The Legend of Question Six - Numberphile

Have you heard of the legend of question six? It's the question that actually stumped a Fields medalist. This is one of the hardest problems ever! Well, it's something legendary. I myself, when I tried to attempt this problem, it took me a year to solve this problem. And when I finally solved it, I'm not, I'm not ashamed to say, I cried. To put question six into context, alright, I'm talking about an international mathematical Olympiad problem. The international mathematical Olympiads is the premier event for young people who are really good at math, they haven't gone to Uni yet, they're under twenty, and it's for young people to go and compete on the international stage, right? And they're fighting for ultimate mathematical glory. Announcer: "A team consists of six kids from each country. "These 'mathletes' are required to answer six questions worth seven points each." So just like the Olympics, they're held in a different country every year. They don't do it every four years, we need the competition. And we're talking about somewhere between 220 to 260 students come from all different countries. And so when you think about it, this is the best of the best. It's about being able to solve awesomely hard problems. But there's one epic problem. Which means a lot to me, because it's from the maths Olympics in my home country of Australia. So we held the math Olympics in 1988. And at that time, there was one particular problem. It was the last problem of the whole event. Over two days, there's three problems each, and on the second day the last problem was problem six. And this one went down as one of the hardest ever. For a long time it wasn't surpassed. Now, there's actually been harder problems that have come up recently. But for a long time, this one was just amazingly difficult. And not only that, it was like, people who did solve it were, like, revered. So for the 260 odd people that competed at this maths Olympics, only eleven people solved it perfectly. And only one person solved it awesomely. And, I have to say, Terence Tao was competing there, and he only got one out of seven. One mark out of a total of seven for this problem, which means he didn't get it right. And Terence Tao went on to win the Fields Medal in 2006. He's awesome, and he's Australian! Can I just say that please? Brady: "And he's from Adelaide." And he's from Adelaide, where I'm from as well, yeah. Brady: "And me!" And you, Brady. I didn't know whether we could mention that. Let's actually, let's actually cut, cut Terence Tao some slack. The Fields medalist, the guy who's, like, coming close to solving the twin prime conjecture. He's actually pretty awesome, because he was only thirteen years old. Brady: "And that's young, is it?" Well, he holds the record as the youngest person to ever have won a gold medal at an international math Olympics. So, it's pretty good. But he still didn't at the last one right. Announcer: "Terry Tao was a toddler when his parents noticed he had a gift. "At a family gathering, they found their two-year-old giving older children a maths lesson." I mean, these are problems that you don't find in a maths book. There's no, you don't look in the back of the book for the answer for these problems. These are really hard problems. And, and in a way, they're actually designed to kind of throw you off if you know high school maths too well. So say, if you really know how to solve quadratic equations, and you see something that looks like a quadratic equation, it almost like throws you down the wrong path. It's almost like, nuh-uh, you're at the international maths Olympics. Question six was pretty special. It was come up by this crazy hardcore West German guy, at a time when there was a West Germany. And he submitted his question to the mathematicians who are in charge of selection. They're given six hours to solve question six. Brady: "These are like the experts, like test, test driving it. They're test driving." Yeah, yeah, yeah, they're test driving it. They're gonna see whether it's gonna make the cut, you know? Because, like, if it's too easy, turf that, right? Well guess what? They couldn't solve it. After six hours. Question six, after six hours, still stood, yeah? So you'd think, hey, this is too hard. But what do they do? They had the courage to submit it. Brady: "They put it on the test." They put it on the test. And in the original copy, they actually put a double asterisk next to it. Just to make sure everyone knew this was hard. Brady: "So don't waste your time on it." Well, not necessarily don't waste your time, just don't beat yourself up if you can't get it. And so here, I've actually got a copy, July 16, 1988. So it's held over two days. This was the second day, and it was the third question on the second day, so therefore, it's question six. Now if you have a look here, the time to answer all three questions: four and a half hours. So that means, on average, you had ninety minutes to solve this problem. And if you can't work out that it was ninety minutes, I think you're gonna struggle with the whole exam. Given the fact some of the best mathematicians in the world were given six hours, and they couldn't solve it, so they gave students ninety minutes on average. Maybe it could be more. Maybe you can finish these ones really quickly, and then you just sit there sweating. And if you have a look at it, it's the shortest one on the page. Look how tiny it is. It says, let a and b be positive integers such that a times b plus 1 divides a squared plus b squared, show that a squared plus b squared divided by a times b plus 1 is the square of an integer. It's not even simple to explain. I mean they have actually covered up part of what, what you need to understand. Okay, let me just write it out for you. Okay, so. So what this says is that first of all, a and b can be elements, they can be 0 1 2 3, blah-blah-blah, they can be any whole number, including 0. Let's stick them into this equation: a squared plus b squared divided by a times b plus 1. Okay, so let's actually do a little bit of a chart for this. So, possibly, it could equal a fraction. So if it does equal a fraction, the question says, no one cares. Really, question six doesn't care. But, if it just so happens that once you divide a b plus 1 into that, and you get a whole number, it then says, oh, that whole number will also be a square. It means that it will be something to the power two. Brady: "Like 25, or-" Yeah. 25, 36, 49, 121. It'll be some whole number multiplied by itself. So only if it's a whole number, it actually will also be a square number. So I suppose, I suppose one way of saying it is that the only solutions to this given whole numbers, when you stick in whole numbers, are fractions and squares. And the thing is, is like, why squares? Perhaps we should find out. So there's the history of the problem explained. And by now, I hope you've cracked out your pencil and paper to try it. For those wanting the solution a bit sooner, the next video will be posted on our second channel, Numberphile2. All will be revealed there. In the meantime, our thanks to the supporter of this video, the Great Courses Plus. This is a huge online collection of video lectures from world experts on every topic under the sun. You can become a master of everything from photography to maybe the Rubik's Cube. Seriously, a forty-six minute lecture from the excellent Art Benjamin on solving the Rubik's Cube. Plus some mathematics on the side. Most people find it a hopeless task to bring the cube back to its proper order. I might finally get this thing solved. Plans start from 14 dollars 99 a month, but you can sign on for a free one-month trial at the great courses plus dot com slash numberphile. There's the address on the screen, or you can click on a link I've put in the video description. Again our thanks to the Great Courses Plus And if you do browse through their seven thousand odd videos, make sure you check out the Rubik's Cube one. That's the one I'm about to watch now. Amazing secret solution to solving this problem is actually a beautiful piece of observation. It is mwah!

Channel: Numberphile

Views: 2,566,551

Rating: 4.7750368 out of 5

Keywords: numberphile, math olympics, International Mathematical Olympiad, terry tao, terence tao

Id: Y30VF3cSIYQ

Length: 8min 45sec (525 seconds)

Published: Tue Aug 16 2016

Reddit Comments

[u/cactus]

The fact that people can solve this in such a limited time, and at such a young age, triggers all my math/intellectual insecurities. Quite humbling, and impressive.

[u/patrick38894]

so frustrating waiting for him to get to the actual question...

[u/[deleted]]

The proof of this theorem (which, as a couple users have already mentioned now, can be found on the wiki page on Vieta Jumping) is fairly elementary and self-contained, although there are quite a few steps. I'm pretty familiar with Infinite Descent but haven't actually seen this technique before so thanks Numberphile.

[u/Bromskloss]

The problem in question:

Let a and b be positive integers such that ab + 1 divides a² + b^2. Show that (a² + b²)/(ab + 1) is the square of an integer.

[u/jamesbullshit]

Im not sure about the geometric picture of this. In the second video, he only showed (without proving) that one can obtain infinitely many solutions by extending lines. (Also he didn't show that this includes all solutions?). But in the proof on wiki, it is done by assuming the quotient is not a square, then derive a contradiction. How do these ideas link up?

[u/Jyben]

The solution

[u/failedgamor]

Can someone explain the whole parabola thing he did?

[u/[deleted]]

Ok, I already posted this as a comment on youtube, but I wonder if anyone on reddit could help me prove a small discovery I made.

Ok, I did a bit of fiddling around with this question after seeing the video, started off in the wrong direction, and found something intresting about this problem which doesn't answer the question. I started off by making a google spreadsheet/times table of the equation and found that for every integer n, there is another integer, n_2 (with n_2 = n^3), which makes it so when a = n_1, and b = n_2, then the quotient of the equation is n_1 squared. However, I also found out that for every n_2 (where n_1 > 1), there is another larger integer n_3, which when paired as a = n_2, b=n_3, also results in the original n_1^2, and found that every n_3 i found can be paired with an n_4 which also results in the original equation (lets call it f(a,b)) also equaling n_1^2. After doing some more tallying in Spreadsheet, and fiddling with wolfram alpha, I discovered the equation for finding the value of n_x for any starting integer n_1 > 1.

I would like to say that even though I don't have proof, this equation matches the pattern of every number I checked, so take this with a grain of salt.

I will have to do a bit of explaining here first, as well as make apologies for it not being clear, i am not used to writing equations with summation as a part of them

1) Let x = the place of the desired co-ordinate in the infinite series C={n1, n_2, n_3, ..., n_x-1,n_x, n_x+1...} (c stands for co-ordinate)(also, the series for n=1 is C={1,0,1,0,1,0...})), where f( n_x, n(x+1) ) = n_1^2.

2) Let n_1 = the first integer in series C.

3) let the function Py(z) be the z'th y-topic, or y-simplex number. (With the added case P0(x) = 1 as long as x>0) Description of what polytopic numbers are: https://en.wikipedia.org/wiki/Figurate_number#Triangular_numbers

4) Let |x| represent the largest whole number less than or equal to a positive real number x

5) Sum f(y), y = j to z is how I write the summation equation, based on the wolfram alpha entry format.

The equation is

nx = (Sum( ( (-1)^y) * P_y(x - 2y) * (n1)^{2(x - 2y} - 1) ), y = 0 to |_x/2| )

Domain = {n_1 > 0, x > 0, | n_1 and x are elements of the set of all integers} Range = {n_x >= 0 | n_x is an element in the set of all integers}

I am pretty certain I typed that in correctly, but I may have made a mistake. If any of you have a better way of testing this than just plugging individual values into Wolfram Alpha and seeing if the values match up, could you try checking it? It would be very appreciated

[u/math_throwaway101010]

C student here. I'd like some help understanding the solution:

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Let a and b be positive integers such that ab + 1 divides a^2 + b^2. Show that (a^2 + b^2) / (ab+1) is the square of an integer.

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Choose integers a, b, k such that a^2 + b^2 = k(ab+1)

So far so good. Does choose integers mean "find three concrete integers"? Or "these numbers follow this rule", like defining a set {(a,b,k) : aEZ and bEZ and kEZ and a>=0 and b>=0 and k>=0 and a^2 + b^2 = k(ab+1)}? I'm assuming the latter.

Now, for fixed k, out of all pairs (a,b) choose the one with the lowest value of min(a,b).

What exactly is min here? I don't think it's the programmer's min.

Label b' = min(a,b), a' = max(a,b).

Again, I have no idea what max and min refer to.

Thus, a'^2 - kb'a' + b'^2 - k = 0 is a quadratic in a'.

I understand that a^2 - kba + b^2 - k = 0 is the same as a^2 + b^2 = k(ab+1), just transformed. I can't jump from that transformation to seeing that the equation holds with the prime'd variables.

Should there be another root, c', the root would satisfy: b'c' <= a'c' = b'^2-k < b'^2 therefore c' < b'

What does 'another root' mean? Another number that could replace a' or b' in a'^2 - kb'a' + b'^2 - k and still have the equation equal 0?

b'c' <= a'c'

If min and max are defined so that b' <= a' then this is true. I don't understand how the rest of that step works.

Thus, c' isn't a positive integer (if it were, it would contradict the minimality condition). But c'=kb'-a',

wait what

so c' is an integer; hence, c' <= 0. In addition, (a'+1)(c'+1)=a'c'+a'+c'+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1 >= 1 so that c'>-1. We conclude that c'=0

Ok, so it's shown that it's an integer, not positive, greater than -1. Has to be 0.

so that b'^2=k.

a^2 + b^2 = k(ab+1)

a^2 + b^2 = b'^2(ab+1)

(a^2 + b^2) / (ab+1) = b'^2

And there's the proof.