The terminology “ninja pair” has gotta be named after the Ninja Report from How I Met Your Mother, right?

I remember working on this problem in summer 2018. We used different methods to show that a solution exists for all integers of the form 8n^2+8n using regular sequences consisting of all integers from 2k^2-2 to 2(k+2)^2-2. You can stitch them together using graph theory to make a really nice honeycomb pattern which is easy to find hamilton paths on (which is equivalent to finding regular sequences)

Imagine the people involved in this just casually writing a proof by induction with 49 base cases lol

3B1B's SoME is probably the best thing for math education since the first KZfaq math channels like OG Numberphile. It has brought out so many fascinating areas of math I had never thought about, and encouraged so many fantastic educators to produce videos like this. Absolutely fantastic job and I'm glad I discovered your channel, and I can't wait to see more content from you!

3:08 I would argue that 1 should be ticked... There is one ordering, and it does satisfy the property, as there are no two consecutive numbers, we can say that the statement is true for this ordering. all the sums of two consecutive numbers are squares. (which is also confirmed later by « any subsequence of a regular sequence is regular ».)

I can't say this is the most accessible maths video for non-maths people, but it does feel exactly like reading a maths paper, and I'd like more content like this!

This is a beautiful, well-crafted example of how a recreational problem generates lines of reasoning; proof construction; and computational techniques - all essential tools for math writ large.

I must say, putting Matt Parker's face in the thumbnail was genius marketing. I thought it was a Numberphile video. I watched the whole thing because math.

For the Dutch IMO selection test, a solution for a particular number was needed was needed for the fourth problem. I instantly knew that there was an easy solution. Although I spended a lot of time on the problem (at least, more than necessary), I managed to solve the problem with 7 points.

Maybe I'm dumb, but I never knew square numbers could be represented like this, as literal squares. This is kinda mind blowing, because now I'm realizing that cubes are the same way but literal cubes. I had no idea! I just thought those were the names we gave them because a square is 2D space while a cube is 3D space. I never realize these numbers actually formed perfect squares and cubes and that's why we call them that. Now, I'm feeling a little angry that this somehow got passed me.

3:00 shouldn't 1 be possible? (As there are no pairs of numbers, every has a sum of a square)

This has been my all-time favorite problem on the numberphile channel! Thank you so much!

Such elegant solutions makes me love maths even more! Thank you for a great video

Wow, I'm so glad this video was reccommended to me! I really hope this video gets the views it deserves (which really ought to be more than the current count of 1709).

Thank you for a great video! You really succeded in making a simple walkthrough of such a complicated proof.

Absolutely brilliant! I've been thinking off and on about this and a related problem for like 5 years. There are few people who could prove this thing as it really required an epic combination of math and computer science skills. The animations were excellent and helped me to understand the proof in a way I never could have by simply reading the original textfile.

Kudos on the production quality of these videos :) Top notch

A beautiful example of what we can do when humans and computers join forces, and everyone does what they are the best at. Obviously, neither humans alone nor computers alone couldn't come up with this solution all by themselves.

Excellent video! I hope you've submitted to 3b1b's video contest; I think he would very much enjoy it.

So good! Thank you for this delightful video. Well done.

The most shocking part about this video is that it doesn't have a "shoutouts to all my patreons" at the end. You absolutely deserve some, fantastic video!

This was a really great watch! Thanks for showing this interesting solution

This was a great video! I can't believe I just found your channel - as a video creator myself, I understand how much time this must have taken. Liked and subscribed 💛

The legend is back!! Glad to see another upload, look forward to watching it (:

Really illustrative and concise. Beautiful solution!

The numberphile video had already given me some intuition as to why this was true (there's ~root 2 - 1 times the square root of n square numbers between n and 2n so every number above 25 has to connect to 2 or more other numbers in the graph, so it never creates another "tail"), but this video gives an intuitive demonstration with high quality visuals. Great video !

What an amazing explanation! Thanks for an awesome watch!

After watching this video, I thought of a way to generate regular sequences (not necessarily from 1 to n though) from any sequence (not necessarily square). So given any sequence (e.g. S = 0,1,2,3,4,5,6,7,8,9 which is not regular), there's a way to generate a generate a regular sequence. I might write about it later when I have some time if anyone's interested. I'm sure someone else has already figured this out, but I couldn't come up with the correct search terms to find it. It involves squaring each element in S and completing the square so to speak and generates anywhere from 1 to 2^(len(S) - 1) different regular sequences.

This reminds me a lot of taking a Hamiltonian path on a graph, with the generated sequences and the spare low numbers being the vertices, which share an edge exactly when their endpoints give a square. I think this breaks down though considering that the sequences have two endpoints, each of which must be used once

I knew the wait for a new video from you was worth it. It was a very interesting one tho!

This is awesome! Really cool and enjoyable to watch!

Perfectly done 👏👏👏

I was suspicious it would be related to cycles, then the end very clearly demonstrated it! Lovely video, amazing work my friend.

watched couple of minutes and I am already loving it.

When you mentioned the two novel techniques at 5:35 I was completely blown away but immediately got them. So cool. I love that feeling. I even said "cooooool" under my breathe but I am very stoned at the moment.

Another awesome and beautifully animated video! 😀

Wow, so much luck involved to make this solution possible. Imagine if the smallest number instead of 49 was some insane number to big for our current computers, or if one of the initial number didn't have a place in the chain or if the 1 ... 3/8 + ninja odd even conditions were too strict to find such a recurrence. I would really love to see all the ideas that failed before this one succeded.

16:42 I was confused on how you know that there's always a way to connect the ends of these sequences (and numbers from 1 to 12), such that the new sequence is also regular, until I realized that the ends are the same and none of numbers repeat.

I am so happy I am one of the first 1k subscribers. This channel will be huge!

Amazing video and great explanation! Thanks! 🔥

Great video with great visualizations!

Why does 1 not count? It seems like for all pairs next to each other (an empty set) you get a square, right?

Great video, I want to ask, how did you make this video, and what tools did you use

For the shift function, couldn’t it result in a non regular series? For example if you have the series s = 17, 19, 6 and then you shift(s,1), s would equal 18,18,7 which is non regular series because 18 is repeated twice. Am I missing something?

great video!

Awesome video, thanks for sharing

Great video! Congrats to Robert Gerbicz!

A mathematical reason why the longer sequences all work and the smaller ones do not is because there is an increasing number of possible square numbers. 1-15 can only reach up to 25 (36 is more than 15+16) But if you have 100, that can go as high as 196 (97+99 or 100+96) This gives an increasing number of possibilities of matches. 51 matches with 13, 30, 49, 70, and 93

Nice vid, you explained the solution pretty well but I would like to see more vids that explain the problem solving methods that lead to the solution

9:40 - That sequence is mirrored in the middle with an extra number at the end. Surely that must be indicative of some special property of the original sequence. Does it have any meaning for sequences generally?

I love how you prove something stronger than just the existence of Regular Sequences of Length n for all n>=99 but you prove the existence of Ninja Pairs of Length n for all n>=99.

Very nice! Here is an intuition why the existence of a solution or a ninja pair is perhaps not so surprising. If we formulate the problem as finding a hamiltonian path in a certain n-node graph, then since the way the graph is defined is very "random", the graph should behave in hamiltonicity as if it was a so-called Erdos-Renyi random graph where every edge in the graph is there independently with some probability. Random graphs start being hamiltonian when every node has around log(n) edges. Here, every node has around sqrt(n) edges (because there are around sqrt(n) square numbers between n and 2n), so we should expect it to behamiltonian for large enough n. Ninja pairs seem to be a very similar story since it is kind of like trying to find two hamiltonian paths in a random bipartite graph. In those graphs, the threshold for hamiltonicity is still around log(n) edges per node. This is not a formal argument, but this type of reasoning is often useful to get an intuition about whether the solution should exist or not. For example, because of this I predict that if you replace second power in the definition of the graph by the third (or any other) power, the graph remains hamiltonian for large enough n.

Loved to see Robert Gerbicz's name in this! Famous fot the Gerbicz check (used in finding big primes).

14:54 the 8 is not highlighted in red. It is above the bottom right corner.

Very Interesting! Great Job!

This is such an interesting and exciting video!

My analytic mind: Wow. So logical. Makes perfect sense. This is AWESOME. My inner child: Wow. Ninja ninja ninja ninja. Ninjas everywhere!!! This is AWESOME.

At 3:06 you color 1 as red, claiming that the sequence "1" is not regular. I would say that this sequence should trivially be regular and also, saying that it is not regular would contradict your later statement that any subsequence of a regular sequence is also regular.

there were some wrong examples in the sequences, like 1+14. But considering the humongous amount of numbers in this I don't blame you one bit. Well done, I was very exited while watching this.

That's amazing!

This is amazing!!!

I thought of another way to create new regular sequences from a given regular sequence. I don't know how useful it is, but it's what I got when he told us to pause. Look at the endpoints of your sequence. Find a number in the sequence, (not right next to it) that adds to a square. Split your sequence at that point. Reverse the side of your sequence containing the endpoint you chose. Example 3 6 19 17 8 1 15 10 26 9 7 2 14 11 5 20 16 See how 3 and 1 make a square? Reverse the numbers from 3 to 8 to get 8 17 19 6 3 1 15 10 26 9 7 2 14 11 5 20 16 On the other side, 16 can make another square with 9, so you can get 3 6 19 17 8 1 15 10 26 9 16 20 5 11 14 2 7 This is cool because it makes new regular sequences with the same numbers, but with different endpoints.

I thought of creating graph containing 15 nodes where each pair of nodes which sums is square of some number is connected by undirected edge. And solution will be any Hamiltonian path in this graph.

Are solutions to consecutive square problem unique from a given starting point (to rule out reversals obv)? If not, which I think is the case, is there a formula for the number of unique consecutive sum sequences for a given n?

Really amazing and interesting proof. Perhaps the same method can be used to solve the Collatz conjecture, as it's also a sequence problem, but idk

do we know anything about if adjacent terms had to add up to cubed number? Is it possible for 1 to n? What is the smallest n such that it works and is there a value N where for any n larger than N there is always a sequence?

Not sure if maybe I just missed it, but how could you ensure that there is a fitting pair of sequences to make any ninja pair work, it seems like you prove only that it has a possible solution, and then you take the assumption that the solution for each number would fit perfectly for the ninja pairs

Perhaps this is covered in the post you link to, but why do you use the shifts from -12 to +12? Could you not go from -24 to 0 so that you only have to place sequences end-to-end? My assumption is that you're not guaranteed to find matches between sequences this way, and the stray integers from 0 to 13 are just there to help find a sequence? I hope my question makes sense and isn't just the rambly mess it kinda looks like to me.

btw If i remember, 3b1b and mathologer did a comp for upcoming youtuber math explainers highlighting their favourites. hope this gets into the mix if they do it again

At 16:00 the sequence: 12 13 3 1 14 2 are not all squares 1 14 is not a square pair?

I wonder if you could use a single row wave function collapse with appropriate rules to figure this out.

Is it possible to make a a regulus in which all neighboring numbers can make a square by addition AND subtraction?

Okay so this was interesting and all but I’m curious as to why n

Great video!

Can you take this construction to walk tour graphs?

This problem reminds me of the Knight's Tour puzzle, where a knight has to visit all 64 squares on a chessboard exactly once. For some solutions, the knight ends on a square that is a knight's move away from the square where it began the tour, allowing it to complete a closed loop. That suggests some questions for the puzzle in this video: Are there paths for which the first number plus the last number gives a perfect square, completing a closed loop (other than the trivial case with n=1, which I would argue isn't a solution anyway)? If such closed loops do exist, is there is a number n such that every number > n has a solution which is a closed loop? * * * * * Researching this further, I can answer my own questions: Yes, there are solutions that are closed loops, or more formally known as Hamiltonian cycles. The first one occurs at n=32. Every n through 85 has a solution which is a closed loop, so it is very likely every n > 85 produces one or more Hamiltonian cycles.

9:50 Is there any significance to the fact that the bottom sequence is (nearly) palendromic?

I remember hearing this problem from a friend, and I was just thinking of the problem and just trying out different combinations in my mind, and then I found a sequence that works! I checked this video to see if it was a different solution than this one, and it turns out my solution was this one - but backwards. 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8.

coould you explain Maryam Mirzakhany's magic wand?

thank you!

well, good effort for something that was just put in a blog post. nice. difficult to get the proof across up to a point but I cant think of a better way. interested in the coding side. yeah, thanks. hope it gets you some good clicks.

A lot of odd leaps are made in requirements for this proof making it more of an arbitrary proof rather than a simple one. (which is then brute forced by a 'computer') Regardless of that premise, it's still an interesting find, be it not a very elegant one.

Amazing video!

Very nice!

16:04 , is it normal it's not a regular sequence ?

Do we know what happens for cubes and other higher powers?

Did Sum(Sum(cycle)) produce a square cycle?

I have another problem I'm sure people will find interesting, let say you have a circle of size c, on this circle there is a pond of size p, and a frog elsewhere of position f as a single point. The frog can only jump at distance j. how many time the frog will go around the circle to land of the pond, from any dry position and with jump longer than the size of the pond?

Is there like an underlying reason why all 98 cases have a solution or is it just like the chance that there isnt a solution is so low that there just happens to not be any impossible cases? Like for example 49n+29, is the formula in the proof the only possible formula and it has some interesting reason for why it works or there’s like 20 to choose from that just happen to be there

Well, not easy as a non mathematician but satisfying. GreatVideo quality

3:45 I wonder if these larger numbers have multiple sequences... and if numbers larger than N will always have more than 4,5,6, 7, 8,... possible sequences...

Great Video!

Excellent video. If only I understood it.

The math was so interesting I watched the video twice, despite that awful ding noise. Usually I leave a video the moment I hear that sound effect.

now that we've found that there's a solution for n >= 25, i wonder if we can find every solution to the problem and see what kind of underlying patterns and classifications these solutions have maybe there are "prime sequences" scattered to infinity that can't be reachable with any generative method

Yes, multiplying by 4, multiplying by 4 and shifting by plus or minus one leads to pairwise different sets. Also true for multiplying by 9 and shifting by plus or minus 1,2,3 or 4. But it's not true for multiplying by 4 or 9. Both contain multiples of 36.

So… next problem. Find a formula that enumerates how many different regular sequences there are for given n.

Brilliant

But how do you know that the missing numbers can always be inserted? Like when 1 to 12 was missing from the new sequence, you just slotted them in. How do you know that always works?

For the next SoME, do this but consider the last number is also paired with the first one