Titu Factoring Troll

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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

4221 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

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ilovepizza2020

12156 posts

ilovepizza2020

#4 h Mar 23, 2023, 10:04 PM • 2 Y

Y by centslordm, Jaxman8

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

5601 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

6876 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

7344 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

2012 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

1138 posts

sixoneeight

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

Y by

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

3446 posts

ihatemath123

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

Y by

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

7585 posts

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

1434 posts

MathWizard10

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

Y by

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

1831 posts

brainfertilzer

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

Y by

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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bachisnotcool7

391 posts

bachisnotcool7

#70 h Dec 29, 2023, 6:01 AM

Y by

Expanding and simplifying: $2x^2+2y^2+2z^2+8x^2y^2z^2-4x^2y^2-4y^2z^2-4z^2x^2-1=2023$
Factorising: $(2x^2-1)(2y^2-1)(2z^2-1)=2023$
Prime factorisation of 2023: $17^2*7$
So
$2x^2-1=17$
$2x^2=18$
$x^2=9$
$x=3$
Doing the same thing will give $y=3,z=2$
so only $(3,3,2)$ as well as permuatations will work

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Leo.Euler

577 posts

Leo.Euler

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

Y by

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

4221 posts

RedFireTruck

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

Y by

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

149 posts

LearnMath_105

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

Y by

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

5601 posts

megarnie

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

Y by

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

149 posts

LearnMath_105

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

Y by

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

329 posts

naonaoaz

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

Y by

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

2251 posts

A_MatheMagician

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

Y by

Post #76 by naonaoaz

@naonaoaz
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

498 posts

Mr.Sharkman

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

Y by

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

Solution

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pie854

243 posts

pie854

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

Y by

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

175 posts

P162008

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

Y by

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

752 posts

Mathandski

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

Y by

Enjoy a non-cutoff related post today.

Subjective Rating (MOHs)

Solution

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

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USAJMO_Qualifier

10 posts

USAJMO_Qualifier

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

Y by

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

607 posts

Ilikeminecraft

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

Y by

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

5601 posts

megarnie

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

Y by

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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