# The Legend of Question Six - Numberphile

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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
Channel Id: undefined
Length: 8min 45sec (525 seconds)
Published: Tue Aug 16 2016

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.

👍︎︎ 45 👤︎︎ u/cactus 📅︎︎ Aug 16 2016 🗫︎ replies

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

👍︎︎ 30 📅︎︎ Aug 16 2016 🗫︎ replies

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.

👍︎︎ 19 👤︎︎ u/[deleted] 📅︎︎ Aug 16 2016 🗫︎ replies

The problem in question:

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

👍︎︎ 13 👤︎︎ u/Bromskloss 📅︎︎ Aug 17 2016 🗫︎ replies

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?

👍︎︎ 11 📅︎︎ Aug 16 2016 🗫︎ replies
👍︎︎ 12 👤︎︎ u/Jyben 📅︎︎ Aug 16 2016 🗫︎ replies

Can someone explain the whole parabola thing he did?

👍︎︎ 2 📅︎︎ Aug 16 2016 🗫︎ replies

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 = n3), 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_12, 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_12. 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_12.

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

👍︎︎ 2 👤︎︎ u/[deleted] 📅︎︎ Aug 17 2016 🗫︎ replies

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

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.

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.

👍︎︎ 3 📅︎︎ Aug 16 2016 🗫︎ replies
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