Powers of 2 (extra) - Numberphile 

To prove loops with four vertices cannot exist, consider such a loop with vertices *a1, a2, a3, a4:* *a1 -- a2 -- a3 -- a4 -- a1 // a4 connects back to a1* Assume *ak* stand for distinct integers, such that two neighboring *ak* sum to a power of two. We may assume *a3 > a1* and *a4 > a2* (if not, shift by two and/or change orientation): *a3 - a1 = (a2 + a3) - (a1 + a2) = 2^{m23} - 2^{m12} (1)* *a3 - a1 = (a4 + a3) - (a1 + a4) = 2^{m43} - 2^{m14} (2)* As they both represent *a3 - a1,* we have "(1) = (2)" and simplify to *2^{m43} + 2^{m12} = 2^{m14} + 2^{m23}* Notice *m43 > m12* (via *a4 > a2* and *a3 > a1* ), so the LHS is a binary representation of an integer with exactly two ones. But then, so is the RHS -- thus *m14 ≠ m23.* By uniqueness of binary representation, one of *{m14; m23}* must equal *m43* -- a contradiction to either *a1 ≠ a3* or *a2 ≠ a4,* respectively.

I wish I watched this video yesterday evening. I watched the original, lost the interest a bit with the bounded solutions. During the evening walk I realized I can prove that a(4) < 6. I got hyped a little bit :) Anyway, my approach: In binary, a power of two is always one followed by arbitrary number of zeroes. So when you add two positive numbers, they have to be a sort of "opposite" of each other to get one followed by only zeroes. For example: 00001101011 01110010101 Leading zeroes can be ignored. There can also be trailing zeroes, but those just multiply the solution. The fun part is that you cannot add a third opposite. So, to create a set of three numbers, you have to pick a negative number. This was done before. But when you want to add the fourth number, it will either be positive and won't be able to form a subset of three, or it will be negative, and sum of two negative numbers is again negative.

Could we see an update to this video explaining why it can't be a closed loop with 4 vertices?

I REALLY LOVE THIS GUY

That's amazing!

So a lower bound of n*log(n), and the 4-cycle thing gives us an upper bound of n*sqrt(n), whereas the total number of pairs is n² (all asymptotically up to a constant multiple). There's much leeway between n*log(n) and n*sqrt(n), but pretty cool progress so far

This is bananas. B-A-N-A-N-A-S.

cheeky unlisted video

What happens if you require no duplicates of powers of 2? No duplicates of any number sums?

Hi neil!❤

This video showed up in my feed 11 hours ago, but it has comments from a month ago. What happened?

Could someone help me here? Why is the generalized form of this n-1? Wouldnt it be n+1?

I wonder if graphs of solutions are planar.

Hello everyone enjoying secret video

Interesting topic, thanks for that. But I cooment for something else: the video was released ome day ago, but it did not show up on my subscription list … strange. I wonder how many videos I missed because of this

Ugh he said "no one knows" now its gonna haunt me to try and prove it.

What the heck was that about!

I hope I found easy solution, why it can't work with 4 distinctive integers. The key is in closer look on how we solve it successfully for 3. 1) for successful solution for 3 you need one negative integer: sum of 2 negative integers can't be power of 2, obviously, for same reason we can't use one negative, zero and one positive, since sum of zero and negative can't be power of 2, obviously, and why you can't use 3 positive integers or 2 positive and zero I will show later. 2) then, if we need one negative integer for solution with 3, we definitely need two negative integers for solution with 4, but obviously sum of two negative integers can't be power of 2 That's the elegant part of proof, the rest is not as elegant... 3) the proof why we need three positive integers is based on fact that if you sum two DIFFERENT powers of 2, you don't receive another power of 2, since 2^n + 2^(n+1) < 2^(n+2), Then we instead focusing on summed integers focus firstly on their sums: we decompose all powers of 2 into all possible couples of different integers, then we easily find two powers of 2 sharing one soluting integer in their couples, i.e. we find in decomposition of one power of 2 a couple with one integer which we can find in a decomposing couple for another power of 2, BUT the other halves of these couples don't sum up to the power of two; if you find more elegant proof that three positive integers don't work as the solution, then it really improves the proof

What?? Please provide some context.