The Legend of Question Six - Numberphile 

It took him 1 year to solve the problem. It took him another year to get to the point of telling us what the problem was

Do you know The Legend of Question Six Me, an intellectual: Why was number six scared of seven?

I came up with a solution in 4minutes....it was wrong.

"And if you can't figure out that's ninety minutes, you're gonna struggle with the whole exam" You didn't have to attack me like that

"Hey, let's give a bunch of teenagers one of the hardest problems ever conceived in mathematics."

I am glad I was at least able to calculate how 4.5 hours for three questions mean 90 minutes for each question on average

"have you heard of the Legend of question six?" no, but I've heard of the Emu war 1932

Why doesn’t anybody give credit to the people who designed these questions. They must be even MORE genius.

Well, it didn't take me a year but it did take me about 10 hours... it is difficult but if you really think about it, all you have to do is press Right button, Right button, B, Right Trigger, Right, Left, Right, Left, Right, Left into your xbox controller and you can beat gta fairly easily. Your welcome.

"Have you heard of the legend of question six?" No but I have heard of thelegend27.

I just solved it, but there wasn't enough room here to type it in so I haven't.

Ye. Terrence may have had 1 out of 7.. But we have to revere the maker of the question, because he made the question AND found the answer for it.

Why isn't the person who designed this problem revered?

Australian maths be like "Oi to the power of mate², carry the roo = shrimps on the barbie."

I think one of the biggest reasons I love this channel is that it's not really a maths channel--it's more of a place that tells stories through difficult questions, and often shows you different ways of thinking about these problems. The stories these professors tell are always super enchanting

The amazing thing is that A handful of participants were able to do it correctly in stipulated time

"Number 6 will shock you!"

4:56 I felt so dumb until I realised he said 90 minutes, not 19 xD

*Takes Simon a year to solve Numberphile: I hope you cracked out your pencils Me: Nah, I'm good

This took me 8 hours. 7 hours and 55 minutes of thinking, and 5 minutes of smashing my computer.

Did you ever hear the tragedy of Question Six the Impossible? It's not a story Terry Tao would tell you

But why did it take 5 minutes to see the question?

I love how this all builds up for 5.5 min, just to get more and more admiration and respect for the problem. Very much enjoyed that!

"[...] one of the hardest problems... EVAH!"

"Three questions per day. It was the third day, and so the third question was question six" Now that is some complicated maths. Let me give my brain some time to process that, I'll get to the rest of the video in a moment.

I was there in 1988 and got one point like Terence Tao! This video inspired me to try this again and after a week of solving I am pretty sure I got a proof...

4 miniutes in "GET ON WITH IT" stop pandering! 5:28

every time this video pops up as recommended I think that the thumbnail is a picture of me

Honestly, delivering a problem as a story like this works amazingly. As always, you deliver a prime product :)

There seems to be a worrying amount of people who don't understand the question and think that supplying one example where (a^2+b^2)/(ab+1) is a square solves the question. The question asks you to show that in *all* cases where the fractions turns out to be an integer, that integer is square. All cases. *Not* one case. All cases. And for people spazzing out about 0 being included in the video, the statement to be proven holds for a=0 or b=0 as well.

"this is one of the hardest problems [pause] *_EVAH_*"

At 0:59, "6 questions worth 7 points each". So the maximum total score is 6 * 7 = 42. I see what they did there.

I really like these kind of videos.

I remember Terrance Tao at Flinders Uni

One of the participants of the 1988 IMO who was able to solve the problem (and win a gold medal with a perfect score) is Ngô Bảo Châu, who would also go on to win the fields medal (in 2010).

~6:00: "a and b can be any whole number, including zero ..." -- huh? Not if the problem states that a and b are positive integers, and it does!

The actual question is discussed starting at 5m10s

Simon is my favourite. Bear in mind that Matt and James have already set the bar astronomically high.

0:00 Did you ever hear the tradegy of Darth Plagueis the wise?

"This question stumped a Fields medalist" *Random RU-vid Commenters who want to feel special*: "Pathetic."

I shouldn't have watched this at 5 mins to midnight, I could be up all night.

Omg just tell me the problem

please check this solution a²+b² can be written as (a²+b²)(1+ab) - ab(a²+b²) and as (1+ab)|(a²+b²) then ab(a²+b²) should be equal to zero In case 1, when a² + b² = 0, the expression (a² + b²)/(1 + ab) simplifies to 0/(1 + ab) = 0, which is indeed a perfect square. In case 2, when ab = 0, the expression (a² + b²)/(1 + ab) simplifies to (a² + b²)/(1 + 0) = (a² + b²)/1 = a² + b². Since ab = 0, it follows that a² + b² = (a + b)², which is a perfect square. Therefore, based on these two cases, it can be concluded that for any values of a and b, the expression (a² + b²)/(1 + ab) is always a perfect square.

"It is about being able to solve awesomely hard problems" should have been my life philosophy and goal.

Lol a and b can't be 0 it says "POSITIVE integers".

I love the passion of this man. Bring him back as often as possible please !

I tried a basic method:- Say that (ab+1) divides (a^2+b^2) and the divisor is +ve integer "k". This solving shall give us an equation:- That a=b^3 or b=a^3 which when put in the main question gives result as a^2 or b^2 which is the perfect quare of an integer.

I have no idea why I'm watching this video, I still count using my hands... it's just giving my flashbacks of high school maths classes

It's amazing that the problem is so simple to understand, but still so hard to solve. Usually when I look at these math olympiad problems there's a bunch of terminology and symbols and stuff that I'm simply not familiar with, so I can't even interpret what the problem asks you to do. But in this case it looks pretty simple, I do understand what the problem asks me to do, but it's still one of the hardest math problems ever conceived. That's fascinating!

Very cool problem! One of my favorites. In fact, one can prove that all solutions can be generated by (k, k^3) for all integers k>0 .(sans order) The argument is surprisingly simple: FIx (a^2+b^2)/(ab+1)=x, and then see that if (a,b) is a solution with a+b minimized, then (xb-a, b) is also a solution with the same x-value (not too difficult compared to the other problems), and if a>b the a>xb-a as well. The only way out is if a=xb and you get the previous solution. (forces k=b^2) Admittedly, this solution exploits a rather modern technique used as Vieta jumping, which basically solves a quadratic in one of the variables. Tells you how much more difficult the problems have gotten these days!

Haha. I like how those problems in number theory are so simple to state...even an 8th grader could under stand what the question asks. But to solve them requires maths of a much larger caliber.

5:29 for those like me that have no tolerance for long winded intros.

At the same time I saw the title of this video, I was in the middle of working through my M2 maths book. I was on Q6.

He has a certain appealing charm.. but can't figure just what it is

It’s not hard to prove if you visualize a*a and b*b as squares on paper and also ab+1 as a rectangle plus one square (the +1 part), where 1 is a fraction of a*a square. This fraction is always a square of an integer because in order to fit (ab + 1) into (a*a + b*b) you need to fit that +1 part into a*a square the whole number of times. That said, the b should be always equal (a*a)*a = a^3 to satisfy the condition of this equation. Try to use b=a^3 to see that it works. It took me about 30-40 minutes to visualize, understand and explain this solution.

1988 Question 6 is the hardest question eva! 2020 Question 6: Am I a joke to you?

Okay, taking a 90 minute break from reading to try. I think I’m going to cry.

3:16 It actually appears in the Niven's number theory book. It's the last problem of the section 1.2

I made a comment earlier, saying that i solved it, and im sure i did. its not a simple a=1, b=2. you have to show that you can solve (A^2 + B^2) / (AB + 1) = X^2 where 'X' is an integer greater than or equal to 1. so, A=1 and B=1 works, but so does A=2 and B=8 (X=2 so X^2=4) you're trying to solve for the sequence of answers. I first assumed that i could make B = A+n, where n is an integer > or = 0 so A and B can be the same so i rewrote the function as (A^2 + (A+n)^2) / (A(A+n)+1) = (x^2) / 1. i wrote is as a ratio cause it made it easier in my brain. then i moved AB+1 to the other side of the equation and tried to solve for X^2 = 4. so, (A^2 + (A+n)^2) = (4)(A(A+n)+1), where 'n' is the value that allows the 2 sides to be equal, from here it was a plug and chug in Desmos graphing calculator to find the intercept between the 2 at which all numbers are integers -> try all values of n = 1 - 10 6 is the only one that works when N=6, the point (2,68) is the intersect so A=2, B=8, X^2 = 4 from this i concluded that B = A(X^2). or you could say that i assumed this was relationship between A and B so (A^2 + (A(X^2))^2) / (A(A(X^2))+1) = (x^2) / 1 from here i thought, what if i make A = X? that seems to hold true for 1 and 2 A = X becomes (X^2 + (X(X^2))^2) / (X(X(X^2))+1) = (x^2) / 1 OR ((X^2) + (X^6)) / ((X^4) +1) = X^2 THEREFORE A = X , B = X^3 when 'X' is an integer > or = 1 and that is the solution for the sequence if X=5 -> (25 + 15625) / (625 + 1) = 15650/626 = 25

I was just watching p*rn and accidentally opens RU-vid and this was in my recommendation , not gonna lie this has more logic and concept than what i was watching before,and even more interesting.

My old math professor at 7:44. Professor Kung at St. Mary's College of Maryland.

Have to love the the enthusiasm Simon has for Numbers :)

I did not try solving it the mathematical way. But if you run this Python program: count=0 for a in range(1,100): for b in range(1,100): for n in range(1,100): if ((a*a)+(b*b))/((a*b)+1)==n: if count==9: break else: print(a,b,n) count+=1 You will come to know the integers that you will get by solving a*2+b*2/a*b+1 are always a perfect square. For example 5, 125, 25 tells us 'a'=5, 'b'=125 and integer=25. 112,30,4 tells us 'a'=112,'b'=30, and integer = 4. p.s: this code is made just in 5 minutes. @numberphile

To the people who keep saying the answer is 1, they're not asking for a solution to the expression, because it already tells you that the possible solutions are all squares. What they're asking is that you prove why that is always the case.

one of my favorite numberphile videos. the storytelling is best of the best

We need more videos with Simon in them. They're always entertaining to watch.

The pairs of integers that fit the equation are x^(2n-1) - (n-2)x^(2n-5) + T(n-4)x^(2n-9) - TT(n-6)x^(2n-13) + TTT(n-8)x^(2n-17) - TTTT(n-10)x^(2n-21) + ... where T(n) is the triangle number TT(n) is the triangle number of the triangle numbers and TTT(n) is the triangle number of the triangle numbers of the triangle numbers and so on. If you substitute n = n - 1 you get the other pair and if the power becomes negative you stop the formula. So if n = 11 you get a=(x^21 - 9x^17 + 28x^13 - 35x^9+15x^5- x) b= (x^19 - 8x^15 + 21x^11 - 20x^7 + 5x^3) cause T(11-4)=28 TT(11-6) = 1+3+6+10+15 =35 TTT(11-8) = 1+1+3+1+3+6=15 TTTT(11-10) =1 and T(10-4)=21 TT(10-6)=1+3+6+10=20 TTT(10-8) = 1+1+3=5. All the coefficients add to either (1,1) (1,0) (0,1) (0,-1) (-1,0) or (-1,-1) so that x = 1 will result in 1.

This problem is hard! But it doesn't compare to problem 6 at Romanian Masters of Mathematics 2016 ! " A set of n points in Euclidean 3-dimensional space, no four of which are coplanar, is partitioned into two subsets A and B. An ABtree is a configuration of n − 1 segments, each of which has an endpoint in A and the other in B, and such that no segments form a closed polyline. An AB-tree is transformed into another as follows: choose three distinct segments A1B1, B1A2 and A2B2 in the AB-tree such that A1 is in A and A1B1 + A2B2 > A1B2 + A2B1, and remove the segment A1B1 to replace it by the segment A1B2. Given any AB-tree, prove that every sequence of successive transformations comes to an end (no further transformation is possible) after finitely many steps." From 113 participans, only one did it completely and took 7 points, and a few scored 1 or 2 points !

What does *"such that ab + 1 divides a² + b²"* mean? That sounds like a clumsy way of phrasing it. I'm already lost and I haven't even finished reading the question!

I love how he takes the time to write down "no one cares" at 6:29 XDD

Reading that question gives me a lot of anxiety remembering how badly I failed my college Trig and Pre-Calc class 😂

I solved this with a great insight and shortcut that took about 45 minutes. I wrote it in the margin of this video, but changed my text view size and lost it.

That competition must have been held on April 1st. What were they thinking?!

I would already cry if I got that question asked in an exam, not only if I were able to solve it. (but I would cry hard if I could solve it!)

Problem 6, as written, does not give the condition that 'a' does NOT equal 'b'. So, let a = b = 1 (a positive integer) => a² + b²/a∙b + 1 = 2/2 = 1 and (1)² = 1, an integer. Solved 1 minute 40 seconds.

I remeber Chau Ngo, Vietnam people 7 point of that problem and he also win Field medal

Parker Square demonstration

5:15 that laugh a professor has when he knows that you will fail at the task. 'well the best mathematicians in the world could not solve this problem in 6hours, well i give you 90 min' you are welcome.

ALWAYS keep on FIGHTING for ULTIMATE MATHEMATICAL GLORY!!!

Timestamps: 0:00 to 5:29 anticipation 5:30 to 8:45 the problem

+Numberphile A small error at 6:04 - a and b cannot be zero as they are positive integers. 0 is not a positive integer.

I'm at 6:27 ... but I'm affraid they will explain the solution. My question: *Spoiler alert?*

Solved it in under 5 mins. Assume a

There's only one negative integer solution to the equation which is -5. The 8 non reducible sets of a and b are (-1,2) (-1,3) (2,-1) (3,-1) (1,-2) (1,-3) (-2,1) and (-3,1) and with these you can Vieta jump to larger absolute values. Like -5(3) - (-1) yields -14,3

4:51 I thought he said 19 minutes and I was like, I thought you were the math guy XD

"That's numberwang!!!"

Two solutions: both a and b equal 1. Or one equals 0 and the other equals 1. Both ways give the fraction of 1 which is the square of itself. Took like 5 seconds of thought... now if the solution of higher squares is required, just involves some intuitive reasoning . Set the fraction equal to the square and reverse solve.

Someone has probably mentioned this already but the problem said a and b are positive integers. So neither are equal to 0, but 0 was included in this explanation. I suppose it doesn’t ultimately come up in the solution but something to make sure to point out for people giving this problem a try.

Crying after solving a problem like this is so understandable. Mathematics is something.

This is one of the hottest problems *_EVAHH_*

Have you ever heard the legend of question six? No. I thought not. It's not a story your math teachers would tell you. It's a mathematics legend. Question six was a question so hard, and so obscure, that it could influence even hardened mathematical scholars to create contests. It required such knowledge of the arcane, that it could even start the ones who cared about it to crying. It could actually cause mathematicians to despair? The Australian mathematics olympiad is a pathway to many numbers some consider to be... unnatural. What happened to it? It became so well-known that the only thing it feared was being solved. Which, eventually, of course, it was. Unfortunately, it was passed to everyone who knew, and it was solved in 1 year. Ironic. It could keep others from passing the test, but not itself. Is it possible to learn this solution? Not from a mathbook.

Me : I am tired of my math study , let's take a break RU-vid : wanna see some brain cells killing math problem ??? Hey RU-vid , for what thing you are taking revenge on me ???

Ok I ckicked on "part two", one does not simply stop at the end of this one.

doesnt positive intagers mean the set 1,2,3,4... not 0,1,2,3,4... that would be a non negative set simon wrote down. ????

(1^2 +1^2)/(1*1+1)=2/2=1, so a solution of the type they request exists. We discard all fraction solutions as contradictory. Alternately, in addition to the solution of (a^2 +b^2)/(a*b+1) being the square of an integer, it could be a prime number or a composite, non-squared number. Starting with a prime number as it is easier, we test (a^2 +b^2)/(a*b+1)=p. (a^2 +b^2)/(a*b+1)=p a^2 +b^2=p(a*b+1) a^2 +b^2=p(a*b)+p a^2 +b^2-p(a*b)=p which implies (a^2 +b^2)/(a*b+1)=a^2 +b^2-p(a*b) a^2 +b^2=[a^2 +b^2-p(a*b)](a*b+1) a^2 +b^2=[a^3*b+a^2] +[a*b^3+b^2]-p(a*b)^2-p 0=a^3*b+a*b^3-p(a*b)^2-p p=a^3*b+a*b^3-p(a*b)^2 = ab [a^2+b^2-p(a*b)] p=a*b*p so p is prime iff a*b = 1, which we already established (within the realm of acceptable solutions) means that p=1 as well. So, now we assume that we have a composite, non squared number, say p is replaced with n*m, where n =/= m. (a^2 +b^2)/(a*b+1)=n*m ... (as above) a^2 +b^2-p(a*b)=n*m and so, (a^2 +b^2)/(a*b+1)=a^2 +b^2-n*m(a*b) ... (as above) n*m=a*b*n*m again requiring a*b to equal 1, and thus a contradiction as n*m = 1 by consequence of our initial test. Therefore, all solutions must be as requested. So what's wrong with this answer?

that evil laugh at the end is him gone completely mad over this at one point for a year.

Forget the scores, we should know who answered Question 6 correctly!

There is a mistake in the vid: a and b are positive integers, none of them can be 0

I paused at 5:30 to do it, 10 minutes. it's 1 guys, a and b are both 1. it does not say they have to be different integers. (but i did just find out that there's no r before the g in 'integers', so thanks firefox spellcheck) (1*1)+(1*1) = 2 and (1*1)+1 = 2, then it's 2/2 =1, 1 is a square root and a square. you have the answer.

This is a trivial problem. Let a=1 and b=1. a^2 +b^2 = 1 +1 = 2. ab=1x1 = 1 and ab+1= 1+1= 2. So the resultant equation is 2/2 = 1 = 1^2 a square number . Took 2 minutes. Note it does not specify a and b must be different numbers.