011}$ such that $\gcd(a_i,a_j) = a_j - a_i$ for any $i$, $j$ such that $1 \le i < j \le 2011$?
8 replies
cyshine
Oct 20, 2011
TheBaiano
4 hours ago
D1036 : Composition of polynomials
Dattier   1
N 4 hours ago by Dattier
Source: les dattes à Dattier
Find all $A \in \mathbb Q[x]$ with $\exists Q \in \mathbb Q[x], Q(A(x))= x^{2025!+2}+x^2+x+1$ and $\deg(A)>1$.
1 reply
Dattier
Saturday at 1:52 PM
Dattier
4 hours ago
number sequence contains every large number
mathematics2003   3
N 4 hours ago by sttsmet
Source: 2021ChinaTST test3 day1 P2
Given distinct positive integer $ a_1,a_2,…,a_{2020} $. For $ n \ge 2021 $, $a_n$ is the smallest number different from $a_1,a_2,…,a_{n-1}$ which doesn't divide $a_{n-2020}...a_{n-2}a_{n-1}$. Proof that every number large enough appears in the sequence.
3 replies
mathematics2003
Apr 13, 2021
sttsmet
4 hours ago
IMO ShortList 2002, geometry problem 2
orl   28
N 4 hours ago by ezpotd
Source: IMO ShortList 2002, geometry problem 2
Let $ABC$ be a triangle for which there exists an interior point $F$ such that $\angle AFB=\angle BFC=\angle CFA$. Let the lines $BF$ and $CF$ meet the sides $AC$ and $AB$ at $D$ and $E$ respectively. Prove that \[ AB+AC\geq4DE. \]
28 replies
orl
Sep 28, 2004
ezpotd
4 hours ago
Arc Midpoints Form Cyclic Quadrilateral
ike.chen   56
N 4 hours ago by ezpotd
Source: ISL 2022/G2
In the acute-angled triangle $ABC$, the point $F$ is the foot of the altitude from $A$, and $P$ is a point on the segment $AF$. The lines through $P$ parallel to $AC$ and $AB$ meet $BC$ at $D$ and $E$, respectively. Points $X \ne A$ and $Y \ne A$ lie on the circles $ABD$ and $ACE$, respectively, such that $DA = DX$ and $EA = EY$.
Prove that $B, C, X,$ and $Y$ are concyclic.
56 replies
ike.chen
Jul 9, 2023
ezpotd
4 hours ago
Non-linear Recursive Sequence
amogususususus   4
N 5 hours ago by GreekIdiot
Given $a_1=1$ and the recursive relation
$$a_{i+1}=a_i+\frac{1}{a_i}$$for all natural number $i$. Find the general form of $a_n$.

Is there any way to solve this problem and similar ones?
4 replies
amogususususus
Jan 24, 2025
GreekIdiot
5 hours ago
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
Valentin Vornicu   67
N Saturday at 7:10 AM by alexanderchew
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
67 replies
Valentin Vornicu
Oct 24, 2005
alexanderchew
Saturday at 7:10 AM
$n$ with $2000$ divisors divides $2^n+1$ (IMO 2000)
G H J
G H BBookmark kLocked kLocked NReply
Source: IMO 2000, Problem 5, IMO Shortlist 2000, Problem N3
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Valentin Vornicu
7301 posts
#1 • 16 Y
Y by Amir Hossein, Davi-8191, MarkBcc168, Understandingmathematics, itslumi, Adventure10, megarnie, RedFlame2112, Leo890, clevereagle, Mango247, cookie130, Dansman2838, PikaPika999, and 2 other users
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
This post has been edited 1 time. Last edited by Amir Hossein, Mar 21, 2016, 7:33 PM
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grobber
7849 posts
#2 • 15 Y
Y by viperstrike, laegolas, Wizard_32, to_chicken, Adventure10, DCMaths, Mango247, cookie130, MS_asdfgzxcvb, PikaPika999, and 5 other users
For any $k\ge 1$ there is such an $n$ with exactly $k$ prime factors.

For $k=1,\ n=3^t$ works for every $t\ge 1$. Take $t$ s.t. for $n_1=3^t,\ 2^{n_1}+1$ has a prime factor $p_2$ larger than $3$. Now take $n_2=n_1p_2$. Then $n_2|2^{n_1}+1|2^{n_2}+1$, and $2^{p_2}+1$ has a prime factor $p_3\not|n_2$. This is because $(2^{n_1}+1,2^{p_2}+1)=3,\ p_2\not|2^{p_2}+1$, and $2^{p_2}+1$ cannot be a power of $3$, since $p_2>3$ (I'm using the well known and easy to prove fact that the only positive integer solution $(a,b),a>1$ to $3^a=2^b+1$ is $(a,b)=(2,3)$). Then take $n_3=n_2p_3$. Just like above, we deduce that $2^{p_3}+1$ has a prime factor $p_4$ which is coprime to $n_3$, and take $n_4=n_3p_4$, and so on. $n_k$ will have exactly $k$ prime factors and will satisfy $n_k|2^{n_k}+1$.
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pluricomplex
390 posts
#3 • 3 Y
Y by Adventure10, Mango247, cookie130
Valentin Vornicu wrote:
Does there exist a positive integer $n$ such that $n$ has exactly 2000 prime divisors and $n$ divides $2^n + 1$?

You can find my paper for a general problem of this problem in This file
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Philip_Leszczynski
327 posts
#4 • 3 Y
Y by Adventure10, Mango247, cookie130
Let $N=2^n+1$. We will assume for the sake of contradiction that $n|N$.

$2^n+1 \equiv 0$ (mod $n$) $\Rightarrow 2^n \equiv -1$ (mod $n$). So 2 does not divide $n$, and so $n$ is odd.

Select an arbitrary prime factor of $n$ and call it $p$. Let's represent $n$ in the form $p^am$, where $m$ is not divisible by $p$.

Note that $p$ and $m$ are both odd since $n$ is odd. By repeated applications of Fermat's Little Theorem:

$N = 2^n+1 = 2^{p^am} + 1 = (2^{p^{a-1}m})^p + 1 \equiv 2^{p^{a-1}m} + 1$ (mod $p$)

Continuing in this manner, and inducting on k from 1 to $a$,

$2^{p^{a-k}m}+1 \equiv (2^{p^{a-k-1}m})^p + 1$ (mod $p$) $\equiv 2^{p^{a-k-1}m} + 1$ (mod $p$)

So we have $N \equiv 2^m+1$ (mod $p$)

Since $p$ is relatively prime to $m$, $N \equiv 1+1$ (mod $p$) $\equiv 2$ (mod $p$)

Since $p$ is odd, $N$ is not divisible by $p$. Hence $N$ is not divisible by $n$. So we have a contradiction, and our original assumption was false, and therefore $N$ is still not divisible by $n$.
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Philip_Leszczynski
327 posts
#5 • 3 Y
Y by Adventure10, Mango247, cookie130
Hmmm... I made a mistake here somewhere but I do not see it.
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Arne
3660 posts
#6 • 4 Y
Y by The_fandangos, Adventure10, Mango247, cookie130
Yes, since there are lots of integers $n$ such that $n$ divides $2^n + 1$, and the statement is true!
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Johann Peter Dirichlet
376 posts
#7 • 3 Y
Y by Adventure10, Mango247, cookie130
There exists a pretty beautiful generalization:

"
Let $s, a, b$ positive integers, such that $GCD(a,b) = 1$ and $a+b$ is not a 2-power.
Show that there exists infinitely many $n \in N$ such that

--- $n=p_1^{e_1} \cdot p_2^{e_2} \cdot p_3^{e_3} \cdots p_s^{e_s} \cdot$ is the canonical factoring of $n$.

--- $n|(a^n+b^n)$
"
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The QuattoMaster 6000
1184 posts
#8 • 3 Y
Y by Adventure10, Mango247, cookie130
Valentin Vornicu wrote:
Does there exist a positive integer $ n$ such that $ n$ has exactly 2000 prime divisors and $ n$ divides $ 2^n + 1$?
Here is a solution that I don't think has been mentioned yet:
Solution
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Arquimedes
8 posts
#9 • 3 Y
Y by Adventure10, Mango247, cookie130
For any i,
2^3^i+1
is divisible by
3^i
(the proof is easy with euler`s theorem+induction and maybe with primitive roots (2 is primitive root modulo 3^i for any i)). Hence, for i=1999,
3^1999
has 2000 divisors and it satisfies the asked in the problem.

it is correct??

please , answer me.

bye

sorry for my english.
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L-b
11 posts
#10 • 3 Y
Y by Adventure10, Mango247, cookie130
Well, the point is to find a number which has exactly $ 2000$ prime divisiors, whereas $ 3^{1999}$ has only one ($ 3$).

But it is a very nice thought to look at powers of $ 3$, when the problem considers powers of $ 2$ (vide grobber's solution, which I do not completely understand yet, but it seems very nice and simple)
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Binomial-theorem
3982 posts
#11 • 5 Y
Y by JasperL, Anar24, Adventure10, Mango247, cookie130
Solution overkilling with Zsigmondy's
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v_Enhance
6877 posts
#12 • 18 Y
Y by Binomial-theorem, WL0410, e_plus_pi, Omeredip, Not_real_name, Takeya.O, TheHerculean11, ZHEKSHEN, Quidditch, HamstPan38825, Msn05, samrocksnature, joseph02, aidan0626, Adventure10, Mango247, MarioLuigi8972, cookie130
Answer: Yes.

We say that $n$ is Korean if $n \mid 2^n+1$. First, observe that $n=9$ is Korean. Now, the problem is solved upon the following claim:

Claim: If $n > 3$ is Korean, there exists a prime $p$ not dividing $n$ such that $np$ is Korean too.

Proof. I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$, which exists by Zsigmondy theorem. Obviously $p \neq 2$. Then:
  • Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$.
  • Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$;
Hence $np \mid 2^{np} + 1$ by Chinese Theorem, since $\gcd(n,p) = 1$. $\blacksquare$

EDIT: The version of the proof I posted four years ago was incorrect. This one should work.
This post has been edited 1 time. Last edited by v_Enhance, May 3, 2017, 1:08 AM
Reason: Wanlin Li pointed out a mistake
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lfetahu
134 posts
#13 • 6 Y
Y by Takeya.O, gghx, Adventure10, Mango247, cookie130, and 1 other user
I feel like it isn't interesting to make any new remark on this problem, but anyway I'm posting my approach too.

If we show that for any positive integer k, there exists a positive integer n with exactly k distinct prime divisors such that n | 2^n + 1, then we are done, since the problem asks us to examine a special case, more exactly k = 2000. Furthermore, we can even show that we can find these n's divisible by a power of 3, which will help us on our proof.

We use induction on k. k = 1, we can choose n(1) = 3, which clearly satisfies the conditions. Assume that k >= 1, and there exists n(k) = 3^t * m, where gcd(3, m) = 1, and m has exactly (k - 1) distinct prime divisors. So, we have n(k) | 2^(n(k)) + 1.
Before generating n(k+1) from n(k), let us look at the number 3n(k), which clearly has k distinct prime divisors. 2^(3n(k)) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1). Since we must have n(k) always odd because of the fact that n(k) = 1 (mod 2), we deduce that 3 | (2^(2n(k)) - 2^(n(k)) + 1), so we have that 3n(k) | 2^(3n(k)) + 1. It is enough to find a prime p, such that p | 2^(3n(k)) + 1 and p doesn't divide (2^(n(k)) + 1), which could guarantee us that p doesn't divide n(k) and consequently, we could generate n(k + 1) = 3n(k)*p, which could clearly work by observing that 2^(3n(k)*p) + 1 = (2^(n(k)) + 1)(2^(2n(k)) - 2^(n(k)) + 1)*A. But, since we can pick up this prime p by Zsigmondy, we are done.

Note that in case of not using Zsigmondy, we can observe that gcd(a^2 - a + 1, a + 1) = gcd(3, a + 1) = 1 if a is not 2 mod 3 and 3 if a = 3k + 2. But if a = 3k + 2, then a^2 - a + 1 is divisible by 3 but not by 9, so we could pick up any prime p that divides (a^2 - a + 1) / 3.
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sayantanchakraborty
505 posts
#14 • 3 Y
Y by Adventure10, Mango247, cookie130
Let $n=\prod_{i=1}^{2000}{p_i}$ and wlog let $p_1<p_2<\dots<p_{2000}$.I dscribe a process on how to construct such an $n$.By the problem we have $2^{2n} \equiv 1\pmod{p_1} \Rightarrow ord_{p_1}{2} |2n$ and so by the minimality of $p_1$ we get that $ord_{p_1}=1$ or $ord_{p_1}=2$.In the first case we get $p_1|1$ which is absurd while in the second case we get $p_1|2^2-1=3 \Rightarrow \boxed{p_1=3}$.Similarly it is easy to note that $ord_{p_2}{2}|2n$ and so by the choice of $p_2$ we get $ord_{p_2}{2}|2*3=6$.Clearly there exists such a prime such that $ord_{p_2}{2}=6$(By Zsigmondy!!)In general we have $2^{2n} \equiv 1\pmod{p_k} \Rightarrow ord_{p_k}2|p_1p_2\dots p_{k-1}$.Clearly there exists a prime such that $ord_{p_k}2=p_1p_2\dots p_{k-1}$(Again Zsigmondy!!).So we are done!!!(In fact by this procedure we may fix any number of primes).
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junioragd
314 posts
#15 • 3 Y
Y by Adventure10, Mango247, cookie130
Since $N=3^k$ all satisfay the condition.Now,it is enough to prove that numbers of the form $N=2{}^3{}^k+1$ have infinitely many primes dividing them,but this is easy to prove,since we have for $n<m$ $2{}^3{}^n+1$ divides $2{}^3{}^m+1$ so suppose opposite.Now,we just need to prove that $a+1$ and $a^3+1$ can't have identical sets of primes for $a>2$,and this is true because $GCD(a+1,a^2-a+1)$ is at most $3$ and $a^2-a+1>3$ for $a>2$ so we are done.
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