The Anti-Parker Square - Numberphile

[@numberphile]

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[@bendisutd.9261]

I am watching this in Germany, and lately the vids from your channel are auto-translated with a creepy AI voice. Can you please deactivate the option?!?

[@mebamme]

Are your titles being auto-translated? The quality isn't great. The German title uses the wrong translation for "square" (the place in a city, not the geometric shape).

[@radhaprabhu7371]

I think numberphile cannot able to solve my problem 😂

[@DavidLindes]

@@mebamme Ein Platz ist nicht ein Quadrat? 😅

[@falkranduhm10]

The Aquaintance's square: A square of numbers in which all the columns rows and diagonals each add up to a different number and all of those numbers are consecutive.

[@josh11735]

The irony of the Anti-Parker Square is that because it’s a wrong answer (but you still gave it a go), it is by definition, still a classic example of a Parker Square.

[@JamesDavy2009]

It's similar in a way to trying to disprove Fermat's Last Theorem.

[@dhayes5143]

I'd almost call it an Inverse Parker Square rather than an 'Anti-'Parker Square.

[@gakulon]

I absolutely adore how enduring the legacy of the Parker Square has been

[@kevinslater4126]

And it's generating new math.

[@jessecook9776]

Rumor has it that if you take a Parker square and collide it with an Anti-Parker Square at high velocity, they annihilate and emit a photon.

[@sweetlane1813]

They emit two photons flying in opposing directions to conserve the momentum. Although I just got to think that some of them may fly into imaginary number direction. Can you spot the imaginary photons?

[@BigMcLargeChungus]

They emit a Parker

[@daniel_77.]

Or you'll get a perfect magic 3x3 square of squares?

[@matthijshebly]

A *mathon, i.e. a particle of pure math

[@rev1601]

@@matthijshebly Not to be confused with a matton, i.e. a particle of pure Parker.

[@AgentM124]

Don't tempt matt. He will reprint his books with "my formerly-friend Ayliean"

[@Olfan]

Nah, he'll be fine. After all, it's not an ((Anti - Parker) Square), but an (Anti - (Parker Square)), so only the squares antagonise one another while Ayliean and Matt can remain friends.

[@U014B]

Friendship ended with Ayliean MacDonald, Steve Mould is my friend now

[@ralphmay3284]

"my dear arch-nemesis"

[@pendarischneider]

Now@@ralphmay3284 may I suggest, instead of ""my dear arch-nemesis"", we go "my dear square-nemesis". It's scary to think where this would go if they moved away from straight lines to curves. 🤪

[@lucromel]

"Parker-Friend".

[@akshatrai9007]

We got Parker square Anti-parker square Can't wait to see standard model of squares

[@Ms.Pronounced_Name]

Standard Model of PARKER Squares

[@VonMoley]

A Parker Cube is something that appears correct from the initial observations but is infact wrong in multiple ways.

[@R_V_]

@@Ms.Pronounced_Name Standard Parker Model of Squares ?

[@adb012]

Or string squares.

[@tapir1073]

Dual Parker Square

[@bananacabbage7402]

The perfect cuboid problem and the magic square of squares are not equivalent but there is a relationship between them. They both require rational points on a congruence number elliptic curve with special properties. In the case of the magic square of squares the requirement is three rational points in arithmetic progression. On the other hand, the perfect cuboid requires three rational points in geometric progression.

[@lucasvignolireis8181]

excellent comment, should be pinned

[@jamesknapp64]

I don't know anything about elliptical curves but I had a strong feeling they weren't equivalent just similar. Glad my intuition was right.

[@kodirovsshik]

leaving a comment here so that I'm notified whenever you guys discuss the smart math stuff

[@diribigal]

​@@jamesknapp64Unfortunately, an "elliptic curve" (useful in number theory) has basically nothing to do with an "elliptical curve" (curve that is or looks like an ellipse/oval).

[@splodeyferret]

@@kodirovsshik I like numbers

Again is almost right, so it all comes to a full Parker cicle.

[@Ms.Pronounced_Name]

Presumably, a Parker Circle is a 5-dimention Rhombus?

[@ZaphodHarkonnen]

Would this be circling the square? 😂

[@mdkc]

Ayliean is trying so hard not to have this named a "McDonald Square". I am here to call that out.

[@IceMetalPunk]

No no, Wendy's is the square patty, McDonald's is the round one 😁

[@Speed001]

Old MacDonald has a sqauare, e^i - e^i = o

[@hoebare]

What if we call it a McDonald X?

[@JMUDoc]

Just when I began to think that Matt Parker had suffered enough.

[@brenatevi]

He suffers for comedy, and he's OK with that.

[@scholberger]

bro had a planet named after him, he'll be fine

[@legygax]

@@scholberger more of an anti-planet may I say

[@hoebare]

@@legygax a Parker planet?

[@Badspot]

Another unsolved problem in recreational math is finding something Euler didn't already do.

[@vincentschult1725]

Hey, I love your videos, but I've noticed that you have auto title translations enabled. This is a feature that you can disable for your viewers via your creator settings. Having your video titles being automatically translated (with whatever AI KZbin is using) results in terrible titles for your international audience. For me, a German native speaker, the title of this video is: “Der Anti-Parker-Platz”, which means “The Anti Parker Spot” or “The Anti Parking Spot”, losing the context of the “Parker Square”. Best regards ^^

[@bluelego4180]

Here in France it's "le carré anti-Parker", which means the "square against Parker"

[@maxine_q]

Not to mention the horrible thing youtube has done without the automatic AI dubbing that has no off switch.

[@lagomoof]

@bluelego4180 That interpretation also exists in English, and is part of the humour of the name. It would be nice to be able to use parentheses in natural language the same as in mathematics because then we could write "anti-(Parker square)" [anti-(carré Parker)?] and then the meaning is described more clearly, even if it loses the joke.

[@bene9127]

I can recommend the "KZbin Anti Translate" Extension for Firefox.

[@MrHeadHaunter]

Square ist eine valides Wort für Platz. aber du hast recht. die automatische Überstzung ist echt lästig

[@Dabbleatory]

To be anti-Parker, wouldn't it have to be correct?

[@donweatherwax9318]

Ouch

[@anathaetownsend1894]

Alternatively, when a Parker and a MacDonald Square co-exist, they are a Perfect Parker Square, which either does not exist, or, is what Matt Parker was trying to create when he discovered the Parker Square.

[@lyrimetacurl0]

​​@@anathaetownsend1894 yes, Parker square has the straight lines correct and MacDonald square has the diagonals correct, so if they are both then we have a true magic square square.

[@michaelbauers8800]

That was my thought.

[@santerisatama5409]

Not really. Like Bishop Berkeley kindly observed in his praise of Analysis, sometimes two wrongs - like Newton and Leibnitz put together - can make a right.

[@wallmenis]

0:53 "I am not looking to start any beef with Matt" *proceeds to start a beef with Matt*

[@rev1601]

4:46 I love how Ayliean missed that in this particular square (as well as in all others with a multiple of 25 as a base) it's not only the diagonals that work, but also the left column. Because to get to this square's magic number of 1875, the left column is missing 361, which happens to be 19^2.

[@guifig]

8:16 oh my god that freehand right triangle is absurd.

[@andrewsutherland7913]

Term one of any decent maths degree is the freehand drawing of polygons in n-dimensions. The dropout rates are atomical.

[@kindlin]

@@andrewsutherland7913 Astronomical? Atypical? I doubt you mean "related to atomic physics".

[@Viniter]

I don't think it's a pointless exercise proving two unsolvable problems are identical. It's essentially equivalent of solving one of them!

[@darraghmooose]

It's amazing to see how far just a single bit can evolve

[@opensocietyenjoyer]

every parker square is also an anti parker square by rotating it by 45°

[@squidsbizarreadventure]

Thinking geometrically!

[@opensocietyenjoyer]

@ no, group theoretically.

[@anathaetownsend1894]

Um, actually, no. In a Parker Square, the horizontal, and vertical lines, consisting of squares, sum to a square, but the diagonal lines do not sum to a square, if I remember correctly. In an Anti Parker Square (a MacDonald Square in honour of the discoverer), the diagonals of the square, sum to a square. Remember, the central number of the Parker or MacDonald Square is X. In a MacDonald Square, the opposite corners are x^2-a and x^2+a or x^2-b and x^2+b meaning that the bottom, middle number is x^2-a-b, which can only be positive when x = 3125 or more.

[@opensocietyenjoyer]

@ uhm, actually yes. in a parker square, the vertical and horizontal center lines add up correctly, so if you rotate by 45° those turn into the diagonals.

[@telhudson863]

@@anathaetownsend1894 The horizontal lines sum to thrice a square which cannot be a square.

[@zaffyr]

The parker square is becoming the new collatz conjecture

[@lyrimetacurl0]

and seems more solvable (similar to the sums of cubes equalling 42 series)

[@Xcyiterr]

I have faith that one day, the legendary crossover of the Perfect Parker Brick will be real

A brick where all the diagonals are nearly whole numbers?

[@xissel]

I wonder, considering there's a negative number there, if there's a complex square of squares that can be derived from anti-parker squares (where the bottom number is imaginary)

[@awebmate]

Finding a magic square of squares and name it "The Parker Square" would be a perfect gift for Matt's square birthday next year.

[@Randomstuff-m7p]

I gave him proof that integer solutions can not exist for Christmas. He just has not figured it out yet.

[@emilyrln]

@@Randomstuff-m7p so Christmas^2 is not an integer?

[@trummler4100]

UPDATE to my other comments: I found that... If you look on one single diagonal (you may call it "Grade -1"), x (the number squared in the middle) always features at least 1 pythagorean Prime Factor; Earliest examples: 5 => [5] 10 => [2, 5] 13 => [13] 15 => [3, 5] 17 => [17] 20 => [2, 2, 5] 25 => [5, 5] 26 => [2, 13] 29 => [29] 30 => [2, 3, 5] 34 => [2, 17] 35 => [5, 7] 37 => [37] 39 => [3, 13] 40 => [2, 2, 2, 5] 41 => [41] List of pythagorean Primes: 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, 109, 113, 137, 149, 157, 173, 181, 193, 197, 229, 233, 241, 257, 269, 277, 281, 293, 313, 317, 337, 349, 353, 373, 389, 397, 401, 409, 421, ... If you look at Diagonal Pairs (the minimum that's required for Grade 0), x always has at least 2 pythagorean Prime Factors; Examples (as outputed from my Java Code, see other comment): 25 => [5, 5] 50 => [2, 5, 5] 65 => [5, 13] 75 => [3, 5, 5] 85 => [5, 17] 100 => [2, 2, 5, 5] 125 => [5, 5, 5] 130 => [2, 5, 13] 145 => [5, 29] 150 => [2, 3, 5, 5] 169 => [13, 13] 170 => [2, 5, 17] 175 => [5, 5, 7] 185 => [5, 37] 195 => [3, 5, 13] If you farther restrict Anti Parker Squares to feature at least 1 "Border Number" to be a Square Number itself (=> Anti Parker Square of Grade 1), the valid x values shrink by a lot less than what I've expected; 25 => [5, 5] 50 => [2, 5, 5] 75 => [3, 5, 5] (no 65) 85 => [5, 17] 100 => [2, 2, 5, 5] 125 => [5, 5, 5] 145 => [5, 29] (no 130) 150 => [2, 3, 5, 5] 170 => [2, 5, 17] (no 169) 175 => [5, 5, 7] 200 => [2, 2, 2, 5, 5] (no 185, no 195) I dared to hypothesize that an Anti Parker Square of Grade n (with 0

[@ilyrm89]

This is the early Christmas gift we didn't know we needed

[@adityakhanna113]

All hail the anti-parker square! The Eulerian texts mention him, "For he shall rise, with diagonals congruent; an epitome of imperfection, announcing the End"

[@dfp_01]

I don't remember this line from Finnegans Wake

[@Becky_Cooling]

'Stop trying to make the Parker Square a thing!' cried Matt in vain

[@ckq]

4:30 Instead of using a spreadsheet. You just want to find an x such that 2x² = x² + x² can be written as the sum of 2 squares in at least 5 ways. then that's equivalent to finding an x² that's only divisible by primes of the form 4k+1 and has at least 5 factors. so 25² works since it has 5 factors. So does 65² etc.

[@tangsolaris9533]

Nice number theory

[@hoebare]

25² has four factors: (5 * 5)² = 5 * 5 * 5 * 5 65² has four factors: (13 * 5)² = 13 * 13 * 5 * 5 also, isn't 2x² always x² + x²? Or did you mean x more generically like 2x² = a² + b²? What am I missing?

[@harvey854]

That's very interesting! Could you please explain why the two problems are equivalent?

[@danielyuan9862]

​@@harvey854 If you have an arithmetic sequence a^2, b^2, c^2, then a and c have the same parity, so we can express then as (n-m)^2 and (n+m)^2. The average of those two is m^2+n^2, and that has to be equal to b^2, forming the Pythagorean triple m-n-b.

[@kindlin]

@@danielyuan9862 Not a single statement in your post is clear to me. I understand the math lingo you threw out there, but I don't get the parity point, why that lets us go -n +m, why the average of those is m and n summed and squared, why that's equal to b^2, nor why that makes a Pythagorean triple. I just found it funny how you lay it all out so succinctly like I'm supposed to understand, and I really only understand the individual words. I could probably figure out a couple of these on my own in a few minutes with a pen and paper, but I made this post instead.

[@alphanimal]

Please disable translated video titles. The German one is wrong. Also I hate seeing German titles on English videos. If I watch an English video I want to see the original title too.

[@KeesAlderliesten]

Set your youtube display language to English. Voila!

[@louisrobitaille5810]

That's a client-side thing. It's not something channel owners control.

[@mathman0569]

@@louisrobitaille5810 creators can turn it off, but it's on by default for some unknown reason

[@TheKastellan]

Im pretty sure there is an option that lets you disable it. YT just has it on by default.

[@hugofontes5708]

​@@KeesAlderliestenvalid answer as nothing was said about seeing English titles in German videos

[@jamesimmo]

Matt Parker’s ‘Love Triangle: 2nd Edition’ is already in the works now…

[@hoebare]

Or possibly Love Triangle 2: Hate Triangle

[@therealbean4372]

Based on the last slide you're basically saying: Find x, a, b in the natural numbers such that: x, x±a, x±b, x±(a-b), and x±(a+b) are perfect squares. This feels solvable...

[@brendanowen7563]

x, x±a, x±b can be found easily as mentioned in the video. I can only find at most one of the x+a+b, x-a-b, x+a+b, x-a-b on top of the original 5 squares.

[@tangsolaris9533]

It’s devilishly difficult!

[@ZeugmaP]

I remember watching the original Parker square video and now I feel old

[@JosiahPlett]

Can I just say I absolutely love how neat Ayliean's handwriting is, it makes the video very pleasant to watch :)

[@trummler4100]

Based on this problem, I've got the following values for each entry of the Magical Square: a^2 // 3*x^2-a^2-b^2 // b^2 x^2+b^2-a^2 // x^2 // x^2+a^2-b^2 2x^2-b^2 // a^2+b^2-x^2 // 2x^2-a^2 a, b and x are natural numbers; x > a, b; In addition to a^2, b^2 and x^2, 2x^2-b^2 and 2x^2-a^2 must also be square numbers. There are an unlimited number of solutions for the two diagonals themselves (see my other comments) . So there are also an unlimited number of cases in which a square number can be found under one of the other terms ("a^2+b^2-x^2", "x^2+b^2-a^2", "x^2+a^2-b^2", "3*x^2-a^2-b^2"). Now, the question I need to leave for "Algebra People" to solve: Are there specific reasons why only one of the terms "a^2+b^2-x^2", "x^2+b^2-a^2", "x^2+a^2-b^2" and "3*x^2-a^2-b^2" can also have a square number? Or to put it another way: Could one put the terms in relation to each other in a form (I lack the concrete terms for the methods) with which one *could* prove that if one of these "edge terms" is a square number, the other 3 cannot possibly be a square number?

[@Rcon_Alef]

I am studying myself whether it is possible to get at least 6 square numbers using the following network of Diophantine polynomials with 4 parameters x1 x2 x3 x4 x5 x6 x7 x8 x9 x1=a^ 2-b^2-2ab x5=a^ 2+b^2 x9=a^ 2-b^ 2+2ab x3=(p^ 2-s^ 2)x1/(p^ 2+s^ 2)+2ps(x9/(p^ 2+s^ 2) x7=2ps(x1/(p^ 2+s^ 2))+(s^ 2-p^ 2)(x9/(p^ 2+s^ 2)) the remaining values are based on the difference of the pairs. Using wolfram alpha, there are no solutions except for the trivial ones forming groups of three numbers within 1000 of each individual value. One of the advantages of this search is that each new square of the number found gives a second square in the mirror side, too. If there is a mythical point with at least 7 squares, then the numbers will be more than hundreds of billions, just a nightmare

[@MathNerd1729]

Great video! :) Btw, I was inspired by Ayliean to play around with this & I think I stumbled on an “Anti-Parker Square” with a smaller center number than the one in the video & only positive entries: A B C D E F G H I Where: A = 1169² B = 13,226,833 C = 2591² D = 12,448,945 E = 2665² F = 1,755,505 G = 2737² H = 977,617 I = 3583²

[@DaviddeKloet]

This also makes an anti Parker square: A = 5^2 C = 17^2 E = 25^2 G = 31^2 I = 35^2

[@BaccarWozat]

"Anti-Parker! Get me pictures of Anti-Spider-Man!"

[@Ultraw]

The Parker Square legacy lives on!

[@bananacat3109]

ULTRAW??????

[@Ultraw]

​@@bananacat3109 yeah that's what it says

[@Bibibosh]

im so glad people like you guys exist.. great interviewer great mather'

[@Crash-yp7ll]

Try the following example of a 'Super Anti-Parker Square of Squares:' 25^2 33^2 52^2 63^2 65^2 16^2 39^2 56^2 60^2 The diagonals AND the center column and row add up to 8,450. Unfortunately, the perimeter columns and rows do not. There are many like this, likely infinite ... .

[@milliams]

The McDonald Conjecture

[@marklonergan3898]

I love how it's not just the square - all of the "close enough" or "off by 1" errors are all getting called the Parker-... 😂 What a legend!

[@BobStein]

And if some things in this category turn out NOT to be erroneous, then we can call it the Parker-Parker-category.

[@jpdemer5]

Instead of sigma values, we can start using "Parker values". As in "Accurate to within 3 Parkers."

[@angeluomo]

The OEIS A002144 overlaps with OEIS A002313 (Fermat's 4n+1 theorem) except for the first member of the sequence, which is 2.

[@fk319fk]

Stay away from this. I checked or rejected every value up to 2^30 or so and found nothing.

[@1lomi901]

I listen to heaters fan sound to fall asleep. You make anti-Parker squares to fall asleep. We're not the same

[@Alvraera]

I have devoted literal weeks of my life toward this problem, and I admit I was a little timorous going into this video that the solution may have been found before I got to it. I don't know about the Euler brick (based on my own research, I don't see any evident connection), but the magic square of squares is all about Pythagorean triangles - in studying this problem, I have learned so much about them that it makes me giddy to talk about to others. The reason your 25^2 and 169^2 integers work is because in order for both diagonals to be equidistant squares with the center, you need to have a product of Pythagorean primes in the center: obviously the primes don't need to be unique, but 65^2 also produces diagonals of 13 - 35 - 65 - 85 - 91 (all squared, of course). You may notice the outside triple is just 13(1 - 5 - 7) and the inside triple is 5(7 - 13 - 17). For a similar reason, 65 is also the first Pythagorean hypotenuse to have two unique Pythagorean triangles: (16,63,65) and (33,56,65). Based on my research, I am becoming increasingly convinced that a truly Magic Parker Square doesn't exist, but I entertain just a tiny glimmer of hope.

[@olbaze]

I think this type of video is very important. These types of videos showcase that maths is more than just "getting to the solution" and "getting the RIGHT" answer. Sometimes, you're wrong. And sometimes, you can't get a solution, but you learn something that's interesting, or that might be useful elsewhere.

[@pacattack2586]

So you briefly mentioned 4x equidistance at 5^8, I suspect you need that as a base because in order to make the sides work they ... also ... need to be... equidistant from the center at a minimum. So any magic square of squares is at minimum 5^8 in the center

[@m802001]

I'm sure Matt Parker is laughing his asteroid off.

[@Cossieuk]

Now what we need to do is get 3 Parker squares and 3 anti Parker squares and make a Parker Rubics cube

[@NoamBJ]

very nice to see an open problem and the path you can take to tackle it

[@jamestappin4741]

Showing that the open problems of the Euler brick and the magic square of squares are the same problem wouldn't be pointless. If they are the same then proving one possible or impossible proves the other.

[@kodirovsshik]

Someone already pointed out these problems are related but not the same

[@ryanlind5239]

Obviously

[@stickfiftyfive]

​@@kodirovsshik someone who?

[@Slindi81]

So you and Matt are friends, so there will not be any beef but he is not happy with you, so I think there will be something. Let's call it Parker-beef.

[@LimeGreenTeknii]

By the way: you titled the video in Spanish "La Plaza Anti-Parker," when I think you meant "El Cuadrado Anti-Parker." Plaza means square as in a public square or shopping center that you go to, literally like how we use the word plaza in English.

[@FoobarDesign]

KZbin does that automatically. And terribly. Yesterday there was a Dutch video about applying a tourniquet (to stop heavy bleeding)... KZbin translate the title in English to 'turnstile applied' which is funny but makes zero sense (a turnstile as in the access gate type is also called a tourniquet).

[@LimeGreenTeknii]

@FoobarDesign Well, it depends on what you mean by "automatically." It was my understanding that KZbin creators have to opt-in to have their video titles appear in multiple languages. When you set that up, it suggests an automatic translation, but you have the option to rewrite it.

[@FoobarDesign]

@@LimeGreenTeknii In that case they should opt out. KZbin is terrible at translating homonyms.

[@AndrewFRC135]

You set out in search of the Anti-Parker Square of Squares, and it leads back to pythagorean triangle primes. Absolute pure poetry 😂

[@androlsaibot]

The auto-translations are comedic. Anti parking space, Auntie Parker 🤨

[@shadowseek27]

one person everything but the diagonals, the other only got the diagonals. with their powers combined theyd be unstoppable

[@acockbur]

They are equivalent. Just rotate by 45 degrees to turn one into the other.

[@aditya.khapre]

0:10 did you try counting the digits of pi?

[@johanneslouw-fn1iu]

funny🤣

[@acaryadasa]

Fun fact. If you put a parker square and an anti-parker square on the same brown paper they will self annihilate.

[@llearch]

Keep a fire extinguisher on hand.

[@T3sl4]

The reaction products include two photons and a neutrino, and by "neutrino" I mean "new KZbin video", when Parker (or Numberphile or etc.) learns of the news. And something something crowdsourced spreadsheet optimizations.

[@markusTegelane]

will we get a reaction video from Matt for this one?

[@stan-bi3hl]

Funnily enough, I've considered a similar direction of meddling with square of squares at night and it did work fast. Granted, I did not launch a spreadsheet. Since middle point square must be the third of the sum, I called the equidistant triple thing a "Parkergorean pair" (u²+v²=2w²). So you "just" need four of them with a single focus point w² and some distances matching and not matching (x + b - a ≠ x + a). Sometime around dismissing even numbers anywhere in the square I fell asleep.

[@WilliamLeeSims]

I love the phrase "bordering on conspiracy theory in terms of math... it's a very strong hunch I have". Let's instead call it "Ayliean's Conjecture" 😁

[@foobar1500]

Smallest x where diagonals are squares and x-a-b is positive - and a square(!) - is x = 845^2, with a = 205,656 and b = 507,000. Unless my brute force reasoning has holes in it there aren't solutions with x

[@kriszfrank]

"I couldn't sleep one night and I thought I would try and do just a little bit of math" is a very very frightening sentence.

[@rtpoe]

Why am I picturing a paper titled "Magic Squares, The Euler Brick, and Pythagorean Primes" in the future?

[@dominiquelaurain6427]

I will replace integer by field extension integer or functions to get : the Golden-Parker squares (entries are squares of Z(sqrt(5)) the Functions-Parker squares (entries are squares of entire functions ,...) and so on ;-) The Functions-Parker squares idea is from last Fermat's theorem for entire functions - checkout "Fermat like equation for meromorphic functions." ....no solution for Z but there exist solutions for n=2 and n=3 for entire functions.

[@FrankHarwald]

OK, I'm going out of my way here: years ago, I've managed to find a way to not just disprove the existence of the Euler brick (over integers or any ring), but because the disprove was algebraic & produced simple equations as further conditions during the process, it would also tell you under which further ring restrictions it might become possible (the conditions aren't complete but strict enough to disprove they could work in the integers). The basic sketch of the proof was by starting with the explicit parametrization of (primitive) pythagorean triples using 2 coprime integers & impose an ordering over them so that you don't double count them. Next add a third parameter because every primitive pythagorean triplet forms a class of linear solutions. This so far is standard non-linear algebra & will be the building block which we will use in repeatedly in multiple places simultaneously (*). Now consider triangle parametrized by an instance of (*) using three parameters on one side of a brick call it a). Now consider a second triangle partly on another side of the brick but also partly on the surface diagonal formed by triangle a) parametrized by another instance of (*) using another three parameters call it b). By looking at the parametrization of b) you'll see that one side which forms the parametric equation for one of cathetus is equal to the parametric equation of a)'s hypothenuse, this means you can link the parametric equations for a) with the one from b) creating a bigger one with more parameters for b) using one of b)'s primitive parameter but expand one of b)'s primitive parameters to only depend on the other primitive parameters from a) forming a larger formula for pythagorean trigangles whose one cathetus is already part of another pythagorean triangle & you might continue this step of linking pythagorean triangles linearly as much as you wanted if you had too & you would receive ever growing chains of formulas for chains of triangles having one side equal to exactly one other triangle, but we don't have to & shouldn't do so here. Instead, the rest of the proof revolves around linking pythagorean triangles using their parametric formulas in the brick in a cyclical fashion throughout all three surface triangles starting at & sharing a single common vertex on the cube called the pivot p (these three triangles will lie in three orthogonal planes & share exactly one point in common, the pivot p) & then go even further & add a second cycle by adding yet one more parametrized pythagorean triangle into this cycle, the reason being that linking three such triangles in a cycle yields a system of equations which still does have solutions while adding yet another edge of a parametrized triangle into this cycle (forming two three-cycles which share a common edge), (which is btw still only a subset of all possible pythagorean triangles inside a brick), already leads to a system of quadratic equation which can be rearranged into a form from which it will be apparent it doesn't have integer solutions, e.g. you don't need to interlink all possible pyth. triangles but only four of them in this specific way. That thing should best be written as a system of multi-parametric quadratic equation next to a visualisation of the triangles on a brick & it's important to both consider how many parameters you need & keep track, how many are independent or interlinked & which pair of parameters really are equal to using one parameter. For example, the three linearization parameters of the pythagorean triangles, which you get when you cyclically link the instanced of them through the equations, can all be accomodates & reparametrized into just using two, because when two of them are set, their cyclical equality will already fix the third one. I've done this years ago, so I hope I haven't messed up anything from the back of my brain.

[@FloydMaxwell]

I would never travel to Hong Kong (or any other CCP controlled) city.

[@joellelabastide3616]

This was so interesting. Also I’m so here for the adorable magical math unicorn genius- so cute

[@OriginalPiMan]

If we ignore the need to sum the outside four lines, then this method produces the orthogonals too. As soon as you've got 4 sets of equidistant squares, you've got a grid that matches everything except the outsides. I'd be interested to see which of those would be the first to then get at least one of the outside lines. And then, if fate would be kind (it probably wouldn't), maybe that would produce the opposite square as fitting too.

[@iamabrawler92]

So... sure, it may still be impossible, but going by that logic, it means the next point could also be that, since x at the center of the square, it's a part of 4 equations in the magic square - therefore, the starter search could have been for a square number with four equilaterally distant sets. Surely, out of all the pythagorean primes, that ought to be a lot rarer than square numbers with just two equilaterally distant sets.

[@brine1986]

i noticed that you can "inherit" equidistant property: e.g. 5×13 squared would have equidistant pairs from both "parents". My pc run search on such combinations for 2 weeks but found only 1 diagonal squares

[@LuxiBelle]

I had this playing in a different tab and then i glanced over and I went "geezus, look at dis dood"

[@galtopel2140]

this is because of complex primes, every same space triplets (x-a, x, x+a) could be created by factoring x to 2 complex integers, when x has more then one Pythagorean prime factors, you could factor x in more than one such way.

[@ryanlind5239]

Aylien is my absolute favorite

[@degalaman]

Here is a smaller example Anti-Parker square (for anyone curious if there was one) 119^2 32785 127^2 22993 145^2 19057 161^2 9265 167^2 32785 = 181^2 + 24 22993 = 152^2 - 111 19057 = 138^2 + 13 9265 = 96^2 + 49 = 96^2 + 7^2

[@degalaman]

If you'd like one extra square: 455^2 1426681 713^2 1015369 845^2 412681 959^2 37^2 1105^2

[@GladionD.Peirce]

Did you know? The Parker satellite reached the sun a few days ago, but not quite....

[@FrankHarwald]

13:00 I support the idea that the problem of finding a Parker Square is isomorphic to the problem of finding an Euler Brick.

[@ryanparker260]

I want to watch this video, but for some reason, it's pushing me away, like some sort of anti-me forcefield. Weird.

[@captainpuffinpuffinson4769]

As much as the idea of an "anti-Parker-square" is Is it not just a rotation of a Parker square?

[@robertpearce8394]

No. The Parker Square works for rows and columns, but not diagonals. This only works for diagonals.

[@_jb_3441]

I love how Matt didn't want it to be called Parker Square only for the whole math community to start using the term Parker Square :D

[@SebBrosig]

i can't contribute to finding a workoing magic square of squares, but I have an opinon on what it should be called: the Parkermost square.

[@jpdemer5]

If it works, it's the Parkerleast square.

[@MsBobsnob]

Im a physicist undergrad with a fall for number theory, especially the cuboid problem. In my final paper, next semester, I will talk about the problem and one small condition I foud that could lead to a solution(although it may be already stablished in another way, as I had a first aproach to pythagorean triples that happened to be a rephrasing of a person named "Dickenson" wich I cant find a reference at the moment) That said, I will try to map what I have done into this problem and, if I find something interessting, I will came back to share.

[@runefjord8446]

3:11 in math, if you think you are getting somewhere and everything is going really well, you are either doing it wrong, or about to hit a wall

[@DanielVCOliveira]

I demand a follow-up on this proving the link between the Parker Square and the Euler Brick

[@tomtiny]

That has to be the coolest way to write "x"

[@joelcooper6441]

i find it interesting that both the magic square of squares and the Euler brick both play well with mods

[@TheWritersMind]

Are there any insights to gleam from the mod 29 magic square that makes all of them add to 0 then? because if we somehow translated that out into the Euler brick problem maybe it could reveal something about the nature of why these things work in different "only x ways" type of things. edit for refernce im referring to (Finite Fields & Return of The Parker Square - Numberphile) 12:14

[@GiornoYoshikage]

There's a problem about multiplicative magic squares in ICPC NWERC 2024 which is awailable on Codeforces. The problem statement mentions the Parker Square!

[@brianlane723]

Could you use this to decompose any magic square into a Parker part and an anti Parker part, like with matrices and hermitian/anti hermitian?

[@GourangaPL]

i know right, once i couldn't sleep and i started to play with numbers and accidently found a pretty accurate method of estimating value of square root of any number, then took me a while to proove it, still needed a bit of help but now got it :)

[@ancientswordrage]

I came to a similar conclusion of the Euler Brick problem when I considered the sets of triples to be vector lengths of a sphere. But Diophantine equations are beyond me.

[@matthewdodd1262]

At least in some fashion, we have another restriction that must be in place for us to know if it works or not.

[@aDifferentJT]

Can you not get an anti-Parker square by rotating any Parker square by 45 degrees?

[@Jimidmih]

Please, I don't know who is responsible for this, but please please if you have any power over this, disable auto translated titles. This video is not about an Anti-Parker-Platz, it's about an Anti-Parker-Quadrat - this title is misleading in a way that's hard to describe. It's the most idiotic "feature" KZbin has ever invented; and it baits people into thinking the videos are in a language that they are not in, dealing with topics that they don't deal with. Worse-Than-Useless. And users cannot disable it on their end.

[@igorszerszunowicz8049]

Moving sofa problem allegedly got solved! You could make a video about it.

[@luciusmeredith4474]

A square of squares is easy if the numbers may be complex. (1 + i)^2 (i)^2 (-1 + i)^2 (1 - i)^2 (0)^2 (1 + i)^2 (-1 - i)^2 (-i)^2 (1 - i)^2

[@Papierkorb2292]

I don't think that's a magic square. For example, adding up the center column gives -2, whereas adding up the center row gives 0. Also, I would say you are only allowed to use nonzero squares, otherwise the problem becomes quite simple.

[@bobbrwon6458]

It is possible to prove this is I'm possible using algebra, you can put elements of the square equal to one element of the square and then it winds up putting two different elenments equal to one another meaning that you can't have 9 unique elements in a 3x3 magic square of square numbers.

[@kodirovsshik]

You don't get to arbitrarily assume that "elements of the square equal to one element of the square", that's forbidden by the problem constraints - all elements must be unique

[@bobbrwon6458]

@@kodirovsshik Sorry should have worded it better what i meant was you can put each element in terms of the other elements.

[@downwithputube]

the primes like 4k + 1 are also involved in the perfect even numbers.

[@CzarKingRex]

Oh wow, what a crossover! We've got Matt and Neal mentioned!

[@NickCombs]

I suggest we go with Rekrap Square

[@rincemind8369]

I have a truly marvelous demonstration of why Parker squares don't exist which this margin is too narrow to contain.