Titu Factoring Troll

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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

EGMO (geo) Radical Center Question

gulab_jamun 7

N 17 minutes ago by happypi31415

For this theorem, Evan says that the power of point $P$ with respect to $\omega_1$ is greater than 0 if $P$ lies between $A$ and $B$ . (I've underlined it). But, I'm a little confused as I thought the power was $OP^2 - r^2$ and since $P$ is inside the circle, wouldn't the power be negative since $OP < r$ ?

7 replies

gulab_jamun

May 25, 2025

happypi31415

17 minutes ago

Projections and Tangents

franchester 43

N 26 minutes ago by StressedPineapple

Source: 2020 AOIME Problem 15

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$ . The tangents to $\omega$ at $B$ and $C$ intersect at $T$ . Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$ , respectively. Suppose $BT=CT=16$ , $BC=22$ , and $TX^2+TY^2+XY^2=1143$ . Find $XY^2$ .

43 replies

1 viewing

franchester

Jun 7, 2020

StressedPineapple

26 minutes ago

IMO ShortList 2002, geometry problem 7

orl 110

N 2 hours ago by SimplisticFormulas

Source: IMO ShortList 2002, geometry problem 7

The incircle $\Omega$ of the acute-angled triangle $ABC$ is tangent to its side $BC$ at a point $K$ . Let $AD$ be an altitude of triangle $ABC$ , and let $M$ be the midpoint of the segment $AD$ . If $N$ is the common point of the circle $\Omega$ and the line $KM$ (distinct from $K$ ), then prove that the incircle $\Omega$ and the circumcircle of triangle $BCN$ are tangent to each other at the point $N$ .

110 replies

orl

Sep 28, 2004

SimplisticFormulas

2 hours ago

Cute NT Problem

M11100111001Y1R 6

N 2 hours ago by X.Allaberdiyev

Source: Iran TST 2025 Test 4 Problem 1

A number $n$ is called lucky if it has at least two distinct prime divisors and can be written in the form:
$n = p_1^{\alpha_1} + \cdots + p_k^{\alpha_k}$ where $p_1, \dots, p_k$ are distinct prime numbers that divide $n$ . (Note: it is possible that $n$ has other prime divisors not among $p_1, \dots, p_k$ .) Prove that for every prime number $p$ , there exists a lucky number $n$ such that $p \mid n$ .

6 replies

M11100111001Y1R

May 27, 2025

X.Allaberdiyev

2 hours ago

China MO 2021 P6

NTssu 23

N 2 hours ago by bin_sherlo

Source: CMO 2021 P6

Find $f: \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$ , such that for any $x,y \in \mathbb{Z}_+$ , $f(f(x)+y)\mid x+f(y).$

23 replies

NTssu

Nov 25, 2020

bin_sherlo

2 hours ago

Prove that the circumcentres of the triangles are collinear

orl 19

N 2 hours ago by Ilikeminecraft

Source: IMO Shortlist 1997, Q9

Let $A_1A_2A_3$ be a non-isosceles triangle with incenter $I.$ Let $C_i,$ $i = 1, 2, 3,$ be the smaller circle through $I$ tangent to $A_iA_{i+1}$ and $A_iA_{i+2}$ (the addition of indices being mod 3). Let $B_i, i = 1, 2, 3,$ be the second point of intersection of $C_{i+1}$ and $C_{i+2}.$ Prove that the circumcentres of the triangles $A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.

19 replies

orl

Aug 10, 2008

Ilikeminecraft

2 hours ago

c^a + a = 2^b

Havu 9

N 2 hours ago by Havu

Find $a, b, c\in\mathbb{Z}^+$ such that $a,b,c$ coprime, $a + b = 2c$ and $c^a + a = 2^b$ .

9 replies

Havu

May 10, 2025

Havu

2 hours ago

An algorithm for discovering prime numbers?

Lukaluce 4

N 3 hours ago by alexanderhamilton124

Source: 2025 Junior Macedonian Mathematical Olympiad P3

Is there an infinite sequence of prime numbers $p_1, p_2, ..., p_n, ...,$ such that for every $i \in \mathbb{N}, p_{i + 1} \in \{2p_i - 1, 2p_i + 1\}$ is satisfied? Explain the answer.

4 replies

Lukaluce

May 18, 2025

alexanderhamilton124

3 hours ago

Orthocentroidal circle, orthotransversal, concurrent lines

kosmonauten3114 0

3 hours ago

Source: My own

Let $\triangle{ABC}$ be a scalene oblique triangle, and $P$ a point on the orthocentroidal circle of $\triangle{ABC}$ ( $P \notin \text{X(4)}$ ).
Prove that the orthotransversal of $P$ , trilinear polar of the polar conjugate ( $\text{X(48)}$ -isoconjugate) of $P$ , Droz-Farny axis of $P$ are concurrent.

The definition of the Droz-Farny axis of $P$ with respect to $\triangle{ABC}$ is as follows:
For a point $P \neq \text{X(4)}$ , there exists a pair of orthogonal lines $\ell_1$ , $\ell_2$ through $P$ such that the midpoints of the 3 segments cut off by $\ell_1$ , $\ell_2$ from the sidelines of $\triangle{ABC}$ are collinear. The line through these 3 midpoints is the Droz-Farny axis of $P$ wrt $\triangle{ABC}$ .

0 replies

kosmonauten3114

3 hours ago

0 replies

inequality

Hoapham235 0

3 hours ago

Let $0 \leq a, b, c \leq 1$ . Find the maximum of $P=\dfrac{a}{\sqrt{2bc+1}}+\dfrac{b}{\sqrt{2ca+1}}+\dfrac{c}{\sqrt{2ab+1}}.$

0 replies

Hoapham235

3 hours ago

0 replies

3^n + 61 is a square

VideoCake 28

N 4 hours ago by Jupiterballs

Source: 2025 German MO, Round 4, Grade 11/12, P6

Determine all positive integers $n$ such that $3^n + 61$ is the square of an integer.

28 replies

VideoCake

May 26, 2025

Jupiterballs

4 hours ago

Centroid, altitudes and medians, and concyclic points

BR1F1SZ 5

N 4 hours ago by AshAuktober

Source: Austria National MO Part 1 Problem 2

Let $\triangle{ABC}$ be an acute triangle with $BC > AC$ . Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$ . The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$ . Let $M$ be the point where $CS$ intersects $AB$ . The line $SF$ intersects the circle $\gamma$ at a point $Q$ , such that $F$ lies between $S$ and $Q$ . Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)

5 replies

BR1F1SZ

May 5, 2025

AshAuktober

4 hours ago

4th grader qual JMO

HCM2001 46

N 4 hours ago by ethan2011

i mean.. whattttt??? just found out about this.. is he on aops? (i'm sure he is) where are you orz lol..
https://www.mathschool.com/blog/results/celebrating-success-douglas-zhang-is-rsm-s-youngest-usajmo-qualifier

46 replies

HCM2001

May 22, 2025

ethan2011

4 hours ago

Registrations Open for IOQM Level Up Test 2025!

oly01230 0

Today at 5:13 AM

Registrations Open for IOQM Level Up Test 2025!

Hello everyone!

Are you a middle or high school student passionate about math competitions like IOQM, NMTC, RMO, and beyond? Do you want to benchmark your preparation, identify your strengths, and get mentored by some of the best minds in Olympiad math?

Then you should definitely check out the IOQM Level Up Test, organized by Narayana Prodigy – a platform dedicated to identifying and mentoring the top young math talent in the country.

- Register here: https://ioqm.co.in/

What is the IOQM Level Up Test?
IOQM Level Up is a nationwide simulation test built on the latest IOQM trends. It is designed for students of Classes 8 to 12 who aim to crack the IOQM, qualify for RMO, and ultimately represent India at the IMO. This is more than just a test – it’s a gateway to deeper mentorship, curated resources, and Prodigy workshops.

Why You Should Participate
- Real IOQM-level experience: Designed by Olympiad experts and past IOQM/RMO mentors
- Detailed test analysis: Understand your strengths and areas of improvement
- Top scorer mentorship: Shortlisted students get access to advanced training sessions
- Access to learning resources: Problem sets, video solutions, and more
- Stand a chance to join the elite Narayana Prodigy program

Test Details
- Test Date: 22 June 2025
- Duration: 3 Hours
- Mode: Online & Offline (select Narayana Centres)
- For: Students of Classes 8 to 12 aspiring for IOQM 2025
- Fees: Rs. 50 for 4 Mock Tests complete with SWOT analysis and National benchmarking Report

- If you're already preparing for Olympiads, this is the test you shouldn’t miss!

About Narayana Prodigy
Narayana Prodigy is the Olympiad division of Narayana Group, known for training some of the brightest students for JEE, NEET, and Science/Math Olympiads. We believe in nurturing academic prodigies from an early stage through workshops, mentoring, and customized learning paths.

Want to Know More?
Explore more about our Olympiad programs and IOQM mentoring initiative:
- https://prodigy.narayanagroup.com
- For queries: support.prodigy@narayanagroup.com

Take the first step toward your Olympiad journey. Register now and Level Up!

0 replies

oly01230

Today at 5:13 AM

0 replies

Titu Factoring Troll

GoodMorning 77

N May 24, 2025 by ray66

Source: 2023 USAJMO Problem 1

Find all triples of positive integers $(x,y,z)$ that satisfy the equation
$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023.$

77 replies

GoodMorning

Mar 23, 2023

ray66

May 24, 2025

Titu Factoring Troll

G H J

Reveal topic tags

USAJMO

G H BBookmark kLocked kLocked NReply

Source: 2023 USAJMO Problem 1

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GoodMorning

826 posts

GoodMorning

#1 h Mar 23, 2023, 10:01 PM • 3 Y

Y by Danielzh, mathmax12, OronSH

Find all triples of positive integers $(x,y,z)$ that satisfy the equation
$2(x+y+z+2xyz)^2=(2xy+2yz+2zx+1)^2+2023.$

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RedFireTruck

4243 posts

RedFireTruck

#2 h Mar 23, 2023, 10:04 PM • 5 Y

Y by brainfertilzer, Vladimir_Djurica, mathmax12, BabaLama, tigeryong

literally just do a lotta stuff until u get (2x^2-1)(2y^2-1)(2z^2-1)=2023, at which point it is easy to see only (2,3,3) and perms work

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vsamc

3789 posts

vsamc

#3 h Mar 23, 2023, 10:04 PM • 5 Y

Y by centslordm, Vladimir_Djurica, Achilles_8055, mathmax12, tigeryong

thonk

factors as (2x^2-1)(2y^2-1)(2z^2-1) = 2023 = 7*17^2, so {2x^2-1, 2y^2-1, 2z^2-1} = {7, 17, 17}, so (x, y, z) is a permutation of (2, 3, 3) and we win

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ilovepizza2020

12154 posts

ilovepizza2020

#4 h Mar 23, 2023, 10:04 PM • 3 Y

Y by centslordm, Jaxman8, ehuseyinyigit

Expand the equation and factor to get $(2x^2-1)(2y^2-1)(2z^2-1) = 2023$ and casework to get $(2,3,3)$ and permutations

Should have been a AMC 10 Problem.

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eduD_looC

6610 posts

eduD_looC

#5 h Mar 23, 2023, 10:05 PM

Y by

did too much busy work with sum of squares to prove that $x \leq 2$ (WLOG $x\leq y \leq z$ ), and then when checking $x=1,2$ , found the factorization. bruh

This post has been edited 2 times. Last edited by eduD_looC, Mar 23, 2023, 10:06 PM

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megarnie

5611 posts

megarnie

#6 h Mar 23, 2023, 10:05 PM • 2 Y

Y by Vladimir_Djurica, The.wrld

Alternate solution: Prove $x + y + z + 2xyz \ge 2xy +2yz + 2zx + 1$ for $x,y,z \ge 3$ . Then bash the cases when $\min(x,y,z) \in \{1,2\}$ . End up with $(2,3,3)$ and permutations.

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v_Enhance

6879 posts

v_Enhance

#7 h Mar 23, 2023, 10:06 PM • 41 Y

Y by megarnie, v4913, eibc, CyclicISLscelesTrapezoid, GoodMorning, centslordm, lrjr24, CoolCarsOnTheRun, IAmTheHazard, EpicBird08, vsamc, bluelinfish, dineshs, YaoAOPS, bobthegod78, Spectator, Sedro, mahaler, rayfish, mathmax12, MarkBcc168, khina, samrocksnature, jacoporizzo, OronSH, Kingsbane2139, Lamboreghini, InCtrl, pikapika007, ChromeRaptor777, Amir Hossein, peace09, aidan0626, vrondoS, fura3334, the_mathmagician, tigeryong, pythonmaster245, cosinesine, jason543, Ilikeminecraft

By Napkin 47.1.5 there is a norm function $\operatorname{Norm} \colon {\mathbb Q}(\sqrt2) \to {\mathbb Q}$ defined by $\operatorname{Norm}(a+b\sqrt2) = a^2-2b^2$ which is multiplicative, meaning $\operatorname{Norm}(u \cdot v) = \operatorname{Norm}(u) \cdot \operatorname{Norm}(v).$ This means that for any rational numbers $x$ , $y$ , $z$ , we should have $\operatorname{Norm} \left( (1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) \right) = \operatorname{Norm}(1+\sqrt2x) \cdot \operatorname{Norm}(1+\sqrt2y) \cdot \operatorname{Norm}(1+\sqrt2z).$ Since $(1+\sqrt2x)(1+\sqrt2y)(1+\sqrt2z) = (2xy+2yz+2zx+1) + (x+y+z+2xyz)\sqrt2$ , this gives the identity $(2xy+2yz+2zx+1)^2-2(x+y+z+2xyz)^2 = (1-2x^2)(1-2y^2)(1-2z^2)$ which kills the problem.

This post has been edited 1 time. Last edited by v_Enhance, Mar 23, 2023, 10:06 PM

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DottedCaculator

7357 posts

DottedCaculator

#8 h Mar 23, 2023, 10:29 PM • 3 Y

Y by ike.chen, mathmax12, centslordm

$(\sqrt2(x+y+z)+2\sqrt2xyz)^2-(2xy+2yz+2zx+1)^2=2023$ implies $(\sqrt2x+1)(\sqrt2y+1)(\sqrt2z+1)(\sqrt2x-1)(\sqrt2y-1)(\sqrt2z-1)=2023$ , so $(2x^2-1)(2y^2-1)(2z^2-1)=2023$

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Amkan2022

2027 posts

Amkan2022

#9 h Mar 23, 2023, 10:33 PM

Y by

Just expand and factor...

After cancel and everything,
$(2x^2-1)(2y^2-1)(2z^2-1) = 2023$

And we know $2013 = 17*17*7$ , so just $(2,3,3)$ and permutations

imo this is not as hard as i would expect a jmo

This post has been edited 1 time. Last edited by Amkan2022, Mar 23, 2023, 10:34 PM

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popcorn1

1098 posts

popcorn1

#10 h Mar 23, 2023, 10:39 PM • 47 Y

Y by GrantStar, sixoneeight, vsamc, CoolCarsOnTheRun, Mogmog8, brianzjk, aidan0626, Bowser498, mahaler, matharcher, I-_-I, OlympusHero, ghu2024, pieater314159, rayfish, tricky.math.spider.gold.1, mathmax12, Ritwin, EpicBird08, centslordm, ike.chen, trigadd123, acegikmoqsuwy2000, Jndd, OronSH, aidensharp, Lamboreghini, anantmudgal09, pikapika007, tani_lex, megarnie, eibc, RP3.1415, juicetin.kim, bobthegod78, Michael1129, resources, ESAOPS, vrondoS, Sedro, Jack_w, GoldenEagle57, ohiorizzler1434, pog, Tuatara, DhruvJha, jkim0656

$50 and two icecream cakes Titu wrote this

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sixoneeight

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sixoneeight

#11 h Mar 23, 2023, 10:49 PM

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so how many points if, while doing algebraic manipulation, i wrote the wrong constant, which didnt affect the factoring, just the last step

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ihatemath123

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ihatemath123

#12 h Mar 23, 2023, 10:56 PM

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I didn't solve this problem; I tried expanding, but forgot about the 2 coefficient in the LHS when doing so, and got a big mess.

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asdf334

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asdf334

#13 h Mar 23, 2023, 10:58 PM • 1 Y

Y by peace09

The motivation for moving the $2$ inside is actually pretty simple: things like $(2x+1)(2y+1)(2z+1)$ usually have geometric sequences as coefficients and so you *really* want that to happen here implying the result

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MathWizard10

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MathWizard10

#14 h Mar 23, 2023, 11:01 PM

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Expand then set $a=2x^2, b=2y^2, c=2z^2.$ You get $a+b+c-(ab+bc+ca)+abc=2024$ which is much like a polynomial f(1) with roots $a, b, c.$ This is the motivation then do the stuff mentioned above

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brainfertilzer

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brainfertilzer

#15 h Mar 23, 2023, 11:04 PM

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for me i tried some false bounding claims that were proven with faulty calcbash. then i realized "wait why am i using calc on a jmo 1" and then i found the factorization after looking at some of the expressions on my scratch paper.

This post has been edited 1 time. Last edited by brainfertilzer, Mar 23, 2023, 11:04 PM

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Leo.Euler

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Leo.Euler

#71 h Feb 9, 2024, 5:52 PM

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Okay now I wonder why I never saw norms :blush:

Let $\mathcal{N}(a\sqrt{2}+b)=2a^2-b^2$ denote the norm in $\mathbb{Z}[\sqrt{2}]$ , and note its multiplicative nature. Move $(2xy+2yz+2zx+1)^2$ to the LHS of the equation. The resulting LHS of rearranges as $\mathcal{N}((2xyz+x+y+z)\sqrt{2}+2xy+2yz+2xz+1) = \mathcal{N}((x\sqrt{2}+1)(y\sqrt{2}+1)(z\sqrt{2}+1)) = (2x^2-1)(2y^2-1)(2z^2-1).$ Thus, we need to find the number of positive integer solutions to $(2x^2-1)(2y^2-1)(2z^2-1)=2023$ . The only solutions are $(2, 2, 3)$ and its permutations, and we conclude.

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RedFireTruck

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RedFireTruck

#72 h Mar 13, 2024, 3:44 AM

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RedFireTruck wrote:

literally just do a lotta stuff until u get (2x^2-1)(2y^2-1)(2z^2-1)=2023, at which point it is easy to see only (2,3,3) and perms work

my goofy ahh ended up getting a 6/7 :rotfl:

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LearnMath_105

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LearnMath_105

#73 h Mar 13, 2024, 6:34 AM

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did you get deducted for the last part? "notice that the only values of x such that 2x^2-1 | 2023 are x=1,2,3." I didn't really see anything wrong so I was wondering what the deduction was for

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megarnie

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megarnie

#74 h Mar 13, 2024, 1:04 PM

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He didn't specify that they satisfy the equation.

@3below It probably would be a 7 if he specified that they satisfied the equation. Although the steps are reversible, he didn't say that $(x,y,z) = (2,2,3)$ satisfied $(2x^2 - 1)(2y^2 - 1) (2z^2 - 1) = 2023$ anyway.

This post has been edited 1 time. Last edited by megarnie, Mar 14, 2024, 1:52 AM

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LearnMath_105

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LearnMath_105

#75 h Mar 13, 2024, 8:38 PM

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So you need to show (2,3,3) and permutations work? or the x=1,2,3 work?

This post has been edited 1 time. Last edited by LearnMath_105, Mar 13, 2024, 8:41 PM
Reason: edit

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naonaoaz

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naonaoaz

#76 h Mar 13, 2024, 8:48 PM

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LearnMath_105 wrote:

So you need to show (2,3,3) and permutations work? or the x=1,2,3 work?

You need to show that the find $(2,2,3)$ and permutations actually work. What he did was just proving that: if $x,y,z$ satisfy the equation, then $(x,y,z) = (2,2,3)$ (and permutations). He never showed that $(x,y,z) = (2,2,3)$ implies that $x,y,z$ satisfy the equation.

In other words, he only showed one direction of the if and only if that the problem wanted. Page 8 of https://web.evanchen.cc/textbooks/OTIS-Excerpts.pdf#page17 has more about this detail.

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A_MatheMagician

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A_MatheMagician

#77 h Mar 14, 2024, 12:44 AM

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Post #76 by naonaoaz

naonaoaz · March 13, 2024 at 4:48 PM (view)helo

LearnMath_105 wrote:

So you need to show (2,3,3) and permutations work? or the x=1,2,3 work?

x 76hellloolo Y 0 hellloolo

@naonaoaz

AoPS User

still would prob get a 6 if this was actual jmo
and plus i think all steps are reversible anways so that is not necessary

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Mr.Sharkman

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Mr.Sharkman

#78 h Apr 11, 2024, 1:38 AM

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Oof I wish all JMO #1's were like this

Solution

Notice that, if we simplify, we get
$(2x^{2}-1)(2y^{2}-1)(2z^{2}-1) = 2023,$ and the solutions are $(2,2,3)$ and the permutations of it.

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pie854

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pie854

#79 h Apr 24, 2024, 8:06 PM

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We can write $2023=-1+2x^2+2y^2+2z^2-4x^2y^2-4y^2z^2-4z^2x^2+8x^2y^2z^2$ , which we can easily factor as $(2x^2-1)(2y^2-1)(2z^2-1)$ . Now it's easy to get $(x,y,z)$ is a permutation of $(2,3,3)$ .

This post has been edited 1 time. Last edited by pie854, Apr 24, 2024, 8:07 PM
Reason: oops

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P162008

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P162008

#80 h Dec 31, 2024, 5:26 PM

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After expanding and simplifying the LHS and RHS calmly, we get

8(xyz)² + 2x² + 2y² + 2z² - 4x²y² - 4y²z² - 4z²x² - 1 = 2023

(2x² - 1)(4y²z² - 2y² - 2z² + 1) = 2023

(2x² - 1)(2y² - 1)(2z² - 1) = 2023 = 7.17²

v_7(2x² - 1)(2y² - 1)(2z² - 1) = v_7(2023) = 1

This implies that one of 3 factors present in the LHS is equal to 7.

v_17(2x² - 1)(2y² - 1)(2z² - 1) = v_17(2023) = 2

Similarly, this implies exactly 2 of the 3 factors present in the LHS are equal to 17.

Now, it's easy to get (x,y,z) = (2,3,3),(3,2,3) and (3,3,2).

This post has been edited 4 times. Last edited by P162008, Jan 3, 2025, 2:55 AM
Reason: Typo

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Mathandski

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Mathandski

#81 h Mar 5, 2025, 3:27 AM

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Enjoy a non-cutoff related post today.

Subjective Rating (MOHs)

Solution

We claim solutions are $(x, y, z) = (2, 3, 3)$ and permutations which clearly work.
$2(x + y + z + 2xyz)^2 - (2xy + 2yz + 2zx + 1)^2 = 2023$ $\Leftrightarrow (\sqrt{2}(x + y + z + 2xyz) - (2xy + 2yz + 2zx + 1))(\sqrt{2}(x + y + z + 2xyz) + 2xy + 2yz + 2zx + 1)) = 2023$ $\Leftrightarrow (x\sqrt{2} - 1)(y\sqrt{2} - 1)(z\sqrt{2} - 1) \cdot (x\sqrt{2} + 1)(y\sqrt{2} + 1)(z\sqrt{2} + 1) = 2023$ $\Leftrightarrow \prod_{\text{cyc}} (2x^2 - 1) = 2023$ $\text{Factoring } 2023 = 7 \cdot 17 \cdot 17, \text{ we are basically done.}$

This post has been edited 1 time. Last edited by Mathandski, Mar 11, 2025, 10:13 PM

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USAJMO_Qualifier

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USAJMO_Qualifier

#82 h Mar 6, 2025, 5:11 AM

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This problem is awesome. Let p(t) = t^3 - at^2 + bt - c be the polynomial with roots x, y, and z. Then the equation in the problem is the same as saying 2(a + 2c)^2 = (2b + 1)^2 + 2023 by Vieta's formulas. This tells us that (-1 + sqrt(2)a - sqrt(4)b + sqrt(8)c)(1 + sqrt(2)a + sqrt(4)b + sqrt(8)c) after subtracting both sides by (2b+1)^2 and using the difference of squares factorization. But now we take a real hard look at that equation... and what do you know... it's just sqrt(8)p(-1/sqrt(2)) * sqrt(8)p(1/sqrt(2)) = 8p(-1/sqrt2)p(1/sqrt2). But that equals 8(x - 1/sqrt2)(y-1/sqrt2)(z-1/sqrt2)(x+1/sqrt2)(y+1/sqrt2)(z+1/sqrt2). By the commutative property of multiplication, that just equals 8(x-1/sqrt2)(x+1/sqrt2)(y-1/sqrt2)(y+1/sqrt2)(z-1/sqrt2)(z+1/sqrt2). Finally (x-1/sqrt2)(x+1/sqrt2) = x^2 - 1/2 by the difference of squares factorization. Therefore, our equation just becomes 8(x^2-1/2)(y^2-1/2)(z^2-1/2) = 2023, which means (2x^2-1)(2y^2-1)(2z^2-1) = 2023. Therefore, 2x^2-1 = d, so x = +- sqrt((d+1)/2) for some divisor of 2023 d. We can ignore the negative solution because the problem says x has to be positive, so x = sqrt((d+1)/2). Notice d is either 1, 7, 17, 289, 119, or 2023. Since x is a positive integer, we need (d+1)/2 to be a perfect sqaure, which means d is 1, 7 or 17. Therefore, x = sqrt((1+1)/2) = 1, sqrt((7+1)/2) = 2, or sqrt((17+1)/2) = 3. The same logic happens for y and z. We now do casework on x. If x = 1, then (2y^2-1)(2z^2-1) = 2023. If y = 1, we get 2z^2 - 1 = 2023, so z = sqrt((2023 + 1)/2), which is not an integer. If y = 2, then 2z^2-1 = 289, so z = sqrt((289 + 1)/2), which isnt an integer. If y = 3, we get 2z^2 - 1 = 119, so z^2 = 60, which gives a non-integer solution. Thus, there are no solutions in the case of x = 1. Similarly, when y or z is 1 there will be no solution. Thus, suppose x = 2. Then (2y^2-1)(2z^2-1) = 289. If y = 2, then 2z^2-1 = 289/7, which yields a non integer value of z. If y = 3, then 2z^2-1 = 17, so z = +- 3, meaning z = 3 since we only need to consider the positive solution. Finally, suppose x = 3. Then (2y^2-1)(2z^2-1) = 119. If y = 2, then 2z^2-1 = 17, so z = 3. If y = 3, then 2z^2 - 1 = 7, so z = 2. Therefore, the solutions are (x,y,z) = (2,3,3), (3,2,3), and (3,3,2). QED.

This post has been edited 1 time. Last edited by USAJMO_Qualifier, Mar 6, 2025, 5:11 AM

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Ilikeminecraft

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Ilikeminecraft

#83 h Apr 15, 2025, 12:00 AM

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Expand everything, and we get:
$\begin{align*} & 2x^2 + 2y^2 +2 z^2 + 8x^2y^2z^2 + 4xy + 4xz + 4yz + 8x^2yz + 8xy^2z + 8xyz^2 \\ = & 4x^2y^2 + 4x^2z^2 + 4y^2z^2 + 4xy + 4xz + 4yz + 8x^2yz + 8xy^2z + 8xyz^2 + 1 + 2023 \\ \implies & (2x^2 - 1)(2y^2 - 1)(2x^2 - 1) = 2025 \end{align*}$ Simple checking gives $(2,3,3)$

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megarnie

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megarnie

#85 h Apr 21, 2025, 11:02 PM

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Here's a way to motivate the factoring from small cases (unfortunately in test I got blocked by this equation forever until bounding $\min(x,y,z)$ and then coming back): Expand and divide both sides by $2$ to get $a + b + c + 4abc = 2ab + 2bc + 2ca + 1012,$ where $a = x^2, b = y^2, c = z^2$ . Now motivated by SFFT, we write it as $(4c - 2) ab - (2c - 1) b - (2c - 1) a - (1012 - c) = 0$
Now multiplying both sides by $4c - 2$ , we have $((4c - 2) a - (2c - 1)) ((4c - 2) b - (2c - 1)) = (2c - 1)^2 + (1012 - c)(4c - 2)$ , after which dividing both sides by $(2c-1)^2$ , we have $(2a - 1)(2b - 1) = 1 + \frac{2024 - 2c}{2c - 1}$ Now multiplying both sides by $2c - 1$ gives $(2a - 1)(2b - 1)(2c - 1) = 2023,$ as desired.

This post has been edited 2 times. Last edited by megarnie, Apr 21, 2025, 11:03 PM

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ray66

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ray66

#86 h May 24, 2025, 7:19 AM

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The norm for $1-x\sqrt2$ is $1-2x^2$ . Because the norms are multiplicative, we get the factorization $-2023=(1-2x^2)(1-2y^2)(1-2z^2)$ . It's impossible for $x,y,z$ to be 0, so we force $1-2x^2$ to be either 7 or 13, giving $(2,3,3)$ and permutations

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