Ryz looool

I just imagine him saying that, the way he did, but with another f-word...

Imagine the scenes 😂😂😂

Top tier comment

Yeah, I thought that was a very confusing "explanation".

Imaginary number solutions

All it is is that each solution that is the 'same' (flipped) can be graphed. The two solutions are just the roots of a quadratic equation- graphed as a porabola. The 'wormhole' is a bit of a misnomer. Basically the two graphs drawn on the paper have parabolas connecting the two solutions. So you have a 3D graph. It's like a tunnel- there are two sets of solutions on either side of the road, and the tunnel is the parabola that joins the two sets of solutions.

Me too,pal

@Daniel Wilkes thank you daniel, you have a way better explanation

This thing escalated real quick.

I *think* he's just referring to how they're connected but only in the z axis otherwise they *look* like unconnected points if you're only charting in x and y. sort of how if you looked at a cross section of an intestine with an MRI or whatever while they digested food, the food would appear to teleport from point to point without traveling through there space in between but it's actually going up and down instead of forwards and backwards or side to side

they didn’t really explain why we had to connect the separate solutions and why (a, a^3) wasn’t enough as a solution.

@@X2Brute ohh

not a wormhole, per se. The problem that the parabola solves is the fact that the function when graphed only in terms of a and b is no longer a function, but we know that the function as a whole should be a function. In two dimensions, it does not pass the vertical line test. That’s because it’s actually a three-dimensional function. The parabola is meant to show where the roots of the function are. What the parabolic nature of the true graph shows is that those two points should have similar properties. Both points intersect their parent graph and are perpendicular to a solution of another graph. That means you can generalize the solutions to draw that line to (8, 30) as well. I think. I have no clue honestly, this is just me speculating.

Sm64 lol

PUs for short

Neumann Lines?

>tfw vieta jumping for 12 hours to build up speed but you realize the distance Mario moves isn't necessarily equal to his speed.

@@Antonio-Russell Or paper mario

Second that

zvezda never came

yeah, i dont like his explanation at all

i agree with this man over here, although i wouldn't pay for it.

@@jaymata1218 Because it is not explanation. What they showed does not correspond to the original question. He even use 0 despite the question clearly said "positive integers".

LOL

Ali what do you mean, it said grunt work, just plug in 0s and 1s in a matrix and stuff and win a medal. /s

karremania

Exactly. Watched a lot of video to be told yeah there are these other solutions you can find too! Cool so did he just draw every single solution? Probably not.

Yeah, the first thing I said when I saw his "explanation" was "How in god's name does this constitute a proof?" Don't get me wrong, I'm sure the people who solved it at the olympiad proved it, but this video doesn't.

probably because they spent all their time on question 6

Time bro

not necessarily failed, just did less well. it wasn't pass/fail.

A lot? You mean 3?

@@Lucas-DX lol u won’t get 1

Not for nothing is the answer to everything.

Yes well think of it like this, The Universe Question answer is a statement of fact, FOR TWO. Duality, Balance, Polarity, Frequency, Spectrum Cell division, galactic wave resonance patterns, photon patterned behavior observed unobserved. So the Forty Two is just a typo or is it?

it feels intentional

You have 84 likes 42*2

Which gets a gold medal and 1st place as in fact does 32 points and 16th place, so have we witnessed a debased metal process that indicates to alchemy being a commutative process? Id rather be happy than right anyway.

I think he wanted to.

I thought it was only me lol

But that would have been NAUGHTY!

Aussi bro he, had to.

HSHAHAHAHAHA YES OMG

...keep watching.

+hama prgasc Im pretty sure he was talking about zvedelina stankova.

I smiled so hard it hurt

Agree, that part was gold!

I almost fell out of my chair at that part!

@@krisx2183 pretty sure any decently educated person has a t least *heard* of Bulgaria

@@omikronweapon yep, especially if you are into mathematics and informatics. This is an Olympics question and it makes it actually normal to have a Bulgarian solving it. It's also normal to have a bunch of Bulgarians commenting under a mathematics video 😁😉

I’m going back to sleep.

I fell asleep in maths class and woke up in the nth dimension

Underrated:

This made me laugh so hard listening to it back ty

😂 everyone who liked has previously experienced this

its just a 2D projection that is part of a 3D solution, calling it a wormhole is extreme

@@srikariyer8544 It’s both extreme and offensive, probably, to worms.

@@chriswebster24 true

i pissed myself laughting

​@@filmamundo9194W T F ☠️

just laughed for 2 minutes and 20 seconds straight

10 seconds later he "works it out" anyway :D

"Alright!"

Qqqqqq

Aaaa

Assuming you understand why every pair has a "connected" pair (explained in the video), you can start from an arbitrary "a" and "b", and vieta jump back to the x or y axis. The solution to the new pair will be a square number (just set "a" or "b" to 0 to see why). Because the new pair has the same solution as the original pair, the solution to the original pair must have also been a square Edit: as IrrelevantNoob pointed out, I didn't show why the vieta jumping has to end at an axis. I added a proof for that 4 comments below this one

maxithewoowoo that's incomplete though... what if there is a counter-solution for which the fraction simplifies to a whole number that _IS NOT_ a square, vieta jumping back towards 0 will not lead you to either the Ox or the Oy axis... :-B

@@irrelevant_noob how would it not reach the x axis or y axis? You just keep jumping back until it does. The funnel shape of the graph ensures that you always end up back at the axes, no matter where you start

maxithewoowoo remember that you're only jumping to points on the ZxZ grid... The two hyperbolas for non-square results will intersect the axes on points that are not on the ZxZ grid. :-B

@@irrelevant_noob actually I see what you mean now. What if the vieta jumping doesnt hit the exact point that the hyperbolas intersect the axes? There is actually a simple solution to this too. Remember that the hyperbolas actually extend through the axes to the negative sides (other quadrants). If vieta jumping doesn't hit a point on the axes, then it will go _through_ one of the axes. That will result with one variable being a negative int and one being positive int. However you can see that this would result in a fraction that is negative, contradicting our initial assumption that the result of the fraction is a positive integer. QED (Also note that we can prove the pairs are always integers, aka on the ZxZ grid, because if one of the vars somehow became a fraction, it would result in a fractional solution)

Same for Bao Chau Ngo who is now a full professor at UChicago. He won two gold medals (one of which he achieved a perfect score like in the video). He later won a Fields Medal! It's crazy how smart these people are.

@@catluong9660 Yup. Fields medal for proving the fundamental lemma for Lie Algebras. He's a great man.

4 times winner of Putnam and perfect scoring in IMO especially in this one is wild ngl 💀

Yeah but it says in the question that you can include zero

@@VSHEGDE1947 No, it doesn't. The question says, "Let a and b be positive integers," and does not mention zero. Zero is neither positive nor negative. Now, you might want to look at how it works with zero to, as the video had said, get a look at how the machinery works before you take it apart. But, that isn't going to be part of the solution since the problem specifically requires a and b to be positive.

Yeah, so just make them both zero. Problem solved if zero is allowed. Which it's not.

@@sprocket454 wot? 0/1 is not a square number...

@@trequor 0 is the square of 0 though

this did not solve the problem though... it was not at all a proof

tfw you realise some people thought this is (not) a solution.

***** well, where's the proof? Only finding solutions doesn't prove the non-existence of a counterexample

I am with you. I want to see more about how mathematicians think, how they examine problems, explore math, solve things. Numberphile has done this before, but never as explicitly as here. More like these!!!

"well, where's the proof? Only finding solutions doesn't prove the non-existence of a counterexample" That was my first thought as well. But after thinking for a bit this solution bumps into any possible integer solution that there is available. Because (0,x) and (x,0) are solutions for any integer x you will get a staircase function starting from any new possible integer and therefore covering the entire 2D space. The video does not explicitly show that but it does by defenition find any possible solution.

I like the way you think. xDDD It's Hilarious

For me, it's a stadium of Problems the size of the Universe :'(

Haha I thought he was going to say it

kkkkkkk

Albert Einstein couldn't count his change....

@@jacobparra6878 haha

@@jacobparra6878 That's really how it is for geniuses, if they never lock on to a concept they never will. They are at the full mercy of their brain.

@Jeffery Lin LOL I haven't noticed that before!

@@jacobparra6878 that made me laugh

Yes and no. It can definitely be helpful to be more well-versed in different types of math, different branches, systems, and theories, I mean the more knowledge you have to bring to a problem the better, in general, for all problems really, not just mathematical problems, but I think truly gifted mathematicians can see/discern a pattern and make the link "discovering" the solution for themselves, even if they weren't aware of other solutions, methods and techniques previously employed to try to solve the problem. This is to say that the greatest mathematicians are "clever" in that they can find/create the solution for a problem without knowing what other work has been done on the problem before they attempted to solve it, i.e. great mathematicians will somehow find the solution for themselves based on the knowledge and intelligence they possess. I mean, isn't that what makes them great, that they can solve problems that others haven't been able to, or be able to find a new, possibly more economical and elegant solution to a problem others have solved, but in a more difficult way?

@@mydogskips2 The person you're replying to already has a strong counterexample of a genius mathematician (Terry Tao) being unable to solve the question.

This video shows the entire set of infinite sets of solutions. The actual proof is one by contradiction (wikipedia has it under "Vieta jumping"), but this is meant to show why the solutions are what they are.

flashtirade It shows how I can generate infinite number of solutions that match the statement, yes. It does not prove in any way that there are no other solutions, not just ones that don't match the statement but even those that do.

While the "wormhole" idea is really neat, it is hard to understand. There is a less beautiful, but easier-to-understand approach. Let's assume there was a solution that wasn't a square. Then you could Vieta jump backwards to find a smaller solution. Then you could jump backwards again to find a solution smaller still, and so on until you got to zero (which, because the solutions are in the integers, has to happen eventually). But you already know the solutions for zero all give you squares. Proof by contradiction.

Alex Potts That's neat direction, it's just not obvious to me that you indeed can jump backwards from each such solution.

That's assuming that vieta jumping is valid for all other inputs, though.

BAHOQISLSOWJW IM WHEEZING WJDJSIWJT THIS COMMENT MADE MY DAY IWHZBSKW

I used to use them in high school but now I hate them because they look like multiplication.

Karl Kastor I know right, I hate how they taught us all these different conventions. Like, for example) they said that cis(z)=cos(z)+i*sin(z) because they thought we were too young to learn about euler's formula.

+SafetySkull >cis(z) Did you just assume my function?

***** I don't think that the cis function has a real-valued value to compare for the sake of a less-than/greater-than result. but i don't think that's what xe had in mind ;P

first i've heard of the cis function. interesting. when were you taught it? We went into eulers formula not long after we were introduced to complex numbers.

So a and b are the variables that the equation uses and that is what he is graphing. He does the graph with the answer equaling 4 but there are many possible solutions. Basically, a and b make the X,Y axis and the answer to the equation makes the Z axis. That's how it is three dimensions.

Chris Doe yeah, but then using X=2 in f(X,y) = (x^2 + y^2)/(xy +1) gives f(y) = (y^2 + 4)/(2y+1). I fail to see how is that a parabola.

The parabola is the connection between the two solutions of the flipped values of a and b

complex numbers I think

First, everyone talking about complex numbers needs to STFU, they are completely irrelevant. Second, Vieta jumping actually does prove the problem, but they failed to explain what Vieta jumping actually is. If you google Vieta jumping it will give you the actual proof. What it basically boils down to is that you assume a minimal solution exists, and Vieta jumping gives another solution that is even smaller. If the number isn't square, than the alternative solution leads to a contradiction.

Engineer >>>> Mathematician

​@@finmat95Mathematicians dig to understand the intrinsic properties to tell the programmers how to program METLAB and instruct engineers and physicists how to use the math they've discovered. What you said was that Engineers were better than Mathematicians for being too lazy to understand fundamental concepts and relying on "plug and chug" methods while having no intrinsic understanding

@@zaclaplant3001 Metlab? what did you do with that program? business with Gustavo?

@@finmat95Hahaha

Why don't you look it out for yourself then?

Why don't you just look up the rigorous solution dude?

If you're a professional mathematician (which you are not) then you should be able to figure it out for yourself.

But that IS the solution!

All you people seem to be missing the point. The video said it would give the solution. It does not give the solution, only a very bland description of (having done some research) Vieta jumping. My ability does not come into this, but I do not claim to be able to solve it. I DID look it up afterwards (and I'm still a bit confused about the idea of Vieta jumping and its link to a proof by descent, but I'll keep trying to learn it). No, I am not a professional mathematician, but I'm certainly well beyond the level the video is at. None of this changes the fact that the vdeo falsely advertises a solution.

the proof comes from that quadratic equation into that "extra dimension". by it's existence, you can work out that all whole numbers generated by the equation are squared. there is a proof somewhere here in the comments section if you're interested.

it's obviously a sketch of a proof. the real proof is to assume that you have a minimal solution that is NOT a square, then vietta jump backwards to find a smaller solution, thus a contradiction. it's on wikipedia

I think I understand what he means by a quadratic equation into a 3rd dimension. It really means you can use the solutions on the a by b graph as the image set of another function which has a domain composed of solutions to a quadratic equation. That latter quadratic equation seems to be the holy grail of this problem. This came to me as I actually found a quadratic equation (2n²+2n+1/n²+n+1) which will produce fraction results of the initial a²+b²/ab+1 equation (for a,b pairs = 1,2 ; 2,3 ; 3,4 ; 4,5 ; etc...). But I may not imply from this that 2n²+2n+1/n²+n+1 = a²+b²/ab+1 because they are separated by another function (or a dimension). I think linking these two expressions together as h(g(f(n))) would have helped me make progress.

No, the jumping between lines is not what was developed in reverberation of this test. That's just Vieta jumping, a technique that was known for who knows how long. It's based on Vieta's formula (François Viète lived in the 16. century) and you can always do that whenever an equation is symmetric in two variables. What he meant in the video is hard to tell from the collage shown in the video, but it was definitely something (completely) different. I suspect it's more about the "wormhole curves" and less about the jumping.

The proof is that the two lines trace out where parabolic equations in the z-axis intersect the x-y plane, and where those intersections pair up, they will always fit into the flipped pair of equations.

@@mycroft16 Thank you, mycroft16.

Thanks, this video didn't even try to give a solution.

No this has nothing to do with physics, this is mathematics

@@Ken-M thanks. All my 3 questions are still unanswered though.

@@swapnilshrivastava116 I also don't understand it completely yet, but the argument must be with the Vieta jumping one does. First of all, it's fair he only included the solutions that are integers because that's part of the assumption. And now the argument seems to be that every time you jump from one line to the other (hence walking either rightwards or upwards on the grid) gives you a new solution, which then turns out to be a square (for some reason I don't understand yet). I think I understand why we find all solutions that way, because the squares are countable and there's a total order of them, right. They go 1², 2², 3², and so on. And now for every square you get those two rays where you can jump from one to the other Vieta style and always get the same square. And these are all the solutions, because they are the only points where the lines intersect the grid. But somehow the second part hardly matters, because you're not really interested in how many solutions there are for a given square, right? So I assume the second part somehow fuels into the first (about why all integer solutions are necessary square), but I really don't see it yet.

1. He's actually wrong, they're connected by hyperbolas, not parabolas. As for the reason, these points are part of the surface defined by the equation z = (x^2 + y^2)/(xy + 1), in this case the x-y graph he plotted is the plane at z = 4. This surface is a paraboloid, and the function's orthogonal planes at fixed x form hyperbolas in the y-z axis. 2. He showed that every time one of the numbers in the pair is 0, the result will always be the square of the other number. In other words, the base case is, for b = 0, the expression result will be a^2. What he needed was a way to show that every other pair that gives the desired outcome, it can be reduced to its base case. Sadly he only proved that it works for n = 2^2 = 4, but it can be generalized. The recent Numberphile video "The Notorious Question Six (cracked by Induction)" goes into detail and explains it really well. 3. It was just an analogy, because it connects 2 apparently unrelated points in the graph like a theoretical wormhole would connect 2 different places in our universe.

its a 2d function that can be graphed on a normal graphing calculator

they exist bc math magic stuff that would take longer than a short video to explain, but he mentions their parabolic shape which makes it that it can only hit the plane at square numbers

Stick: z = (x^2 + y^2)/(x*y +1) Into google. It's actually a 3D function if the solution (4) is set as a variable. This is because they wanted to find other possible solutions which follow the same rule of being a square of an integer.

+Untitled oooohhhhhh, thank you :)

yes

[Later edit: please note that the attempted proof mentioned here is wrong, since the ordering of Q is not monotonic. See further replies below for a better approach.] You're right, this is not sufficiently explained... The rigurous proof is by contradiction, basically like so: if there are some such solutions, you could get smaller and smaller (closer to zero)... But that's not really possible, since Q is countable, so if there were *_any_* then there'd have to be a _first_ , but out of that supposed first you'd be able to get to an even smaller one.

Basically, we vieta jump from any pair (a,b) all the way back to one of the axes, giving a new pair (0, k). Because the new pair has the solution k^2, we know that the original pair also had the solution k^2. But how do we know that the vieta jumping always hits one of the axes? The video shows that vieta jumping always results in another pair of integers with the same solution. If a vieta jump somehow didn't end exactly at an axis, it would have to go through the axis, making one of the variables negative. This would result in a negative solution, but we assumed our original pair had a positive solution! So this is not possible, proving that vieta jumping must always reach one of the axes exactly. But you are right, this was not sufficiently explained in the video

@@maxithewoowoo Yes, to prove the theorem you need to perform the "vieta jumping" downwards. Unfortunately the video shows an *increasing* sequence, which is misleading. Also, you need to show that the sequence generated by vieta jumping: 1. always produces solutions with the same k, 2. is strictly decreasing, 3. is bounded by 0 from the bottom, and no, it is *not* obvious, because we have the a,b > 0 assumption, but only for the initial values. It does not guarantee that the elements of your vieta sequence will be positive, this has to be proven. Because the sequence is bounded by 0 and strictly decreasing, in a finite number of steps it has to reach 0. Then, because of point 1. above, it is easy to show that the original number is a square (of the last non-zero element in the vieta sequence).

@@krzysztofmichalak642 for your point 1., vieta jumping by definition will result in the same k. Because you are jumping to another point on the curve, and that curve represents all solutions for a given k. You're right that a rigorous proof would require you to prove that vieta jumping would be strictly decreasing. From the graph it seems obvious, but to rigorously prove it you can simply show that the graph is a hyperbole rotated 45 degrees clockwise (you can do this by rotating it 45 degrees counterclockwise using a rotation transform). And as for showing that it is bounded at 0, you can simply show that via the proof by contradiction I explained in my earlier comment in this thread

You can prove by remainder theorem .you can see a little bit above where I have given answer.

0 is an integer, end of story.

@@RN-uo2vo true, but it says 'positive integers' in the text

I agree, 0 is not positive it is neutral

That threw me off too. 0 was definitely not considered a positive integer when I did algebra.

0 with any other integer will always satisfy this. That is why to force that realization the question will doesn't say whole numbers.

When?

Some of the greatest mathematical minds are/were Bulgarians. They are especially great at euclidian geometry.

this is the legacy of great Soviet influence! BRING BACK THE EASTERN BLOC!

your point of view is fair enough, but ask yourself this: Would you rather live in a less restricted world, or in a more scientifically-oriented one?

so we're not a banana country after all ...

By the way we didn't use graphical solutions, the way we did it was to brake it down to quadratic formulas and gain a multitude of answers which can be plotted to form the graphical answer. #Ihatemath

моля ти се не лъжи

Well, even though engineering and math are related, they aren't the same thing. Engineering might apply math, but I don't believe it doesn't have any application of this particular problem yet. However, math is only pure math without application. Eviltech's point is valid, I would say.

You are right. He hasn't solved the problem in the video. Perhaps that was intentional, and he's hoping you'll have a go now. When I tackled the problem I didn't use this amazing graphical insight. For me it was more natural to show that for a value of k (I also started with k=4) I could generate an infinite sequence of solutions, and I could generate those solutions using a simple recurrence relation. Then I wanted to know, why that recurrence relation. That was when I realised that for a given value of, say, b, there were two possible values for a: which led me to a quadratic equation. That equation allowed me to explain the recurrence relation. Then I asked, why doesn't it work if k is not a perfect square? It turns out there's a really neat simple answer, which I haven't seen in any of the articles on this which I subsequently googled. But I can't believe that no-one else has found it, given how simple and elegant it is!

@unknowning unknown he found a bunch of solutions, and showed that they were linked by a neat graph, but we don't know (from this) if that's even all the solutions. In reality there are more rigorous proofs that show that those are all the solutions, but that proof is not here. He would have to show that there exist no integers a, b such that the result of the fraction is an integer which is not a perfect square.

for every pair of a and b it equals some number z. or in order words a 3d function. like f (x,y)

no, you're right. They made a mistake.

And how many great research mathematicians have you produced?

depends how you define zero O_o

Adam Dunkley It's like saying 1=2 is about how you define 1 and 2.

Welll it is an non-negative iterger...

Andrew Wolf That doesn't make it positive.

That specific competition may have defined zero as positive. It was the 80s after all, so it wouldn't have been the weirdest thing to happen at the time.

TLDR: to satisfy divisible constrain, we need integer a b satisfy a=b^3 and the equation yields b^2, thus 'is the square of an integer'

Time travel 😂

(a,b) first super pair (2,8) ----> (8,30) ---->(30,112)---->(112,418) ...(b,b^2-k/a), k=4 is there any other k?!

Jane Za all squares work in this way

True... Remarkably enough I could follow the train of thought of this solution asif it were my own, right up to the point where he started graphing the solutions and describing the generic formula, that was quite surprising and unexpected indeed... nice...

The solution to the equation is being treated as a variable (Let's just say X) which is a square number (Which is stated in the original question), so since you have two lines which both give one particular solution of X (4, or 2 squared), then there must be other lines for other potential solutions of X. These other lines need to be shown in the third dimension (which is the X dimension, which is for other values of X), so you end up with 3 dimensions (a, b, and X). The quadratic equations are related to the fact that, in the third dimension, a parabola is formed of all the values of X (Since X needs to be a square integer). Not sure if I explained this clearly enough, but hopefully it helps!

Untitled Thanks for the explanation. But where is the third point to define the quadratic function? I'm missing that information. You only got two points.

TheTruthSentMe You have a and b which both make the two dimensional lines for the solution of X = 4, and then you have X, which makes the actual parabola. If this was a 2D quadratic (Just Y = X squared), then you'd have X as your Y (The height) and the a,b function as your X squared. Since this is in 3 dimensions, your a's and b's occupy a plane (The sheet of paper) instead of single points. I recommend pasting this into google: z=(x^2 + y^2)/(x*y + 1) This should give you the 3D parabola, where the z is your X, and x and y are a and b. It should look more familiar from the top (yellow) down if you ignore 1 half, since it stretches into the negative.

Untitled Thank you again for your effort. I think I understand a little better now.

I posted this above but in case you didn't see it: Look at this: (a^2 + b^2)/(1 + ab) = 4 implies a^2 + b^2 = 4 + 4ab implies a^2 + b^2 - 4 - 4ab = 0 implies b^2 - 4ab + (a^2 - 4) = 0. Let a = 2 and we get b^2 - 8b = 0 Hence (2,0) and (2,8) are connected by a parabola in the sense that their b coordinates are roots of the parabola x^2 - 8x = 0.

No.

Yes, you should feel ashamed. JK LOL.

just think of it this way. they solved a problem that has no benifit on human society. unless maybe one day it so happens to create a warp drive

And you've posted a comment that has no benefit on human society. Unless one day it so happens to convince the rest of the world that everything must benefit society in some way.

Jackson Percy if Tao barely solved it, you should not feel bad at all :)

I read this comment whilst watching this video at 2 in the morning…

@@NoriMori1992 same

it’s 2.48 in the morning

@@ksjdoiwfehfueu It's literally the exact same time for me wtf

as a great youtube creator once said "see what happened here? I lost my focus" -Dunkey

Yeah, it felt like he completely lost the thread of the topic, and/or forgot who his audience was. I'm kind of pissed off that I still have no idea what the solution is. If it's too advanced to explain then they shouldn't have bothered bringing it up.

​@@Xezlec I do not know if you (or anyone else) will check this comment, but the solution is as follows (in many words rather than a paragraph, to explain why it is like it is): Disclaimer: A and B are interchangeable, so for this explanation if I say "if A is number X, then B is either number Y or number Z" then this is the same as "if B is number X then A is either number Y or number Z". Imagine you have (A^2 + B^2) / (A*B + 1) = N^2 Then for every possible positive integer N (whole number) there are an infinite set of combinations of A and B, but the relationship between A and B is not random. If A is number X, then B is either number Y or number Z. To start of with explaining it, let's assume that you have decided on what A is. In this situation N^2 will always be equal to A^2 for exactly two values of B. This N^2 we will call K, and K=A^2 If we say A is 2 then we have this equation: (2^2 + B^2) / (2*B + 1) = 2^2 --> Let's multiply by (2*B + 1) on both sides. 4 + B^2 = 8*B + 4 --> 4 goes against 4. We subtract by 8*B on both sides. B^2 - 8*B = 0 --> This is the same as: B * (B - 8) = 0 --> For this to be 0, B has to be either 0 or 8. So you have both (2, 0) = 2^2 and (2, 8) = 2^2 that gives the square of the integer 2. However, now we have found 8, and so we can easily calculate the next combination that also gives the square of 2. We say that A is 8 and have this equation: (8^2 + B^2) / (8*B + 1) = 2^2 --> Let's multiply by (8*B + 1) on both sides. 64 + B^2 = 32*B + 4 --> We subtract by (32*B + 4) on both sides. B^2 - 32*B + 60 = 0 --> This is a quadratic equation which is easy for us since we know one part is (B - 2). The other part is therefore (B - 30). if A is 8 and K is 4, then B has to be either 2 or 30. We have found a new solution for 4, namely (8, 30). "Why is K=4?" you might ask. Well, that is because we are still on the line of infinite solutions to find K=4, however A=8 of course has "it's own K" so to speak, where 8 is the base number. You can therefore find two values of B if you say that A=8 and K=A^2=64. Try it yourself, and you will find K=64 for (8, 0) and (8, 512). A quicker way than doing quadratic equations exists. For any given numbers where (A, B) = K the next combination of numbers will be (B, B*K - A). So from (8, 30) = 4 the next combination would be (30, 30*4 - 8) = 4, so (30, 112) = 4

Basically each pair of points is connected by a parabola, so that proves the whole numbers will be squared because y=x^2. It’s just that these parabolas extend onto the z axis. This might be wrong, but this is my best guess.

but that's not all the solutions

Lyan Villacorta that’s why it took him a year lol

It's more to illustrate the idea that the information is there, but it is hidden beneath a layer of obvious stuff. It's not just out in the open and plainly visible. On top would have worked just fine, but he's creating a physical metaphor for the difficulty of observing the solution.

It works as a parabola, if you made a parabola like that, its gradient would have to be negative, which is just making it more complicated

Cause its "hidden" lul

putting the "wire" on the front of the paper would be showing the parabola extending into the -z axis instead of the +z axis

yeah, too bad he used 0 a lot. Or maybe it's a math wizz-kid thing and zero counts because it didn't state strictly positive or something similar.

I don't know a lot about english in math but in french if you don't state "strictly positive" you can use zero as a positive integer because zero is neither positive nor negative

I don't know if this can be considered true in English or not, but historically, I've always seen that if you want to specify a list that contains all positive integers and zero, you'd describe that as "all non-negative integers." How most English speaking mathematicians would interpret the word "positive" as either being strictly positive or ambiguous enough to include zero, I'm not sure. I'm certainly not a mathematician, so I can only go by my experience... this video does contain an English speaking mathematician, however, and he thought using zero is fine, so maybe people with a fuller understanding of math have good reason to interpret it that way. Not sure...

No, Chanasak is right. Even without stating "strictly positive", 0 is not a positive integer, mainly because you can hardly call it positive.

If I'm correct , it should say non-negative integer if you want to include 0 (at least thats how it is taught in Germany and from what I can gather also in the English speaking world) btw: the English speaking world calls non-negative integers "whole numbers"??? (which confuses Germans (and French?) because a "whole number" is a integer to us and you us are talking about natural nubers with 0 ... )