A Magic Square Breakthrough - Numberphile

Matt Parker is back with Multi-Magic Squares and news of a mathematical breakthrough. Extra footage and Matt's warning is at: • A warning about huntin... --- More links & stuff in full description below ↓↓↓
Original Parker Square video: • The Parker Square - Nu...
More Parker Square stuff: • Parker Square on Numbe...
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[@numberphile]

Extra footage and Matt's warning: kzbin.info/www/bejne/aIXakmCwlLyDe5Y Original Parker Square video: kzbin.info/www/bejne/l4C3kJV9YtuKr8k More Parker Square stuff: kzbin.info/aero/PLt5AfwLFPxWKcowaX3-U4lSRUnX5E1eGw

[@MathGuyDan]

How would I go about contacting you regarding an update to this update? This result has been improved since Nov 2nd of 2024.

[@klaxoncow]

I have found a Parker square that works. Not only that, but it works for all possible dimensions as well. Albeit that there's repeated numbers, admittedly. The solution you've all been looking for: 0 0 0 0² 0² 0² 0³ 0³ 0³ 0 0 0 0² 0² 0² 0³ 0³ 0³ 0 0 0 0² 0² 0² 0³ 0³ 0³ All rows sum to zero. All columns sum to zero. All diagonals sum to zero. And it works for any power. Where do I claim my $10,000? (I kid, of course. I know this is cheating and doesn't obey the rules. Zero is a repeated number here. But it does make me wonder what the rules are for including a zero in there? Is that allowed? Because I'm wondering if you dropped a zero in the middle, for example, then you only have to control two numbers and not three for each row / column / diagonal. I wonder if using a single zero in the right place helps or hinders the quest...)

[@Bibibosh]

I think the diagonal rule is wrong because, even a 8x8 magic square, there would be 8x2 diagonals.

[@KailLabs]

​@@klaxoncoweven if you could get that to work, I think it falls into a similar category as duplicate numbers. Since every working magic square sums to 3 times the center number, you would need to sum everything to zero requiring that every number would also need it's negative on the opposite side of zero.

[@addymant]

2:26 I enjoy the running joke Matt Parker has on this channel of pretending he's recalling something and instead is just reading it off

[@RichConnerGMN]

nice pfp

[@gakulon]

In his deep thought he just happens to look at an off-camera sheet of paper with the answer. It happens to the best of us 😔 Also, off topic but I think you have a nice profile pic as well

[@noahbleeker3671]

​@@RichConnerGMNyou and gakulon shooting ur shot💯

[@mattparker-2]

i mean we all knew it works for mod 3,026,892

[@addymant]

thanks! I like y'all's too!

[@tremapar]

The $10k promised Parker prize should be a check for like $9998.50

[@alistairmackintosh9412]

Or have the cheque ID# with one digit wrong.

[@JavierSalcedoC]

Better yet, preemptively add the same 1,000% compounded inflation every 14 years but daily

[@Living_Murphys_Law]

Let's go, Numberphile main plot progression

[@kurotoruk]

Oh no, I was trying to stick to side missions!

[@oz_jones]

Heh, plot

[@parzh]

12:58 "We now have a breakthrough that's answered all our questions except for «Does a 3-by-3 magic square of squares exist?»" One might say that that's a Parker breakthrough

[@BobJones-rs1sd]

I was thinking the same thing. Better yet, define it as the PARKER PROOF: A proof that succeeds in proving every single case except the one you actually were initially trying to prove. Edit: Just want to say I mean this in no way to throw shade on all the work done in this proof, which is a fascinating and amazing contribution. It's just hilarious that it's so on-brand for these Parker-related discoveries.

[@andersjjensen]

@@BobJones-rs1sd With his sense of humor I think he would be totally on board with having that proof type of proof named after him.

[@minamagdy4126]

A Parker proof of the Generalized Riemann hypothesis would be such a troll (how dare you not prove it for the zeta function). Sadly, it might not be possible

[@BobJones-rs1sd]

@@andersjjensen Oh, I'm sure Matt would probably fine with it. My edit to my comment was referencing Nick Rome and Shuntaro Tamagishi, the authors of the actual paper Matt cites here. I don't know how they'd feel having their work characterized as a "Parker Proof."

[@josh11735]

The Parker Square saga continues!

[@halocemagnum8351]

Omg are you the PowerPoint guy? It’s been so long!

[@josh11735]

@ O_-

[@eggtimer2]

??? Add the word "saga" to anything and it's worth a comment?

[@josh11735]

@@eggtimer2 Sorry it was a Parker Comment :(

[@AanotherAardvark]

​@@eggtimer2certainly more than putting "???" at the start of your own comment. Just go troll someplace else.

[@youarentreadingthisareyou]

They secretly solved it, but they're gonna wait until Matt's 45th birthday in December to finally reveal the magic square of squares.

[@tobyfitzpatrick3914]

Why not his 49th birthday..?

[@darksnowman7192]

Because his 45th birthday is in the year 2025=45²

[@francescoalexgiacalone878]

Or the day before that, in true Parker Square fashion

[@connoranastasio]

Magic Squares getting an update patch in 2025 is wild

[@ckq]

Because 2025 is a square?

[@MatttNguyen2]

Random fact: 2025 is the sum of all numbers in the multiplication table from 1 to 9 (yes its just a way to say 2025 = (1+2+3+...+9)^2 but its cooler to think this way)

[@onecupofconsciousnessplease]

@@ckq And it's the year when Matt turns 45, which is the square root of 2025.

[@petrospaulos7736]

and also (20+25)^2 = 2025

[@superj1e2z6]

Closer and closer for Parker to be famous in some obscure mathematical book with the papers kinda stemming from the classic numberphile video.

[@joehopfield]

The vast power of the collective maths genius at Matt‘s disposal truly boggles the mind. As long as he uses it for good and not evil.

[@backwashjoe7864]

Even if he tried to use if for evil, I'm sure that Matt would end up as a Parker Villain. :)

[@Awkwerp]

My first thought near the beginning when you showed the bigger squares was "I wonder if they have a minimum size required for each power- like maybe squares minimum is four, and cubes minimum is 7, etc". Really cool to see that end up being exactly whats being studied.

[@NemisCassander]

I feel like the proof of existence has to hinge on the degrees of freedom. As stated in video, for an n x n square, you have 2n+2 requirements and n^2 - 1 options. And I think it's that -1 that is causing an issue with n = 3, because the two expressions are equal at n = 3. Interesingly, the relaxation of one diagonal constraint makes the number of options greater than the number of requirements. And this has been found to work. I'm coming at this from an OR perspective, as well as DoE, where degrees of freedom are crucial for analysis of systems. That said, this is all intuitive, and I haven't done rigorous analysis.

[@rahulkumar-hf2jz]

We got Parker Square Update before GTA 6 😭

[@katherinek6166]

That is the best mathematical result ever. It answers all of our questions, but leaves just a finite space of unknown to work out as a treat.

[@acelm8437]

You could call it a Parker result

[@M4DA.]

The first square also works in modular arithmetic. Actually for every square where there are two different sums it should be possible to find a modulus in which the square is correct. The first square works in mod 3, 9, 11, 27, 33, 99, 297, 349, 1047, 3141, 3839, 9423, 11517, 34551, 103653.

[@mathguy37]

it also works in mod 1

[@wicowan]

ahh trivial response

[@josenobi3022]

yeah idk what he was talking about

[@bunnyrape]

8:09 *difficultiness* classic Parker word

[@mazza420]

gosh it was eight year ago….. time really does feel like its flying at times

[@andrewdunbar828]

"difficultiness" is my favourite new word of the day

[@aikumaDK]

You know how Fermat's last theorem has the restriction that it's 'only' true for when the exponent is greater than 2? At this point, I'm inclined to think magic squares are only possible for _n x n_ grids with n>3

[@neilwoller]

I have a truly marvelous demonstration of this proposition that this youtube comment is too narrow to contain.

[@sebas31415]

​@neilwoller make a paper about it then so that @Numberphile can make a video on it!

[@AlexM-xj7qd]

🤣​@@neilwoller

[@leefisher6366]

A magic square that works raised to every power? SImple... 1 1 1 1 1 1 1 1 1 A magic Parker square.

[@FrankHarwald]

Another case: 0 0 0 0 0 0 0 0 0

[@wwysota]

@@FrankHarwald Doesn't work for the power of 0

[@qazsedcft2162]

You can't repeat numbers in a proper magic square.

[@legiondarywarrior39]

@@qazsedcft2162 all of those 1s are in a different secret base that completes the square

[@byGDur]

@@qazsedcft2162 i knew there had to be an exception for that one. I think he does not mention it here, but makes sense

[@unvergebeneid]

I love how Matt subtlety tries to redefine "Parker Square" to mean "a 3×3 magic square of square numbers" instead of _almost but not quite_ that. I see you Matt! Nice try!

[@EdwardWhite-q2c]

Cheah Xu Heng's surname is almost certainly Cheah and not Heng. Cheah is the transcription of the surname 謝 (via cantonese) used in Singapore and Malaysia.

[@sandekv]

Translating names is so confusing, especially when the systems are different. "Did whoever romanized it also fix the ordering, or did they leave it the way it was?"

[@deltaangelfire]

I was going to ask if a 9x9 magic square sudoku is around the corner, but then I remembered every sudoku is a magic square, and some even include diagonals >.>

[@nanamacapagal8342]

Same argument about magic square of squares. Or cubes. Or any positive integer power. Or any real number power.

[@fadhielmq]

But sudoku has repeated number. As only use 1 - 9.

[@DeadBen.]

Every time there is an update to the Parker Square, I hope it will cover the restrictions to the grid pattern that the Pythagorean theorem demands. If the grid has no repeating values and first row A, B, C; second row D; E; F; and third row G, H, I, then it is not possible to have a grid in which B>C, E>I, and D>G. Though, it “may” be possible if DC, E>I, and D G > B > I > D F > A > H > C > D G > E > I A > E > C B > C G > H

[@johnchessant3012]

I don't mind Numberphile's filler episodes, but I love it when they seriously advance the main plot like this.

[@VanceMorris]

I think this is the most important area of mathematical research today. Thank you for the news and congratulations!

[@IceMetalPunk]

So the paper has proven upper bounds on the possible lower bounds of square sizes? Honestly, my favorite part of this is that it's actual newly discovered maths brought on by a KZbin channel. If it weren't for the original video, and Tony's video, and then Brady's questioning, and then a viewer's watching it and pondering it... the new maths wouldn't have happened. I love the impact you guys are having on the world at large!

[@danielbergman1984]

I love the fact that a Parker Square really is a thing and that it's still a work in progress to find a complete magic square of squares.

[@matttrybus925]

Matt parker is the goat. Truly ushering in a modern mathematics. I dont think I would be in engineering right now without channels like numberphile and Matt's and james' channel

[@gui1521]

I'm a simple man : I see Matt, I like

[@kwanarchive]

Kind of funny how the question of power-of-2 magic squares can be proven to go right down to a 4x4 square, but 3x3 is still difficult, in a similar way to how the Poincare Conjecture was proven from 4 and up, but 3 remained unproven for the longest time. Can we pique Grigori Perelman's interest into this problem? And also that guy with the nVidia cluster who found the current largest Mersenne Prime?

[@denelson83]

For the not keen-eyed among you, you'll notice that the Pfefferman square has the added property that it contains every natural number from 1 to n² exactly once.

[@joemroz1033]

Rather exactly what the not-keen-eyed among us would NOT notice

[@ridinkulous]

_The Return of the King._

[@cordial001]

Parker Squares. The best thing to ever come out of mathematics.

[@charbelboutros8497]

Way to up-stage the Superbowl!

[@Nethershaw]

That owl is getting lazy.

[@charbelboutros8497]

So the Owl is not Superb 😂

[@joemroz1033]

A Parker Owl

[@aukir]

5:30 Matt, I'm sure you know, but others might look into why we call computers "computers" in the first place. It used to be a human profession.

[@electricmaster23]

never have i clicked on a video so quickly. was literally looking at this again yesterday.

[@Ryan_gogaku]

This seems a lot like Fermat's Last Theorem, both in terms of being sums of powers, but then also how larger cases like n > 4 were easier to solve than the small case.

[@cykkm]

A square of much smaller square numbers with the same defect as the winning solution is in Lee Sallows, The lost theorem, The Mathematical Intelligencer, 19(1997), n°4, 51-54. All rows, columns and _one_ diagonal sum up to the square number 147²=21609. I don't know Parker's definition of "nice and small" numbers; this reference may be irrelevant. 127² 46² 58² 2² 113² 94² 74² 82² 97²

[@delibirdite]

Funnily enough I've found proof of a working parker square but alas it is too large to fit in this comment section

[@MarcRidders]

The square from Cheach Xu Heng with 7 times the magic sum 194481 is nice, but in 1997 Lee Sallows published a 3x3 square with 7 times the magic sum 21609. This square is constructed by squaring the numbers of the 1st row: 127, 46, 58; 2nd row: 2, 113, 94; 3rd row: 74, 82, 97

[@DeGuerre]

I just noticed that Lee Sallows' magic square is Cheach Xu Heng's magic square with all entries divided by 3 (modulo symmetry).

[@Parborway]

12:05 this is the kind of mathematical insight that keeps me coming back to this channel.

[@tassiehandyman3090]

You and I had the same thought, my friend... if Matt was looking for an epitaph, "134 is bigger than 3..." is quite profound, eh..? 😂🇦🇺👍

[@Peter_1986]

I just heard Matt mention the Parker Square in this video, so this video is already perfect.

[@ghostkr3676]

9:38 ofcourse! why not. Bro had his hands in every math related thing

[@awebmate]

Hope i will find it in time for Parker's square birthday, and send it as a present! I will off course name it "The Parker Square"!

[@acelm8437]

Getting the 3x3 magic square of squares in a square year would be the most satisfying math result of all time

[@twiddle7125]

No one thinks it's more than coincidental that the squares that work have a power of 2 dimension? In this video he showed a 4x4 that works, and 8x8 that works, and a 128x128 that works, all powers of 2. This is striking a chord with my intuition...

[@JamesDavy2009]

That property is shared with the number of dimensions in which hypercomplex numbers are possible. I'm talking quaternions, octonions, etc.

[@BetzalelMC]

My father(b. 1930’s) taught me a neat math / square trick: only works for numbers ending in 5, thus say 15^2 would be always ends in 25, then take the 10’s digit and multiply by next highest number (1+1=2, thus 1x2=2) thus answer for 15x15= 225; another example: 55•55, (ends 25) so 5+1=6•5=3025, easy to do in your head quickly.. works with hundreds but multiplication becomes harder in head with 2-digit numbers but it will work for say 115, 13225 no? (11•12=132) just for funsies!

[@SocietyIsCollapsing]

Figuring out that 128x128 one by hand sounds utterly diabolical to me.

[@QuarkTwain]

How in the world did someone find that in 1905??!!

[@alexpotts6520]

I'm assuming the old squares from Pfefferman and the other guy were using some sort of ingenious number theory rather than just trial-and-error brute-forcing like the modern search for a 3x3 square of squares. You know, like how we know how to look for even perfect numbers, they're super-predictable, but odd perfect numbers are a (possibly nonexistent) mystery and we can only apply brute-force searches.

[@Just4FC]

Weird, just last night I watched the older videos once again, and the last one was "The Parker Square" one, and today I get this one!

[@Scientificus]

Awesome video Matt! Somehow I feel that disproving the existence should also merit the prize.

[@dean244]

The proof stopped just before what you wanted? What a Parker paper!

[@markbakker5214]

Here is how you solve every Rubik's cube, except for the 3x3

[@kailomonkey]

I think I found one! No wait, it's only a parker square.

[@bhavyapal]

You know you are a math enthusiast when you are one of the first viewers of a numberphile video

[@MelindaGreen]

Reminds me of the 4-color conjecture which was solved for the longest time on all surfaces except for the one we really cared about.

[@bilalbaig8586]

A research area that Matt has a close, personal connection with.😂😂😂

[@coloneldookie7222]

12:50 Using the numbering of each square, isosceles configurations (1,6,8 or 2,6,7 or similar), all pre-squared numbers have to end in one of five sets of last digits: {1,9}, {2,8}, {3,7}, {4,6}, and {5}. Respectively, the results of any number per set's last digit after squaring are: {1}, {4}, {9}, {6}, and {5}. Then, because of the isosceles configurations that don't share cells (such as 1, 6, 8 and 2, 4, 9), the remaining 3 squares (as exampled, 3, 5, 7) exists outside the missing difference between their row/column mates but must add up to the difference along the way. Not speaking of the cells outside of the two isosceles triangle of cells, adding up each pair in a column and row is limited in its last digit even further. If lettered a to i instead of numbered 1-9 to avoid numerical confusion, we should be taking the difference of squares instead of concerning ourselves with adding squares. a^2 - b^2 = f^2 - i^2 Then, if we take (a+b) and swap it with (f-i), we should get a correlative value to what should be used for c, e, and g. This analysis alone should limit down the search parameters more than any other I've seen being used (though, I can't say I've read EVERY limiting system being used).

[@andersjjensen]

A 4*4 square that work to the power of 4 would also be insanely cool! In fact, anything that follows n*n to the power of n is just extra mega magic.

[@MathFromAlphaToOmega]

If you want to use modular arithmetic, you could have entries 1,2,3,4,5,6,7,8,9 and work mod 1. :D

[@darraghmooose]

This is basically Matt's Riemann Hypothesis

[@rosiefay7283]

1:53 But this is all even numbers, so why not halve them all?

[@javen9693]

Squared

[@pepebriguglio6125]

Yes, I don't get it either. I mean, every sum is going to be ¼ of the original sums 🤷‍♂️

[@mewtone4781]

Yep, and also in the other near-magic square the numbers are all multiples of 3, so I'm also confused :/

[@nbooth]

If you divide by 2, the number will no longer be square.

[@nbooth]

​@@mewtone4781Because then the number wouldn't be square. You can only factor out squares.

[@bananatassium7009]

MATT IS BACK

[@isavenewspapers8890]

Now this is quite pleasing to see.

[@davecorry7723]

5:39 _"We've since invented computers."_ The twentieth century in four words.

[@Clyntax]

I understood Brady's question at 4:55 differently. He wanted to know if there is a magic square that your can raise to any power and it still works. And that is impossible. There has to be some biggest value v among the numbers. There will be a power p where v^p is larger than all other numbers combined. Therefore when raising the magic square to the power of p there will be a row/col/diagonal, which's sum is smaller than v^p. But as v^p is itself part of that magic square, it can't be a magic square.

[@ancientswordrage]

An n-dimensional magic square of squares relates to co-ordinates on an n-sphere, and high dimensions are notorious for being easier in some ways to move and manipulate things, which feels intuitive compared to d=1 to 3.

[@geoffstrickler]

I would posit that even sided magic squares are slightly easier because the diagonals don’t share any elements. So no element is used by more than 3 lines, and the majority are used in only 2 lines. In an odd sided square, the center element must appear in 4 lines.

[@Arsenniy]

wow thats cool

[@fuseteam]

Matt has accepted the parker square :D

[@williamn1055]

Return of the Square King!

[@squibble08]

i love how matt has become the magic square guy by sheer failure. honestly it feels kinda on brand

[@D4N1CU5]

The 8x8 and 128x128 (and I would guess any n=2k x n=2k square) are probably easier to find solutions for because there is no number that has to work in 4 different directions. Instead, you end up with 4 numbers in the center that only have to work in 3 directions.

[@juliuszkocinski7478]

Friendly reminder to anyone wanting to seek it: I had (I thought) brilliant idea for finding them using Gaussian primes / pythagorean triplets. But man, people searched. There is no 3x3 squared magic square with number in the center smaller than ~10^24

[@JouvaMoufette]

I think you got previous footage pasted in there twice; it's echoy

[@numberphile]

I was trying for a flashback "in the past" effect! :)

[@JouvaMoufette]

@numberphile I was also thinking that might have been intentional for that reason, yeah

[@pepebriguglio6125]

The flashback effect worked perfectly. I could've sworn it could confuse no one 😅😅

[@TarenNauxen]

More like PTSD 😅

[@wbfaulk]

3:40 "I'm not going to try to pronounce this." Immediately tries to pronounce it. It's a real Parker pronunciation-avoidance.

[@Matthew-bu7fg]

Given the research didn't come to a conclusion about magic squares of 3 which is what we care most about, it sounds like a rather Parker Square Research Paper.

[@koenth2359]

Matt pretends that he pretends that he dislikes him being dissed for creating the Parker Square

[@aarondavis5386]

One thing I'm noticing is that the 3x3 squares tend to fail in the diagonal and to nxn squares that work have n being an even number I suspect another factor making even nxn squares work is that there is no center number which must fit into 4 equations

[@bergerniklas6647]

I woulnd't say the case n=4 counts as easy if we needed Euler to find them...

[@MichaelBerthelsen]

The 128x128 magic square from 1905 is a HYPER-magic square...!😱😍

[@MoneyChanger02]

Petition for a working 3x3 Magic Square of Squares to be named an Imperfect Parker Square

[@YellowBunny]

Using the powers of modular arithmetic I have found some marvellous magic squares mod 1 that this comment is too short to contain.

[@cubesquared2291]

Never clicked a video so fast!

[@matheuscastello6554]

behold, the castello square: 1 1 1 1 1 1 1 1 1

[@rickharold7884]

thats so awesome, love it

[@mathewgriffiths1870]

If there isn't one we need to give Matt £10,000

[@Matthew-bu7fg]

did you ask Matt about his upcoming 45th birthday celebrations now we are in the year which is the square of his age?

[@doctornefario]

Every magic square works for d=0

[@ares395]

Or does it...?

[@terrance_huang]

how about d=-1? 🤔

[@rahulkumar-hf2jz]

How about magic cubes ?

[@lonestarr1490]

Yeah, we should try and write some down! Just let me get some paper and... oh. But in all honesty, I'd assume they're pretty much the same. The number of options grows much faster than the number of constraints. So from a certain point onwards, you should be guaranteed a solution exists. And yes, the same argument applies to all finite-dimensional cubes.

[@JamesDavy2009]

@@lonestarr1490 In a cube, the four triagonals also need to share the same sum and in an odd number of elements per side the centre element shares eight lines.

[@elinope4745]

I imagine that there is an abacus trick that is simple once you learn it that helps finding and fabricating magic squares.

[@funkdefied1]

Here we go again

[@David_Last_Name]

Ok, so now we need a magic square that works for powers 1, 2, 3 and 4 at the same time.

[@justinhoffman5339]

You heard it here first: a Parker square with large negative numbers is better than one with small positive numbers :P

[@lars3509]

I just want to add that I actually managed to find an even smaller magic square of squares. Behold: the 1x1 magic square of squares 923^2 works. It sums to 851929 in all directions. There may or may not be more solutions.

[@robertolson7304]

Its columns and rows. We moved beyond the the stone age.

[@markstyles1246]

Petition to make the $1000 Parker Prize actually $1003.14...

[@bertblankenstein3738]

If I decide to spend sometime solving the Parker Square, I'll quietly send any winners to Matt.

[@bertblankenstein3738]

I'm sure it will take a big weight off some shoulders.

[@stephenlund4972]

My conjecture is that the Parker square will have a solution when Matt Hair=0.

[@FloydMaxwell]

It does seem to be impressive what they did so long ago. However, the numbers can be pre-calculated (i.e. calculated only once) and then checking is just a matter of a bunch of additions. A bunch of additions is perhaps boring but not computationally impressive.