Catch a more in-depth interview with Ben on our Numberphile Podcast: kzitem.info/news/bejne/jqp91J1viJR7iqA
@bradensorensen966
3 жыл бұрын
What if you make a slanty square... INSIDE a slanty square?
@krishdevi6433
2 жыл бұрын
Is there an equation (or equations) for the sequence/s of odd numbers that are the result of the sum of two squares? (Not including 0^2 + n^2) For example: 5 13 17 25 29 37 41 45 53...? Where you put in 'n' and it gives you the number in the sequence?
@tarynleffler2606
2 жыл бұрын
@@krishdevi6433 I would also like to know this.
@dont5014
2 жыл бұрын
WOWW YOU DON'T READ MY PROFILE PICTURE 😶😶❌❌
@numberphile
4 жыл бұрын
This was filmed BEFORE the lockdown but edited during it! :) - Brady
@stevemattero1471
4 жыл бұрын
What a great topic! Why 4k+3 and not 4k+1?
@MrPictor
4 жыл бұрын
@@stevemattero1471 Watch mathologers's videos.
@anantkerur557
4 жыл бұрын
4k+3 is equivalent to 4k+4-1 which is the same as one less than a multiple of 4 [ 4k-1]. Note that these 'k's are different
@jojojorisjhjosef
4 жыл бұрын
Illegal maths meeting.
@yaminireddy5157
4 жыл бұрын
Ahh ,i was just about to ask. :)
@howsjimmysocool
4 жыл бұрын
I was yelling at my computer asking why the area wasnt being solved using pythagoras - and then he surprised me with it being a proof for pythagoras...
@oldcowbb
4 жыл бұрын
exactly, i was about to comment "this is a wasted chance to use pythagoras "
@eve8372
4 жыл бұрын
Haha same here!
@AngryDuck79
4 жыл бұрын
Me too. Pythagoras jumped out at me about a minute in and he took eight minutes to get around to it lol
@kezzyhko
4 жыл бұрын
Yep, started writing a comment already, then decided to check if there is already such a comment
@saint_n9ne
4 жыл бұрын
Was looking for this comment...
@itwasinthispositionerinoag7414
4 жыл бұрын
Rubik's cube in the background feeling all superior with its extra dimension
@darksnowman7192
4 жыл бұрын
Never thought to meet a kripperino on numberphile
@claymournesden8705
4 жыл бұрын
Continuum transfunctioner
@praveenanookala4457
4 жыл бұрын
Another agadmator fan!
@ffynloparnell1888
4 жыл бұрын
Klein bottle next to it feeling superior with its fourth dimension
@chronyx685
4 жыл бұрын
dark snowman kripp was not carried by annihilan battlemaster this game
@MozartJunior22
4 жыл бұрын
8:52 It's amazing how Brady has developed a mathematician's mind after all these years of doing these vidoes. This is exactly the question a mathematician would ask
@inigo8740
4 жыл бұрын
When I first found the channel, I had no idea he wasn't a maths guy, he really seemed to know. Of course after having watched many videos and having learned about the channel, I can now tell a bit he isn't originally a math guy. But you can also see he's getting a bit of a hang on it.
@Liggliluff
4 жыл бұрын
I wished we were given an answer to that question.
@alessandrofelisi6037
4 жыл бұрын
@@Liggliluff They become "less sparse" as you go up! In fact, they tend to "fill" all the natural numbers, in a certain sense.
@jamieg2427
4 жыл бұрын
He's been filming Numberphile long enough that some of his first viewers could have gotten a masters in math twice over.
@RipleySawzen
4 жыл бұрын
@@alessandrofelisi6037 Do you have a list of these somewhere or a proof of this?
@craigmcqueen7992
4 жыл бұрын
7:03 That was a real pleasure to see how such an elegant proof of Pythagoras' theorem just popped out like that.
@EnteiFire4
4 жыл бұрын
I really like the one without algebra, where you rearrange the four triangles to make two rectangles, where one on the top left corner horizontally, and the other is on the bottom right vertically. The rest of the square is made of a square of side "a" and a square of side "b".
@svz5990
11 ай бұрын
@@dont5014why the chicken kfc borgor are you everywhere?
@edwardstennett4794
4 жыл бұрын
Ben Sparks is by far my favourite KZitem mathematician. His knack for explaining things in a way that's easy to understand for pretty much anyone makes maths so much more accessible. I regularly rewatch his videos - would love to see him do even more videos on Chaos.
@fredresz7773
4 жыл бұрын
Edward Stennett Man I love all of the people on this channel! ALL of them! They’re all so fun to watch and enjoy math with.
@apothecurio
2 жыл бұрын
I feel sometimes the guests can speak pretty dryly. Ben Sparks is NOT one of these guests. Not by any means whatsoever.
@dont5014
2 жыл бұрын
WOWW YOU DON'T READ MY PROFILE PICTURE
@adamqazsedc
2 жыл бұрын
He is, a math teacher
@CursedJoker
4 жыл бұрын
I'm sure that if you go and film Matt Parker long enough, he'll come up with some "kinda possible" Squares.
@nanamacapagal8342
4 жыл бұрын
3.
@quinn7894
4 жыл бұрын
*cough* *cough* classic Parker square
@jackwilliams7193
4 жыл бұрын
well done
@caleblewis8169
4 жыл бұрын
Parker squares
@dont5014
2 жыл бұрын
WOWW YOU DON'T READ MY PROFILE PICTURE 😶😶❌❌
@chrismcdonald6195
4 жыл бұрын
"We're getting square numbers because we're drawing squares." FINALLY - something on Numberphile I kinda know already!
@caleblewis8169
4 жыл бұрын
What's a square
@rexlapis3126
3 жыл бұрын
Quadrilateral
@rameshshinde4488
3 жыл бұрын
Hi
@dont5014
2 жыл бұрын
WOWW YOU DON'T READ MY PROFILE PICTURE 😶😶❌❌
@svz5990
11 ай бұрын
@@dont5014because your name says Don't!
@midwinter78
4 жыл бұрын
As soon as I saw the square-in-a-square diagram, I started yelling "that's the square on the hypotenuse!"
@GuyNamedSean
4 жыл бұрын
I realized that as well! I also roundabout found my way toward the theorem behind what numbers are and are not candidates.
@loganstrong5426
4 жыл бұрын
Same! Which made it pretty obvious to me that any square you can make has an area that is the sum of two squares.
@criskity
4 жыл бұрын
Me too, and I was wondering why he decided to go the messier route by subtracting the areas of 4 triangles. Pythagoras is right there to begin with!
@fredresz7773
4 жыл бұрын
Logan Strong Stumbling across little gems like this and the comment from GuyNamedSean above is what really deepened my love for math!
@lamusicadepedrovicente
4 жыл бұрын
yes! it got me a bit nervous they not using that to find the area
@Dude-sr4ji
4 жыл бұрын
7:20 why do I feel like I just got rickrolled by Pythagoras?
@dovahseod
3 жыл бұрын
You got Pythagorasrolled
@TangoWolf09
4 ай бұрын
Pythagorolled
@Peetzaahhh
4 жыл бұрын
5:41 in and I'm bewildered how Pythagoras didn't come up EDIT: 7:24 oh it’s because he’s proving it
@matthewlyons6544
4 жыл бұрын
Peetzaahhh this was exactly what I was thinking. In my head I was shouting Pythagoras, but then realised the reason it wasn’t referenced was because it was being proved!
@pickles974
4 жыл бұрын
@Peter Attia when did you stop learning maths and how old are you? I'm asking a bunch of people in the comments because I'm assuming people who are amazed by this video must be about 11 years old or younger.
@mhr6780
3 жыл бұрын
@@pickles974 🙄
@OwlyFisher
3 жыл бұрын
@@pickles974 rude. not all of us intuit maths
@adamqazsedc
2 жыл бұрын
@@pickles974 yknow how this video isn't _just_ about The Pythagorean Theorem.
@waltercisneros9535
4 жыл бұрын
7:10 "We prove Pythagoras" *drops mic*
@chrisbersabal102
4 жыл бұрын
3:10 i thought i am the only who does that pen cover thing
@hamiltonianpathondodecahed5236
4 жыл бұрын
xD I didn't notice
@sparkytheteacher
4 жыл бұрын
@@hamiltonianpathondodecahed5236 neither did I...
@fasjdays
4 жыл бұрын
Thats exactly what I said too! I'm not the only one after all
@DouglasZwick
4 жыл бұрын
At 11:20, I was really hoping he was going to circle the number on his screen with that marker.
@stevejobs5488
4 жыл бұрын
That moment made me go 😬
@AdamBomb5794
4 жыл бұрын
That one 3Blue1Brown video of which coordinates are on a circle just popped into my mind.
@zackszekely6618
4 жыл бұрын
Because you can't have an odd number of threes 😃
@programmingpi314
3 жыл бұрын
also that video hints at another way of finding if a square is possible or not.
@RajSingh-qp9st
3 жыл бұрын
8837666846
@cfgauss71
4 жыл бұрын
The real question we want answered: where does that ladder lead to?
@Bibibosh
4 жыл бұрын
Arthur Clay hahaha !!!! Nice observation
@labiadh_chokri
4 жыл бұрын
It leads to z axis
@ChocoHearts
4 жыл бұрын
It leads to Dennis, of course.
@trueriver1950
3 жыл бұрын
Clue: This was filmed just before lockdown, when Covid-awareness was rising. It's the emergency escape in case the other person coughed unexpectedly. Hold breath while outclimbing the viral aerosols and on exit breathe out before inhaling. Luckily we subsequently thought of using masks.
@MrCreeper20k
4 жыл бұрын
I’d be interested to see this extended into 3D. Might be a little more tedious than insightful though
@HeavyMetalMouse
3 жыл бұрын
In 3D... Assume (0,0,0) is a vertex, and lattice point (a,b,c) is a vertex (with integers a,b,c >=0). The other two vertices on the cube 'adjacent' to the origin in the other two directions would need to be of the form (x,y,z) and satisfy ax + by + cz = 0 (perpendicular to (a,b,c) and (x^2) + (y^2) + (z^2) = (a^2) + (b^2) + (c^2) and x,y,z in the Integers At this point, I'm not entirely sure what method to use to show when you can find two suitable lattice points satisfying those conditions. But if you do, then you get the other four for free, as they're just adding together the vectors, and adding integers always yields integers. If you want to know what *integer* volumes of the cubes are possible, then you also are restricting your search to cases where sqrt(a^2 + b^2 + c^2)^3 is an integer, which only happens when sqrt(a^2 + b^2 + c^2) is an integer. In which case, your solution set is some subset of the cube numbers. However, all cubed integers are, by definition, formable on lattice points (just take the orthogonal points), therefore any solution that could theoretically be formed by a 'tilted' cube must also be formed by a non-tilted cube. Therefore, if you only want integer cube volumes, the solution is a trivial "All integers of the form s^3, where s is a positive integer.", as as any tilted cube on the lattice points must have a either a volume in that set, or a non-integer volume.
@danjtitchener
4 жыл бұрын
Occurs to me that this is similar to the infinite forest problem, when it was asked which trees could you see!
@clockworkkirlia7475
4 жыл бұрын
Oooh, I don't know that one! It sounds interesting.
@timh.6872
4 жыл бұрын
If each grid point has a line orthagonal to the plane (representing a tree trunk), and you stood near the origin, can you see the horizon? If so, how much?
@anandsuralkar8376
4 жыл бұрын
Right
@stefanjoeres7149
4 жыл бұрын
Is area 51 possible?
@stydras3380
4 жыл бұрын
51 is 3*17 and the power of 3 is odd, so no :0
@MattiaConti
4 жыл бұрын
President send me a message to eliminate this comment as soon as possible
@brokenwave6125
4 жыл бұрын
Area 51 was an inside job
@Jivvi
4 жыл бұрын
@@brokenwave6125 if it was an outside job, it would be area ∞-51.
@robo3007
4 жыл бұрын
@@stydras3380 Proof that mathematics was invented by the government to cover up their secrets!
@MarcDittner
4 жыл бұрын
I agree with Ben, this is also my favourite proof of Pythagoras.
@ffggddss
4 жыл бұрын
"Which numbers are possible?" Ans: All the 2-square numbers; i.e., any number that's a sum of two squares. 0, 1, 2, 4, 5, 8, 9, 10, 13, 16, etc. Not 3, 6, 7, 11, 12, 14, 15, etc. Reason: once you draw one side of the square, the rest is determined (but allowing reflection across the initial side). That side must connect a pair of grid dots, the square of whose separation, s, is always a sum of two squares : s² = ∆x² + ∆y². But the square's area *is* just A = s² = ∆x² + ∆y² And of course, ∆x & ∆y are always integers. PS: The method he uses to prove Pythagoras is, I believe, due to James A. Garfield, when he was schoolteacher, before becoming 20th president of the US. PPS: The characterization of the 2-square numbers is based on characterizing primes in the ring of complex integers. [If you don't know what a mathematical ring is, don't pay it any mind - it isn't necessary; it just might help a little if you do know.] Warning: This gets a bit heavy, which is probably why it isn't in the video, so proceed at your own risk! Real primes can be sorted into 3 classes, modulo 4 [when dividing any integer by 4, the remainder is one of: 0, 1, 2, or 3; equivalently, 0, ±1, or 2]: There are no primes that are 0 mod 4. (i.e., no multiples of 4 are prime!) There's only 1 prime that's 2 mod 4; 2 itself. All others are ±1 mod 4 [I.e., 1 or 3 mod 4]. 2 can trivially be written as a sum of 2 squares: 2 = 1 + 1. Any number that is -1 mod 4, cannot, because all squares are 0 or 1 mod 4, so any sum of two of them can only be 0, 1, or 2 mod 4; never 3 == -1 mod 4. So among real primes, only 2, and the +1 mod 4 primes, can be written as a sum of 2 squares. It so happens that all +1 primes can be written as a sum of 2 squares - I'm not recalling the proof of that at this time. [I invite anyone who knows how to do that, to show it here!] So among the complex integers, the +1 primes are composite, being factorable into a product of 2 complex integers: p = a² + b² = (a + bi)(a - bi) while the -1 primes remain prime, because any product of 2 complex integers must be a conjugate pair in order for the product to be real; and such a product is necessarily a sum of 2 squares, which in turn, cannot be -1 mod 4. Now, the _coup de grace._ For complex numbers, the squared modulus [modulus = its "length"] of a product of them is the product of their squared moduli: w = u + vi; z = x + yi; wz = (ux-vy) + (uy+vx)i |w|² = u² + v² ; |z|² = x² + y² |w|²|z|² = |wz|² ; that is, (x² + y²)(u² + v²) = (ux-vy)² + (uy+vx)² . . . [This can be verified by simply expanding both sides.] Thus showing that a product of a pair of 2-square numbers is again a 2-square number. Now consider the prime factorization of any positive integer, N. Factor out any squares; that is, any prime, p, raised to a power ≥ 2, can be factored into p times an even power of p, which is thus p times a square. You now have N = one big product of squares, which itself is a square, times a product of single, distinct primes. If any of those distinct primes is -1 mod 4, N cannot be written as a sum of 2 squares; if none of them are -1 mod 4, N can be written as a sum of 2 squares. Thus, 3, 7, 11, 19, 23 cannot, being -1 primes; but neither can 6, 12, 14, 15, 21, 22, 24, 27, or 28, because of their prime factorizations. 0, 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, and 29 can each be written as a sum of 2 squares. Fred
@alexcerullo3143
4 жыл бұрын
ffggddss Fred
@josevillegas5243
4 жыл бұрын
Thanks for this! I think I mostly get it, except for the second to last (penultimate) paragraph: "If any of those distinct primes is -1 mod 4..." Cam you explain? So far, I get N has been factored into N = p*q*...*r*(a*b*...*c)^2 = pq...rA^2 where p,q,...r are primes and a,b,...c can be any integers, and A = a*b*...*c. Let pq...r be shorthand for p*q*...*r A^2 is a 1-square and is it trivially a 2-square since A^2 = 0^2 + A^2? Then using the fact that the product of a pair of 2-square numbers is itself 2-square, pq...rA^2 is a 2-square iff pq ..r is a 2-square? I think that's where my confusion arises because I don't know how modular classes behave under multiplication. As you mentioned the primes p,q, ..r have to be +/-1 mod 4. Does your conclusion (the penultimate paragraph) hinge on which mod 4 class the pq...r product is in? If you multiply two +1 mod 4 numbers, you get another +1 mod 4 number: (4k+1)(4j+1) = 4m+1 for some m But also if you multiply two -1 mod 4 numbers, you still get a +1 mod 4 number: (4k-1)(4j-1) = 4n+1 for some n To be exhaustive, if you multiply a -1 mod 4 number by a +1 mod 4 number, you get a -1 mod 4 number: (4k-1)(4j+1) = 4o-1 for some o So what would happen if p,q,..r had an even number of -1 mod 4 primes? E.g. if p,q,...r was just 3 and 7. Their product is 21 which is a +1 mod 4. Thanks for reading this far! Looking forward to your response and hopefully understanding what's going on. I'm really curious.
@mufasao6776
3 жыл бұрын
The Fred at the end is so funny to me. No QED, no square or symbol, just "Trust me, I'm Fred" lol
@danielstephenson7558
4 жыл бұрын
Ben Sparks' videos are some of the most watchable videos. The Mandelbrot videos. the Golden Ratio and the Chaos Game are among my favourites.
@Cardgames4children
4 жыл бұрын
It's also my favorite proof of the Pythagorean theorem. It's so simple and intuitive.
@Bibibosh
4 жыл бұрын
We need part two. This guy is awesome!!!!! The idea is awesooooooome!! This channel is _________!!!!!!!!
@dascha78
4 жыл бұрын
...awes(49*o)me
@Filipnalepa
4 жыл бұрын
... Numberphile?
@Danilego
4 жыл бұрын
7:08 Oh man, I wasn't expecting that! It must be the simplest proof of pythagoras
@wj11jam78
4 жыл бұрын
This video was excellently done, because in the first few minutes I had essentially watched the whole thing. The information was presented in a way which meant that I could easily jump ahead, and figure out the formulas and proofs on my own, without the explanation. It made all the math behind the problem jump out at me. As soon as I saw the triangles, I knew Pythagorean theorem was coming, so I tried it out, and the whole thing solved itself. I'm not the best at math, especially algebra (though I do love geometry), so props to this guy. Really intuitive way of teaching this.
@nymalous3428
4 жыл бұрын
This was unreasonably interesting for me. I find myself compelled to make a spreadsheet with [a,b] possibilities...
@mikedrewson5545
4 жыл бұрын
Wow, I did not expect this to be a proof of Pythagoras. This is why math is amazing.
@anantkerur557
4 жыл бұрын
At 4:03, You can get numbers that are of the form a²+b², so 5 = 2²+1², 9 = 3²+0², and so on, but you can't write three as such. Edit: Yes!! I never knew such a simple problem could be so intricate and advanced!
@judychurley6623
4 жыл бұрын
but you cant have a side of 0 units...
@tomwakefield1726
4 жыл бұрын
@@judychurley6623 the two numbers are the sides of the triangles which creates the slant. If you have a 0 it just means that the triangle is just a straight line, so there's no slant
@praveenanookala4457
4 жыл бұрын
Nice
@judychurley6623
4 жыл бұрын
@@tomwakefield1726 it's the length of the sides of the triangles that are squared.
@renpnal229
4 жыл бұрын
@@Seven-ez5ux The actual proof of the fact that a number can be expressed as a sum of two square if and only if its prime factorization contains no primes of the form 4k + 3 raised to an odd power.
@sumdumbmick
4 жыл бұрын
@9:00 I love that you describe this as a problem that you personally have no intuition about. So often I see mathematicians and scientists talk about intuition as something that is universal, and so if they don't have a good intuition about something they're highly likely to write the entire human race off as having no intuition about it, which is astoundingly solipsistic really. So it deserves mention and respect that you did not fall into that pattern at all but demonstrated recognition that you are but one of many minds, and just because you lack knowledge or intuition about something does not imply that others necessarily would as well. To be perfectly clear, I also have absolutely no intuition about this particular thing, but I quite expect that some people do.
@snowingbook
4 жыл бұрын
Klein Bottle in the background : **exists and Ben doesn't mention it** Clifford Stoll : YOU HAVE SINNED, MORTAL
@justinjustin7224
4 жыл бұрын
How to draw a square of area 3 using a grid of equally spaced dots: 1. Draw the smallest square you can on the grid. 2. Define the shortest distance between dots to be sqrt(3). 3. Laugh at the problem giver for not clearly specifying units.
@whatno5090
4 жыл бұрын
Another fun way to figure this out, is that you know that for any such "slanty square" lying anywhere in the real plane, you can fix one of the vertices on a lattice point and rotate the square about that point; if (and only if) somewhere along the way, one of the nearest vertices hits another lattice point, then you can do a slanty square of that area. This means that we can reduce the problem to finding lattice points on the circle of radius s, where s is the side length of the square; and s = sqrt(A). But the equation for the circle of radius r is x^2 + y^2 = r^2, so of course, this means we need to find integer solutions x^2 + y^2 = A!
@JM-us3fr
4 жыл бұрын
This is why understanding which primes are sums of two squares is important. 3Blue 1Brown does an excellent video on this, showing why these are the only numbers that can't be expressed in this way.
@cee_jay_0
4 жыл бұрын
but lets ask the opposite question: for which integers can you find a pair of multiple solutions, like 0²+5² = 25 and 3²+4²=25. up to 1000 there are 6 integers, that can be written as the sum of two squares in 3 different ways, and i haven't found any number above that qith more pairs yet. and i haven't found any pattern in them either. here's the list: 325 425 650 725 845 850
@willnewman9783
4 жыл бұрын
There is a known formula given an "n" that gives how many pairs of a,b have a^2+b^2=n. 3blue1brown derives this formula in his video about pi/4=1-1/3+1/5-...
@Benlucky13
4 жыл бұрын
I got really excited with the first few numbers in the string because they're adding the digits of pi after the initial 3. so 3 to 6 is '3', 6 to 7 is '1', 7 to 11 is '4', and 11 to 12 is '1'. unfortunately the pattern breaks after that, was hoping this would be another one of those odd ball "why the heck does pi show up here" strings. 3141 is still a fun coincidence, though.
@timh.6872
4 жыл бұрын
Actually, you can get to Pi from this fact! That divisibility rule he shows can be used to count how many grid points are at a distance sqrt(n) from the origin. If you add up all the counts for n from 1 to some large integer R, you approximate the area of a circle of radius R. Using that 4k+1, 4k+3 only if odd rule, you can rearrange the count into the sum 1 - 1/2 + 1/3 - 1/4 + ... times 4 R², which means the alternating sum is equal to π/4. 3Blue1Brown has a more in depth walkthrough, I think it's called "Approximating Pi with Prime Numbers", but I might be wrong there.
@tamirerez2547
4 жыл бұрын
3= 2^2 + i^2
@anantkerur557
4 жыл бұрын
That leads to a interesting question - what if we allowed Complex numbers?
@sergey1519
4 жыл бұрын
@@anantkerur557 it is easy to see that it is equivalent to just searching for a²±b²=c solutions
@diogor379
4 жыл бұрын
@@anantkerur557 My intuition would be that instead of a grid of dots we would have a space of dots to work with. The area 3 square would have to be "lifted" in space by one of its corners into the complex axis I think.
@alansmithee419
4 жыл бұрын
You can't travel a distance i on the square grid, making this not applicable.
@alfeberlin
4 жыл бұрын
@@diogor379 Intuition then fails us because even if we drew the imaginary part in a third dimension, the squares stretching into this would appear to grow with a growing imaginary part, but mathematically they should shrink.
@ThemJazzyBeats
3 жыл бұрын
This is a lot like the "Pi hiding in prime regularities" video of 3b1b, where one of the things he does in that video is check if a number can be expressed by the sum of 2 squares
@CaptainSpock1701
4 жыл бұрын
I'm looking at this and the whole time I'm thinking hang on guys, why not just use Pythagoras? It's so obvious. Then "... do you realise we just prove Pythagoras?" - *Mind = Blown* Wow! Simple proof. Going around the complete oposite way as what I was expecting. Great work guys. Always love your videos!
@MrNacknime
4 жыл бұрын
It's literally only about which distances of points exist, having squares around just complicates the issue. And because of pythagoras, you can have all distances of the form sqrt(a^2+b^2), a,b>=0. And then your square sizes are just squares of these numbers, so a^2+b^2.
@chinareds54
4 жыл бұрын
One small quibble... I would say a>0, b>=0, because a 0x0 is not a square; it's a point.
@brian23fink
4 жыл бұрын
What about the opposite: slanty squares that bound, instead of being bounded by, a non-slanty square?
@5eurosenelsuelo
4 жыл бұрын
This guy is simply on another level
@EebstertheGreat
4 жыл бұрын
This is a direct corollary of Fermat's theorem on the sum of squares, which states that a prime number p is the sum of two integers squared x² + y² if and only if p ≡ 1 (mod 4).
@notoriouswhitemoth
3 жыл бұрын
@7:30 not only does it prove the Pythagorean theorem, it proves it in the same way the Pythagoreans discovered it in the first place.
@TeddSpeck
4 жыл бұрын
Just came from the Periodic Table video on arsenic to this one. First scene in this one? A green wall. Coincidence?
@tzisorey
4 жыл бұрын
"And how big is that square?" _16 uni...._ "Don't worry, it's not a trick question" ..... _Well now I'm not sure..._
@gabrielmello3293
4 жыл бұрын
As a current engineering student, the moment he tried to calculate the area of the square, I was yelling in my head "just use the damm pythagorean theorem". A few minutes later I remembered how I used to watch numberphile way back in middle school when I still didn't know the pythagorean theorem and all the heavy math I know now, and only then I could appreciate the beauty and the art in the video. This video is intended for people like 14 year old me who didn't know that much math, but absolutely loved these kinds of problems. Thank you for keeping up the making of videos that motivate and introduce math outsiders into such a beautiful discipline.
@numberphile
4 жыл бұрын
He also deliberately held back pythagoras for the reveal later on!
@gabrielmello3293
4 жыл бұрын
@@numberphile Yes, that's what I was trying to say.
@elvest9
4 жыл бұрын
Are these related to the Parker Squares?
@benzeh4769
4 жыл бұрын
MORE BEN!!! I LOVE THIS MAN
@robertveith6383
3 ай бұрын
Stop yelling in all caps.
@russellthorburn9297
Жыл бұрын
So many questions popped into my brain while watching this(some of which were answered): 1. What method can I use to check to see if any number works? 2. Is there a visual pattern going on here? 3. Do the numbers that don't work get farther apart? 4. Are there any examples of where 3 in a row don't work or is 2 in a row the maximum? 5. Are there any other examples of where two integers in a row don't work? 6. Can this problem be extended into 3d space as well? This, by the way, is how math should be taught. Rather than simply dumping the pythagorean theorem on children (Here kid. Use this.) it would be far better to give them this type of problem to investigate which in turn leads them to the pythagorean theorem.
@Fritzafella
4 жыл бұрын
I remember watching a video by 3blue1brown giving a proof for something where youd draw a circle with its center on the origin, and only with an integer radius and refering to where these circles intercect at an (integer,integer) pair. This is essentially the same question just with a slightly differnet spin but it was different in the way that the proof did not use pythagoran or 4k-1)^(2n-1) where k and n are integers, but used the fact that certain prime numbers can be writen in a+bi form that makes them not prime , for example, 13. 13x1 is the only real number pair but in imaginary there are four related ones. And they are (3+2i)x(3-2i) you get 9+6i-6i-4i^2 or 13 (also cuz 3^2 + 2^2) but 3,7, 11 etc cannot be writen as regular primes or as complex primes either. Ill see if i can find the video cuz if you like numberphile ull like 3blue1brown.
@grandexandi
4 жыл бұрын
I asked this question as a comment on a Numberphile video years ago. I'm going to go ahead and presume this video was made in response to that one comment of mine, of course. In which case, thank you! I love it!
@macronencer
4 жыл бұрын
I was asked to prove Pythagoras' Theorem during a university entrance interview for Cambridge in 1982, and this is the way I did it! I think the interviewer liked my approach, because he said he enjoyed geometric proofs. :) I didn't pass the entrance exam so I ended up at Southampton... but I often remember this. Another interview I sat was for Nottingham, oddly enough, where many of Brady's videos are made. In THAT interview, I was asked to integrate e^x.sin(x).cos(x) while they watched, which was WAY harder and made me sweat a bit!
@Satokaさとか
4 ай бұрын
Looking at the animation at 11:54 I think the answer to Brady's question would be more sparse, because as you go on the angles available becomes more and therefore the numbers.
@dewaard3301
3 жыл бұрын
I hope Brilliant will still be around when my boys grow up.
@KuroroSama42
7 ай бұрын
When he asked to find the pattern to the slanty squares, my first thought was "well you Pythagoras to find the hypothenuse, then square that..." Turns out I was spot on but going from his conclusion instead lol
@kemikao
4 жыл бұрын
Nice! Now, what volume cubes can you make in a "dotted" lattice?
@trogdorstrngbd
4 жыл бұрын
My off-the-cuff conclusion is that only the cube numbers are possible. Proof (?): The volume of a cube with all integer-component edge vector is V = (a^2 + b^2 + c^2)^(3/2). Since a^2 + b^2 + c^2 and V are both integers, a^2 + b^2 + c^2 must be a square number and V a cube number (no other kind of integer can be raised to the 1.5th power and get another integer). EDIT: I'm assuming you want V to be an integer. Interestingly, this ambiguity wasn't present in the 2D case since restricting the edge vector components to be integers there automatically forced the area to be an integer as well.
@timh.6872
4 жыл бұрын
I think it's possible, and in fact, a volume of 3 is peanuts (a = b = c = 1). The trick is figuring out whether there's any discernable pattern. I suspect 4 dimensions and quaternions would be easier to work with.
@trogdorstrngbd
4 жыл бұрын
@@timh.6872 Can you elaborate? If all sides have a length of 1, the volume is 1. If one corner is at (0,0,0) and an adjacent one at (1,1,1), the volume is 3^1.5, which isn't an integer.
@jhnoor9705
3 жыл бұрын
For the past two years, I have been studying an area of math known as googology. Googology is basically the study of very large numbers and the notations that is used to express them. When you study googology in depth, you can see that the so-called scientific notation which we usually use to express large numbers is actually incredibly weak in comparison to many of the commonly used notations in googology such as Knuth's up-arrow notation, fast-growing hierarchy, Bird's Array Notation and Bashicu Matrix System, although it may not seems weak at all to an average person. This is mainly because we don't need numbers much larger than those that can be made using exponents in real life. For example, the mass of Sun is approximately 2*10^30 kg and the number of subatomic particles in the universe is approimately 10^80. Below I will show you the formal definition and some examples of expression in Knuth's up-arrow notation: a^^^^...^^^b with n uparrows = a^^^...^a^^^...^^a^^^...^^a... ...a^^^..^^a with (n-1) uparrows between each successive a's Where a^b is the same as a raised to the power tower of b First, we will take a review of addition, multiplication and exponentiation. We had all studied in school that addition is repeated counting, multiplication is repeated addition and exponentiation is repeated multiplication, mathematically: a+b = a+1+1+1...+1 (repeated b times) a*b = a+a+a...+a (repeated b times), and a^b = a*a*a*a...*a (repeated b times) Now, we will start with the double up-arrow operator, which is better known as tetration. a^^b (read this as "a tetrated to b") = a^a^a^a^...^a (b times) (a power tower of a's b terms high) Keep in mind that exponents are always right-associative, so a^b^c^d is the same as a^(b^(c^d)) For example, 2^^3 = 2^2^2 = 2^4 = 16 2^^4 = 2^2^2^2 = 2^2^4 = 2^16 = 65536 2^^5 = 2^2^2^2^2 = 2^2^2^4 = 2^2^16 = 2^65536 = 10^(log10(2)*65536) using rules of logarithm = approx. 2*10^19728 (a number WITH 19729 DIGITS!!!!!!) 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987 (approximately 7.6 trillion for short) 10^^^3 = 10^10^10 = 10^(10 billion) = 1 followed by ten billion zeroes Now do you see how powerful tetration is in comparison to scientific notation? But that's not the end of the story. Just like how tetration is repeated exponentiation, pentation is repeated tetration, which is normally denoted as triple up arrows (^^^) In summary, the n arrow operator is repeated (n-1) arrow operator After that, I suggest you to learn the definition of fast-growing hierarchy, which is basically like this: f_n(a) = (f_(n-1))^a(a), when n is a successor ordinal, or in other words, f_(n-1)(f_(n-1)(...(f_(n-1)(a))...)) (nested n times) When n is a limit ordinal, f_n(a) is defined as f_(n[a])(a), where n[a] is the n-th element of the fundamental sequence of the ordinal n For the definition of successor and limit ordinals, you can search it yourself in Googology Wiki Now, here's an equation for you. Given that 1/f_x(100) is the amount of DNA I inherit from my mom, try to find the ordinal x. The ordinal x here is known as my DNA ordinal Here's the approximate value of x in Bashicu Matrix System: (0,0,0)(1,1,1)(2,2,2)(3,3,3)(3,3,0)(4,4,1)(5,5,2)(6,6,2)(7,7,0)(8,8,1)(9,9,2)(10,9,2)(11,9,0)(12,10,1)(13,11,2)(13,11,2)(13,11,1)(14,12,2)(14,11,1)(15,12,2)(15,11,1)(16,12,0)(17,13,1)(18,14,2)(18,14,2)(18,14,1)(19,15,2)(19,14,1)(20,15,2)(20,14,1)(21,15,0)(22,16,1)(23,17,2)(23,17,2)(23,17,1)(24,18,2)(24,17,1)(25,18,2)(25,17,0)(26,18,1)(27,19,2)(27,19,2)(27,19,1)(28,20,2)(28,19,1)(29,20,2)(29,19,0)(30,20,1)(31,21,2)(31,21,2)(31,21,1)(32,22,2)(32,21,1)(33,22,2)(33,21,0)(34,22,1)(35,23,2)(35,23,2)(35,23,1)(36,24,2)(36,23,1)(37,24,2)(37,23,0)(38,24,1)(39,25,2)(40,25,2)(40,25,1)(41,26,2)(41,22,1)(42,23,2)(42,23,2)(42,23,1)(43,24,2)(43,23,1)(44,24,2)(44,23,0)(45,24,1)(46,25,2)(47,25,2)(47,25,1)(48,26,1)(49,27,0)(50,28,1)(51,29,2)(52,29,2)(52,29,1)(53,30,0)(54,31,1)(55,32,2)(56,32,2)(56,32,0)(57,33,1)(58,34,2)(59,34,2)(59,34,0)(60,35,1)(61,36,2)(62,36,2)(62,36,0)(63,37,1)(64,38,2)(65,38,2)(65,38,0)(66,39,1)(67,40,2)(68,40,2)(68,40,0)(69,41,1)(70,42,2)(71,42,2)(71,42,0)(72,43,1)(73,44,0)(74,45,1)(75,44,0)... ... For the definition of Bashicu Matrix System, you can search it up yourself in google So can you please help me to analyze my DNA ordinal?
@rossholst5315
Жыл бұрын
You can draw a square of 3 if you add in another axis for a z dimension. As the length of the diagonal of the cube is root 3.
@tonypiff
4 жыл бұрын
and now i have to watch every ben sparks video.
@zorm_
Жыл бұрын
There is an easier way to work out the size of the middle square : using the pythagorean theorem, we know that the square of the hypotenuse is equal to a^2+b^2. But the square of the hypotenuse is exactly what we're looking for
@m.rohwer6989
4 жыл бұрын
This channel is for people that think amazon prime is about numbers. I love it
@laurak1545
4 жыл бұрын
These mathematicians are all so nice and funny and entertaining... and most of us would never realise if it wasn't for numberphile :)
@denelson83
4 жыл бұрын
Neil Sloane's A022544. That's also 3Blue1Brown's favourite proof.
@jamespalmer2620
4 жыл бұрын
3 videos in a row with Bath professors! Exciting times.
@RaichuKFM
Жыл бұрын
This has probably been noted already, but you *could* do a square of length three if you had a 3D grid of points spaced one apart. Because then the square (0,0,0), (1,1,1), (2,0,0), (1,-1,-1) has side lengths of sqrt(3). That could also get you six, using a vector like . It's the same basic idea, but now you can do sums of three squares, rather than just sums of two squares. That still can't get every number, though; there's no way to do seven, for instance. But if you had a 4D grid, then you could use the dots to find a square with any positive integer area you wanted.
@Chalseu___
4 жыл бұрын
I wish I learnt Pythagoras that way, a lot more natural and intuitive imo
@SuperYoonHo
2 жыл бұрын
thanks so much
@nice3294
4 жыл бұрын
Does this have something to do with 3blue1brown's video on pi hiding it prime regularities? His video talks about how many grid points show up on circles of square roots of numbers and the same pattern showed up and he goes into some detail about why certain numbers hit a certain amount of grid points.
@MoreParksLessParking
4 жыл бұрын
It should be noted that it is possible to construct a square with an area of 3 using only lines and dots; just not in 2 dimensions. If you move into the 3rd dimension then you can first construct a (1,1) square that has length of root 2, then construct a right angle triangle with one unit of the new dimension that gives a hypotenuse with length of root 3. Make this the length of a square and you now have area of 3 obviously.
@willmunoz1638
3 жыл бұрын
13:02 We spend so much time getting our bodies into shape. Me: *laughs while eating second breakfast.*
@electraelpindrai1964
4 жыл бұрын
yay, classic numberphile
@thesushi1947
4 жыл бұрын
I imagined this like a 3b1b video where he sees if any lattice points land on a circle, except the circle has the radius of the square root of the area and the only way for that circle to land on a lattice point is for the radius to be the square root of a sum of two integers because the points always have integer coordinates
@RhejMacTavish
4 жыл бұрын
Really well put together video; nice one :)
@numberphile
4 жыл бұрын
Thanks
@wattsvanes4613
4 жыл бұрын
Now I’m curious, can u draw a square of area unit 3 on a some sort of curved surface?
@vrj97
4 жыл бұрын
I mean more simply any integer of the form 4n+3 cannot be written as the sum of the two squares because the squares mod 4 are 0, 1, 0, 1 and so the sum of two squares is only ever going to be 0, 1 or 2 mod 4. So the numbers that can't be written as a sum of two squares are at least linearly spaced.
@vrj97
4 жыл бұрын
If you're not familiar with what mod means, note that any integer can be written as 4k, 4k+1, 4k+2 or 4k+3 for some other integer k. If I want to see what remainder is left when I square one of these numbers and divide it by 4, note that for (4k+r)^2 I get 16k^2+8k+r^2 =4*(4k^2+2k) + r^2 and so (4k+r)^2 leaves the same remainder as r^2 when dividing by 4. (More generally, (zk+r)^2 leaves the same remainder as r^2 when dividing by z, for the same reason. And even more generally, (zk+r)^n leaves the same remainder as r^n when dividing by z (consider the binomial theorem, perhaps)). Then realize that 4k+r + 4k' + r' leaves the same remainder as r+r'. This mod notation is just sort of short-hand for these facts. So what this shows is that if I square any two integers, add them and then divide by 4 the remainder is the same as taking the original two numbers and first taking their remainder when divided by 4, squaring just those remainders and adding them and then taking the remainder of that sum. And so then you should be able to convince yourself that no number of the form 4k+3 can be written as the sum of two squares.
@bucketsaremyfriend
4 жыл бұрын
Damn, slawnty squahs are more interesting than I ever imagined...
@alvkarthik2018
4 жыл бұрын
i was thinking about drawing root 3 square and i actually came up with an idea but we have to use a extra tool " compass" . draw a circle of radius 2 units, center at origin. now we can draw a 60 degree line ( numberphile made a video on that. "Euclid's big problem".) originating from center and touching the circle and now from that point drop a line which is perpendicular to the x- axis and that perpendicular line is root 3 units. from trigonometry.
@theoriginalstoney
4 жыл бұрын
Question: I noticed there's a secondary pattern happening which is not quite so obvious. If you take the "most diagonal" offset - which gives the smallest area of the inside square (so for 2x2 square, this is [1,1], for 3x3 square, this is [1,2], for 4x4 = [2,2], 5x5 = [2,3], etc. and then repeat that, you find the area of the square goes up in a very regular pattern: 1, 2 (+1), 5 (+3), 8 (+3), 13 (+5), 18 (+5), 25 (+7), 32 (+7), ... Where does that pattern arise from? I've been sitting here staring at diagonal squares and wondering...
@abramthiessen8749
4 жыл бұрын
All the possible square areas are all just the hypotenuses of Pythagorean triples. However, on a triangular grid, you can make a line of length square-root 3 because the distance is u^2+uv+v^2. Though of course, you need a different measure of area to make a regular shape with 3 units of area.
@Joe-un1tl
3 жыл бұрын
What an incredible way to show the proof of the Pythagorus theorem. I found a very non rigorous proof of it using similar triangles and an inscribed rectangle.
@galgrunfeld9954
4 жыл бұрын
The bigger t he square is, the more points there are in the square that surrounds it (the kind that was used in the video, the "a+b" square), so if you imagine an 2D X-Y axies system, the angle between the triangles and the X axis can become smaller and smaller, because there are more and more points the "tip" of the triangles can "anchor" to. That allows for a selection smaller and smaller distances, and therefore sizes, to be subtracted from the bigger square. As the area of the square approaches infinity, the angle approaches 0, which makes those sizes approach 0 - e.g they're infinitesimal. I haven't worked out the math, so I don't know if it's possible to prove from these arguments that any number-sized square could be created. *Except for a finite list of numbers* Edit: finished watching the video: *nope* xD
@evanfortunato2382
4 жыл бұрын
I'm hype for number theory next semester
@stealthis
4 жыл бұрын
11:54 spirograph. You never know where math will end up.
@thehiddenninja3428
4 жыл бұрын
Before I watched it all, I figured it out. The area of a square is the square of its side-length, obviously. But If the corners are all on points, then the side length, by pythagoras's theorem, is sqrt ( a^2 + b^2), where a and b are the two whole numbers being the vertical and horizontal distance between the two points. So if we're sticking to the grid, a and b can only be whole numbers, so a^2 and b^2 can only be square numbers so the area of the square is sqrt (a^2 + b^2)^2 = a^2 + b^2 is the sum of any 2 square numbers.
@pierreabbat6157
4 жыл бұрын
All numbers that can be done, except 0, when written in binary, end in 01 followed by some number of zeros. The first number of that form that cannot be done is 21 (10101).
@jonathanallan5007
4 жыл бұрын
Removing a trailing zero from a binary number is dividing by two so these numbers are of the form (4k+1)*2^x where k is the number encoded by bits to the left of the 01 and x is the number of trailing zeros. As an example consider 1010001000 this splits to 10100 01 000 which means it is (4*20+1)*2^3 = 81*8 = 648. As such, numbers like this which cannot be done are those for which the 4k+1 part cannot be done. Taking 1010001000 again 81 can be done, so 648 can. So the most interesting ones are those that end 01 (i.e. numbers of the form 4k+1 which are not expressible as the sum of two squares).
@jonathanallan5007
4 жыл бұрын
...and these, in turn, must have an even number of prime factors of the form 4k+3 with odd exponents!
@clockworkkirlia7475
4 жыл бұрын
Fascinating, intuitive... and Pythagoras! Aha!
@jainamshah7996
4 жыл бұрын
Can you please make a video on Pick's Theorm i think that this quetion can also be solved by it. Please make a video on it
@TheOfficialCzex
4 жыл бұрын
I'm liking the Klein bottle on the shelf.
@EternalLoveAnkh
3 жыл бұрын
I'm working on this currently, but expanding it. I have two questions regarding this. The first is: if you wrote all the possible squares in order, is there a formula that would give you a possible square, such as the 4th possible square is 5? RJ
@AB-Prince
2 жыл бұрын
a graph of the numbers that cannot be the area of an integer grid square is linear up to 10,000. which would suggest they're roughly evenly distributed.
@davidwilkie9551
Жыл бұрын
A variation of "Tangency Space" in which condensation not aligned with cofactor 2, ie duality condition, floats in Quantum-field emptiness of Absolute Zero-infinity.., => 2-ness ^2 "squares" inclusion-exclusion prime-cofactor probability density-intensity, numberness-frequency. ("You just look at it", and see self-defining log-antilog, Holographic Conformal Field Number Theory)
@ianflanagan209
4 жыл бұрын
If you have parabolas y=n*x^2 and x=n*y^2 what is the largest square you can make within the bounds the parabolas intersection where n is a whole number?
@RedValkyries
4 жыл бұрын
It's a common theme now that systems are more than just the sum of its parts. I wonder if it may be an unexplored realm of mathematics how this could be formally represented. This video inspires the question of how can you represent numbers. I mean, we now 3 cannot be a square, but we could represent it as a rectangle !_!!_!!_! or as an L - SO what does the different ways in which a number can be represented show about them. And how the ways these shapes interact with one another say about parts interacting within systems
@idkthatxool749
Ай бұрын
The graph is beautiful
@doubledarefan
4 жыл бұрын
Overlap the triangles for more square possibilities. E.g. 3x8 triangles in an 8x8 square. How big would the square-in-the-square be?
@waygonedon
4 жыл бұрын
Do constructible points come into play here? Is there a geometric parallel?
@everybird3530
4 жыл бұрын
i think there is an easy way for checking if a square can be made , you can just put one end of a line segment the size of sqrt(area of square) and you rotate the other end and if it touches a point on grid it can be made into a square
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