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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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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
USAMO 2001 Problem 2
MithsApprentice   54
N 3 minutes ago by lpieleanu
Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ=D_2P$.
54 replies
+1 w
MithsApprentice
Sep 30, 2005
lpieleanu
3 minutes ago
German-Style System of Equations
Primeniyazidayi   1
N 39 minutes ago by Primeniyazidayi
Source: German MO 2025 11/12 Day 1 P1
Solve the system of equations in $\mathbb{R}$

\begin{align*}
\frac{a}{c} &= b-\sqrt{b}+c \\
\sqrt{\frac{a}{c}} &= \sqrt{b}+1 \\
\sqrt[4]{\frac{a}{c}} &=\sqrt[3]{b}-1
\end{align*}
1 reply
Primeniyazidayi
an hour ago
Primeniyazidayi
39 minutes ago
gcd nt from switzerland
AshAuktober   5
N an hour ago by Siddharthmaybe
Source: Swiss 2025 Second Round
Let $a, b$ be positive integers. Prove that the expression
\[\frac{\gcd(a+b,ab)}{\gcd(a,b)}\]is always a positive integer, and determine all possible values it can take.
5 replies
AshAuktober
2 hours ago
Siddharthmaybe
an hour ago
Shortlist 2017/G1
fastlikearabbit   92
N an hour ago by Ilikeminecraft
Source: Shortlist 2017
Let $ABCDE$ be a convex pentagon such that $AB=BC=CD$, $\angle{EAB}=\angle{BCD}$, and $\angle{EDC}=\angle{CBA}$. Prove that the perpendicular line from $E$ to $BC$ and the line segments $AC$ and $BD$ are concurrent.
92 replies
fastlikearabbit
Jul 10, 2018
Ilikeminecraft
an hour ago
strange geometry problem
Zavyk09   0
2 hours ago
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and internal bisector $AD$. Let $AD$ cuts $(O)$ again at $M$ and $MO$ cuts $(O)$ again at $N$. Point $L$ lie on $AD$ such that $(AD, LM) = -1$. The line pass through $L$ and perpendicular to $AD$ intersects $NC, NB$ at $P, Q$ respectively. Let circumcircle of $\triangle NPQ$ cuts $(O)$ at $G \ne N$. Prove that $\angle AGD = 90^{\circ}$.
0 replies
Zavyk09
2 hours ago
0 replies
100 Selected Problems Handout
Asjmaj   35
N 3 hours ago by CBMaster
Happy New Year to all AoPSers!
 :clap2:

Here’s my modest gift to you all. Although I haven’t been very active in the forums, the AoPS community contributed to an immense part of my preparation and left a huge impact on me as a person. Consider this my way of giving back. I also want to take this opportunity to thank Evan Chen—his work has consistently inspired me throughout my olympiad journey, and this handout is no exception.



With 2025 drawing near, my High School Olympiad career will soon be over, so I want to share a compilation of the problems that I liked the most over the years and their respective detailed write-ups. Originally, I intended it just as a personal record, but I decided to give it some “textbook value” by not repeating the topics so that the selection would span many different approaches, adding hints, and including my motivations and thought process.

While IMHO it turned out to be quite instructive, I cannot call it a textbook by any means. I recommend solving it if you are confident enough and want to test your skills on miscellaneous, unordered, challenging, high-quality problems. Hints will allow you to not be stuck for too long, and the fully motivated solutions (often with multiple approaches) should help broaden your perspective. 



This is my first experience of writing anything in this format, and I’m not a writer by any means, so please forgive any mistakes or nonsense that may be written here. If you spot any typos, inconsistencies, or flawed arguments whatsoever (no one is immune :blush: ), feel free to DM me. In fact, I welcome any feedback or suggestions.

I left some authors/sources blank simply because I don’t know them, so if you happen to recognize where and by whom a problem originated, please let me know. And quoting the legend: “The ideas of the solution are a mix of my own work, the solutions provided by the competition organizers, and solutions found by the community. However, all the writing is maintained by me.” 



I’ll likely keep a separate file to track all the typos, and when there’s enough, I will update the main file. Some problems need polishing (at least aesthetically), and I also have more remarks to add.

This content is only for educational purposes and is not meant for commercial usage.



This is it! Good luck in 45^2, and I hope you enjoy working through these problems as much as I did!

Here's a link to Google Drive because of AoPS file size constraints: Selected Problems
35 replies
Asjmaj
Dec 31, 2024
CBMaster
3 hours ago
Centrally symmetric polyhedron
genius_007   0
4 hours ago
Source: unknown
Does there exist a convex polyhedron with an odd number of sides, where each side is centrally symmetric?
0 replies
genius_007
4 hours ago
0 replies
Geometry problem
Whatisthepurposeoflife   1
N 4 hours ago by Royal_mhyasd
Source: Derived from MEMO 2024 I3
Triangle ∆ABC is scalene the circle w that goes through the points A and B intersects AC at E BC at D let the Lines BE and AD intersect at point F. And let the tangents A and B of circle w Intersect at point G.
Prove that C F and G are collinear
1 reply
Whatisthepurposeoflife
5 hours ago
Royal_mhyasd
4 hours ago
Conditional geometry
Orestis_Lignos   16
N 6 hours ago by Just1
Source: JBMO 2023 Problem 4
Let $ABC$ be an acute triangle with circumcenter $O$. Let $D$ be the foot of the altitude from $A$ to $BC$ and let $M$ be the midpoint of $OD$. The points $O_b$ and $O_c$ are the circumcenters of triangles $AOC$ and $AOB$, respectively. If $AO=AD$, prove that points $A$, $O_b$, $M$ and $O_c$ are concyclic.

Marin Hristov and Bozhidar Dimitrov, Bulgaria
16 replies
Orestis_Lignos
Jun 26, 2023
Just1
6 hours ago
(Theorem, Lemma, Result, ...) Marathon
Oksutok   0
Today at 11:48 AM
1.(Świerczkowski's Theorem) A tetrahedral chain is a sequence of regular tetrahedra where any two consecutive tetrahedra are glued together face to face. Show that there is no closed tetrahedral chain.
0 replies
Oksutok
Today at 11:48 AM
0 replies
cute geo
Royal_mhyasd   4
N Today at 11:07 AM by Royal_mhyasd
Source: own(?)
Let $\triangle ABC$ be an acute triangle and $I$ it's incenter. Let $A'$, $B'$ and $C'$ be the projections of $I$ onto $BC$, $AC$ and $AB$ respectively. $BC \cap B'C' = \{K\}$ and $Y$ is the projection of $A'$ onto $KI$. Let $M$ be the middle of the arc $BC$ not containing $A$ and $T$ the second intersection of $A'M$ and the circumcircle of $ABC$. If $N$ is the midpoint of $AI$, $TY \cap IA' = \{P\}$, $BN \cap PC' = \{D\}$ and $CN \cap PB' =\{E\}$, prove that $NEPD$ is cyclic.
PS i'm not sure if this problem is actually original so if it isn't someone please tell me so i can change the source (if that's possible)
4 replies
Royal_mhyasd
Today at 7:54 AM
Royal_mhyasd
Today at 11:07 AM
Nice Collinearity
oVlad   10
N Today at 10:30 AM by Trenod
Source: KöMaL A. 831
In triangle $ABC$ let $F$ denote the midpoint of side $BC$. Let the circle passing through point $A$ and tangent to side $BC$ at point $F$ intersect sides $AB$ and $AC$ at points $M$ and $N$, respectively. Let the line segments $CM$ and $BN$ intersect in point $X$. Let $P$ be the second point of intersection of the circumcircles of triangles $BMX$ and $CNX$. Prove that points $A, F$ and $P$ are collinear.

Proposed by Imolay András, Budapest
10 replies
oVlad
Oct 11, 2022
Trenod
Today at 10:30 AM
Macedonian Mathematical Olympiad 2019 problem 1
Lukaluce   5
N Today at 10:17 AM by AylyGayypow009
In an acute-angled triangle $ABC$, point $M$ is the midpoint of side $BC$ and the centers of the $M$- excircles of triangles $AMB$ and $AMC$ are $D$ and $E$, respectively. The circumcircle of triangle $ABD$ intersects line $BC$ at points $B$ and $F$. The circumcircle of triangle $ACE$ intersects line $BC$ at points $C$ and $G$. Prove that $BF\hspace{0.25mm} = \hspace{0.25mm} CG$ .
5 replies
Lukaluce
Apr 20, 2019
AylyGayypow009
Today at 10:17 AM
DE is tangent to a fixed circle whose radius is half the radius of (O)
parmenides51   1
N Today at 10:15 AM by TigerOnion
Source: 2017 Saudi Arabia JBMO Training Tests 2
Let $ABC$ be a triangle inscribed in circle $(O)$ such that points $B, C$ are fixed, while $A$ moves on major arc $BC$ of $(O)$. The tangents through $B$ and $C$ to $(O)$ intersect at $P$. The circle with diameter $OP$ intersects $AC$ and $AB$ at $D$ and $E$, respectively. Prove that $DE$ is tangent to a fixed circle whose radius is half the radius of $(O)$.
1 reply
parmenides51
May 28, 2020
TigerOnion
Today at 10:15 AM
An equation from the past with different coefficients
Assassino9931   13
N Apr 27, 2025 by grupyorum
Source: Balkan MO Shortlist 2024 N2
Let $n$ be an integer. Prove that $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.
13 replies
Assassino9931
Apr 27, 2025
grupyorum
Apr 27, 2025
An equation from the past with different coefficients
G H J
Source: Balkan MO Shortlist 2024 N2
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Assassino9931
1371 posts
#1 • 1 Y
Y by PikaPika999
Let $n$ be an integer. Prove that $n^4 - 12n^2 + 144$ is not a perfect cube of an integer.
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GreekIdiot
270 posts
#2 • 1 Y
Y by PikaPika999
Similar to HMO 2025 p4.
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Assassino9931
1371 posts
#3 • 1 Y
Y by PikaPika999
@above Indeed, the technique from there is useful (and popular), but the very essential similarity is with this JBMO Shortlist 2020 problem.
This post has been edited 1 time. Last edited by Assassino9931, Apr 27, 2025, 1:05 PM
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Sadigly
229 posts
#4
Y by
Also AZE BMO TST P3
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MR.1
138 posts
#5
Y by
GreekIdiot wrote:
Similar to HMO 2025 p4.

what is hmo?
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Sadigly
229 posts
#6
Y by
MR.1 wrote:
GreekIdiot wrote:
Similar to HMO 2025 p4.

what is hmo?

Hellenic Math Olympiad
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MuradSafarli
112 posts
#7
Y by
Similar to Greece Math olympiad 2025
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Tamam
21 posts
#8 • 1 Y
Y by Lahmacuncu
$n^4-12n^2+144=(n^2-6n+12)(n^2+6n+12)$ from Euclidian Algorithm we can deduce that $gcd(n^2-6n+12,n^2+6n+12)=(n,12)$
If $n$ is even let $n=2k$ we will get that $16(k^4-3k^2+9)=m^3$ from here we know $v_2$ of $LHS$ is 4 which is not divisible by 3
If $n$ is divisible by 3 let $n=3k$ we get $9(9k^4-12k^2+1)=m^3$ since m is divisivible by 3 $LHS$ should be divisible by 27 which yields a contradiction.
So both the terms in the product are perfect cubes.
Now let $a=n-3$ which is even. Let $(n-3)^2+3=n^2-6n+12=b^3$. So $a^2+4=b^3+1=(b+1)(b^2-b+1)$ Since $b\equiv 3 (mod 4)$ there exists a prime congurent to 3 modulo 4 such that $p|b^2-b+1|a^2+2^2$ but from this we get $p=2$ which is a contradiction. So no solution.
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MuradSafarli
112 posts
#9
Y by
My solution in the tst
Attachments:
BMO 2024 SL N2.pdf (76kb)
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dangerousliri
932 posts
#10 • 3 Y
Y by MuradSafarli, farhad.fritl, ehuseyinyigit
This problem was proposed by me (Dorlir Ahmeti, Albania).
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megarnie
5611 posts
#11
Y by
Suppose the expression was a cube for some $n$. Factor as $(n^2 - 6n + 12)(n^2 + 6n + 12)$.

Claim: $n^2 - 6n + 12 $ and $n^2 + 6n + 12$ are relatively prime.
Proof: Suppose otherwise. Firstly, if $p$ divides both $n^2 - 6n + 12$ and $n^2 + 6n + 12$, then $p$ divides $12n$, and since $p$ divides $12(n^2 - 6n +12)$, $p$ divides $144$, so $p$ is $2$ or $3$. But if $p = 3$, then $3 \mid n$, but this means $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are both $12 \equiv 3 \pmod 9$, so their product has a $\nu_3$ of $2$, so it is not a perfect cube. Thus, $\gcd(n^2 - 6n + 12, n^2 + 6n + 12)$ is a power of $2$. Since the gcd is not $1$, it is even, so $2$ divides $n$. Since $n$ and $n - 6$ are even and different mod $4$, $8 \mid n(n-6) = n^2 - 6n$, so since $8 \mid 12n$, $8$ also divides $n^2 - 6n + (12n) = n^2 + 6n$. Thus, $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are $12 \equiv 4 \pmod 8$, so their product has a $\nu_2$ of $4$, so it is not a perfect cube. $\square$

Thus, $n^2 - 6n + 12$ and $n^2 + 6n + 12$ are perfect cubes. Taking modulo $7$ gives that $n^2 - n + 5$ and $n^2 + n + 5$ are both in $\{-1,0,1\}$ modulo $7$. Multiplying by $4$ and taking mod $7$ gives $(2n-1)^2  - 2$ and $(2n+1)^2 - 2$ are in $\{-4,0,4\}\pmod 7$.

Hence $(2n-1)^2$ and $(2n+1)^2$ are in $\{-2,2,6\}\pmod 7$, and since $-2$ and $6$ are not QRs modulo $7$, they are both $2$ modulo $7$.

Thus, $(2n+1)^2 - (2n-1)^2 = 8n \equiv 0 \pmod 7$, so $7 \mid n$, but then $(2n-1)^2 \equiv 1 \pmod 7$, absurd. Therefore, no such integer $n$ exists.
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eg4334
636 posts
#12
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First, factor it as $(n^2-6n+12)(n^2+6n+12)$. The only prime factors they can share are $2$ or $3$, because they must divide $12n$. If they are even, then $n=2a$ so we need $16(a^4-3a^2+9)$ to be a cube which is impossible because the parenthesses is odd. Similarly, if they are divisible by $3$ then $n=3a$ but $9(9a^4-12a^2+16)$ cannot be a cube by $v_3$ analysis. Therefore for the sake of our problem they are relatively pirme. Then let $a^3=n^2-6n+12, (a+3x)^3 = n^2+6n+12$, because their difference must be a multiple of $3$. Then $12n=9a^2x+27ax^2+27x^3 \implies n = \frac{3a^2x+9ax^2+9x^3}{4}$. Put this into the latter equation to get $16(a+3x)^3 = (3a^2x+9ax^2+9x^3)^2 + 24(3a^2x+9ax^2+9x^3)+192$. Its easy to see that when $x \geq 2$, this is impossible by size constraints because the right side is so big. When $x=1$, we have $12n = 9a^2+27a+27$ which is impossible by $\pmod{4}$. Therefore there is no such $n$,.
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ektorasmiliotis
110 posts
#13
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in HMO: x^2 +108= is not a perfect cube
This post has been edited 1 time. Last edited by ektorasmiliotis, Apr 27, 2025, 7:48 PM
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grupyorum
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#14
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Set $n^4-12n^2+144=k^3$. First, I prove $k\equiv 0,3\pmod{4}$ are impossible. Suppose $4\mid k$ and $k=4u$. Notice that $v_2(LHS)=4$ for $v_2(n)\ge 2$, so $v_2(n)=1$ necessarily. Setting $n=2t$ for $t$ odd, we find $16t^4 - 48t^2 + 144 = 64u^3\Rightarrow t^4-3t^2+9 = 4u^3$ which is impossible by parity. Likewise, if $k\equiv 3\pmod{4}$, then $n^4-12n^2+144\equiv n^4\equiv 3\pmod{4}$, a contradiction.

In the remainder, we will rule out $k\in\{1,2\}\pmod{4}$. Suppose first $k\equiv 1\pmod{4}$. Note that
\[
(n^2-6)^2 + 10^2 = k^3 - 8 =(k-2)(k^2+2k+4).
\]Since $k-2\equiv -1\pmod{4}$, there is a prime $q\equiv 3\pmod{4}$ such that $q\mid \alpha^2+\beta^2$ for $\alpha=n^2-6$ and $\beta = 10$. It is well known that $q\mid\alpha,\beta$ in this case, but $q\mid 10$ is clearly impossible.

Lastly, suppose $k\equiv 2\pmod{4}$. We have
\[
(n^2-6)^2 + 9^2 = k^3 - 27 = (k-3)(k^2+3k+9).
\]Once again, $k-3\equiv -1\pmod{4}$, so there is a prime $p\equiv 3\pmod{4}$ dividing both $n^2-6$ and $9$. At the same time, $3\mid n$ is impossible: otherwise $v_3(n^4)\ge 4$, $v_3(12n^2)\ge 3$ whereas $v_3(144)=2$ gives $v_3(LHS)=2$, not possible for a perfect cube.
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