the equation Ramanujan couldn't solve!!

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Күн бұрын

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@MichaelPennMath 11 ай бұрын

Head to squarespace.com/michaelpenn to save 10% off your first purchase of a website or domain using code michaelpenn

@NH-zh8mp 11 ай бұрын

I have a question : let I be a n-rectangle in R^n and f : I -> R be Darboux integrable. Suppose P_m be a uniform partition of I, which means P devises each edges of I to m intervals of the same lenghth. How to show that the Riemannian sum with respect to P_m is convergent to integral of f on I ? Please help, thanks.

@MathFromAlphaToOmega 11 ай бұрын

In fact, it turns out that the Diophantine equation x^2+A=2^n for fixed A has at most 2 solutions, except when A=7. Pretty weird...

@bamdadtorabi2924 11 ай бұрын

Do you have a proof of this fact? I imagine it would use some algebraic number theory for the UFD part of the proof.

@franzlyonheart4362 11 ай бұрын

Thanks for that interesting addition. Is the proof as easy as the one shown here, or is it more complicated?

@pierrecurie 11 ай бұрын

A=0 has an inf # of solutions (n even, x = 2^(n/2))

@mrlimemil 11 ай бұрын

@@bamdadtorabi2924 It's mentioned on the wikipedia page for this function. en.wikipedia.org/wiki/Ramanujan%E2%80%93Nagell_equation

@MathFromAlphaToOmega 11 ай бұрын

@@franzlyonheart4362 It's much, much more complicated. There's an article by Beukers titled "On the generalized Ramanujan-Nagell equation", where he proves that there is at most one solution except when A=7, 23, or 2^k-1 for k at least 4. From skimming it, it appears to use a lot of hypergeometric series, p-adic numbers, and factorization in imaginary quadratic fields.

@praharmitra 11 ай бұрын

Ramanujam seems to be the absolute GOAT of making insane conjectures without proof. Most of them end up being proven correct after a significant amount of effort. Really shows how much intuition he seems to have about mathematics.

@idjles 11 ай бұрын

He probably saw the answers before he finished reading the question, wrote down the answers and went on to something more interesting.

@ssl3546 11 ай бұрын

And most tragic of all, he was killed by the low quality of early 20th century British food.

@extreme4180 11 ай бұрын

well in ancient india, ppl used a small blackboard thing called slate to write with chalk which was later rubbed off ..i guess ramanujan used the same for writing proofs and so he erased them...

@charleyhoward4594 11 ай бұрын

maybe so, but he still couldn't prove them - now could he ...

@charleyhoward4594 11 ай бұрын

@@extreme4180 guess they couldn't get a piece of paper and pencil, Uh ?

@franzlyonheart4362 11 ай бұрын

35:40, the k was actually correct, the 2 is incorrect. In the equation right above that, it also says 2, which also is false. That should also be an exponent k, rather than the exponent 2 as written. There are quite a few calculation errors in the whole video today! And they're being glossed over, because many steps have been skipped over. Well, it already is a very long video today, so going cleanly through all steps probably isn't KZbin-friendly.

@ffggddss 11 ай бұрын

x² + 7 = 2^n. Pre-watch: ("Ramanujan" implies it must be Diophantine, right?) Just using trial-and-error, move the "7" to the right [x² = 2^n - 7], run through some powers of 2, and see what makes the RHS a square. I quickly get these: (x, n) = (1, 3), (3, 4), (5, 5), (11, 7). I can see how a general solution could have eluded him, though...or determining whether these are the only (+ve) integer solutions. Post-watch: Oh I see, it wasn't that he couldn't solve the equation, but that he couldn't prove there were only those few solutions you gave. Fred

@HagenvonEitzen 11 ай бұрын

13:24 Well, at that stage, we might still have y + omega = omega^a * omegabar^b and (y + omega)bar = omega^b * omegabar^a for othrecombinations of a and b with a+b = n-2. But a short check shows that we must have a=0 or b = 0: If both a and b are positive, then y+omega and y + omegabar are bothis divisible by omega omegabar, i.e, by 2. Then their sum y + omega + y + omegabar = 2y+1 = x is also even. But we already know that x is odd. Alternatively, the difference of y+omega and y+omegabar is sqrt(-7), which has norm 7, hence the factors cannot have a common factor of even norm (such as omega or omegabar). EDIT: Aha, I startedt commenting just a few seconds too early

@abi3135 11 ай бұрын

16 omega's

@IAmTheFuhrminator 11 ай бұрын

Hi Michael! I noticed quite a few calculation mistakes this time, and while you pointed some of them out in your post-edit, some of them weren't caught. This plus the skipped calculation steps for "trivial" calculations made the video slightly harder to follow, and I ended up having to look up the paper you referenced to get the full picture. If you were rushing at the end to try and meet a "KZbin-friendly time limit" as someone else in the comments said, then I would much prefer seeing a proof like this in all of its glory with all of the calculation steps even if it's 1 hour+ long. Sometimes the best math just can't fit into a 37-minute video! Thanks for all of the great work; I'll be looking forward to the next video!

@MyOneFiftiethOfADollar 8 ай бұрын

We are all brokenhearted that you had to waste your precious time to read the resource provided in the video. Use all your well intentioned advice to produce a video on this topic that meets your fastidious demands. I look forward to viewing said video on YOUR channel.

@IAmTheFuhrminator 8 ай бұрын

@@MyOneFiftiethOfADollar dude chill, I was just commenting to let him know what would have made the video more enjoyable for me. He doesn't have to do anything I asked for, but if people don't give feedback to content creators then they don't know if they are producing videos that people will keep enjoying. So sorry not sorry 🤷

@59de44955ebd 11 ай бұрын

The name of this norwegian guy was actually "Nagell", no trailing "e" (and therefor also not pronounced french/italian).

@jimm1037 11 ай бұрын

This is so similar to an equation I have been looking at: Integer solutions to 2^n - n = x^2. The only solutions appear to be n = 0, 1 and 7 but I cannot prove it.

@pnintetr 7 ай бұрын

I could check that all the solutions other than (n, x) = (0, 1), (1, 1), (7, 11), if exist, would require n ≡ 7 (mod 8). Checking n ≥ 0 is trivial. Even integers (≠ 0) are easily eliminable since we need (2^m - 1)^2 < x^2 < 2^(2m) for all integers m ≥ 1. Odd integers are difficult to approach, although I suspect Pell's equation might help. We can at least reduce n ≡ 1 (mod 2) to n ≡ 7 (mod 8) by gradually checking parity.

@koga2960 3 ай бұрын

@@pnintetr I reached the same results, but I also noticed that analysing n = 7 mod8 for sufficiently big n will force x = floor(sqrt(2^n)). Do you know of any interesting ways to look at the growth of 2^n - floor(sqrt(2^n))²? If we show it grows quicker than linear (although I'm not even sure that is true) then the problem would be over

@koga2960 3 ай бұрын

I checked all the values of n from 0 to 400 and only 0, 1, 7 work in that range, as well. Really makes it seem like they're the only answers

@anksssssssss 11 ай бұрын

ramanujan never needed to prove anything , he thought was waste of paper for and paper was very expensive for him , he just give answers and theorums and leave proving to future mathmathecians as their life goals

@demenion3521 11 ай бұрын

i think it's worth pointing out that omega_bar is also an element of your ring since omega_bar=1-omega which is clearly an element of Z[omega]

@krisbrandenberger544 11 ай бұрын

@ 27:48 The sum should have a two in the front, and the b_n^k term should be b_n^j.

@siquod 11 ай бұрын

When I saw the thumbnail, I had the intuition to adjoin the square root and use a linear recurrence or something, but then I thought "If Ramanujan couldn't solve it, what chances do I have?" Turns out I was on the right way, even though it gets a bit more involved after that start.

@goodplacetostop2973 11 ай бұрын

37:01

@Macieks300 11 ай бұрын

It actually sounds like a classic Ramanujan. He just made a conjecture he didn't care to prove which turned out to be true.

@saifidinrahmonculov-kv6xp 11 ай бұрын

You're a super mathematician

@allanjmcpherson 11 ай бұрын

I'm a bit confused by the step at 36:10 where we've substituted b_12 and b_13. That gives us a term of (-1)^(j+k-12) that appears to simplify, but I don't understand how it does.

@franzlyonheart4362 11 ай бұрын

He's making A LOT of calculation mistakes in those steps. You can see that he's rushing it and glancing at his notes all the time. I suppose he didn't want the video to become that long, in order to remain somewhat KZbin friendly. The only way to "get it" is to sit down and recalculate it yourself, using the video more like a guideline than fact, and not skipping all the steps he skipped, and then you'll avoid (and fix) all these calculation errors. And perhaps in addition using the original paper with the proof that he references at the beginning.

@Alex_Deam 11 ай бұрын

I think that k-12 should be k-j, which means that term becomes (-1)^k, but since k is odd this reduces to -1

@hassanalihusseini1717 11 ай бұрын

I must admit I did not understand the proof. But anyway it was interesting to see the limited numbers of solutions. Can it be proven for other numbers than 7 (I mean for x^2 + p = 2^n, with p€N, or p prime)?

@justinharper6909 11 ай бұрын

Anything can be proven in maths. Otherwise it would not exist.

@MrMessimm 11 ай бұрын

Michael Pennis my favourite matematician!

@carlpeterkirkebo2036 11 ай бұрын

As far as I know there is or was no Norwegian mathematician with the name Nagelle. And that name does not sound very Norwegian. There was a Norwegian mathematician with a similar name; Trygve Nagell. He worked in this field.

@koenth2359 11 ай бұрын

4:00 To me it wasn't immediately clear that for an arbitrary complex number ω the definition Z[ω] = {aω+b with a,b integer} would define a ring. E.g. for ω=e^πi/6 and x in Z[ω], Im(x) would always be an integer multiple of 1/2, whereas Im(ω^2) would be sqrt(3)/2. So for this ω, the given definition of Z[ω] would not be a ring, because it is not closed under multiplication. For ω = (1+sqrt(7)i)/2 however, we have ω^2 = (-6 + 2sqrt(7)i)/4 = ω-2, so no problem in this special case. Also, the assertion that elements have unique factorization is not clear to me. Still any composite integer would factor in more than one way, would'nt it?

@reijerboodt8715 11 ай бұрын

These are good observations/questions, answered by a first course in ring theory. To help you on your way: Let R be a ring without zero-divisors, that is no elements a and b such that a*b = 0. 1) What we actually mean when we write R[ω] is the set of all expressions of the form a_0 + a_1 ω + a_2 ω^2 + ....., with a_k in R, and where only finitely many of the a_k's are non-zero. This always defines a ring with multiplication given by expanding brackets and collecting like powers of ω. 2) In the cases where ω satisfies a polynomial equation of degree n, that is P(ω) = 0, this collapses down to { a_0 + a_1 ω + ... + a_{n-1} ω^{n-1} | a_k in R for all 0

@koenth2359 11 ай бұрын

@@reijerboodt8715 sorry but i did not mean to ask any questions, it was more a comment on the way it was presented.

@Sandsteine 11 ай бұрын

The comment i searched for

@vikramanbaburaj525 6 ай бұрын

Wow..sounds like a poem. Absolute pleasure!!!

@devondevon4366 10 ай бұрын

n = 7 x =11 answer x^2 + 7 = 2^ n x^2 = 2^ n - 7 x^2 is odd (since 2^ n is EVEN and 7 is 0DD and EVEN - 0DD = EVEN OR vice versa) Hence x is odd Possible ending 1 * 1 = 1 3 *3 = 9 5* 5 = 25 7 * 7 - 49 9 * 9 = 81 So x^2 either ends ( or the last digit) with 1 or 9 or 5 Hence, 2^ n ends ( or last digit) is either 8 (7+ 1), 6 ( 7+ 9 =16) or 2 ( 7 + 5 =12) So we are looking for a 2^ n, which ends with last digit is either 8, 6 or 2 e.g. of 2^n = 2, 4 8, 16, 32, 64,128, 256, etc.. And when subtracting 7 from any of these numbers, it must be a square Let's try 128 (which the last digit is 8 . Recall it has to be either 8, 6, or 2) 128 -7 = 121 which is 11^2 11 is odd , so we are on the right track Answer 2^n 128 and 128 =2^7 hence n=7 and x =11 ansewr

@mcalkis5771 6 ай бұрын

Seeing Michael smile at the beginning is the highlight of the video.

@llchan 11 ай бұрын

I have been watching the abstract algebra series in the Math Major channel for the past 2-3 months. I didn’t completely understand the entire proof of this video, especially when it’s claimed that the solutions for b[n]=-1 has to come from n=4k+1. Anyway, it’s nice to see how ring theory can be applied to solve number theory problems. This is a really good example to show how the two areas are tied to each other.

@Gandarf_ 11 ай бұрын

Behind the scene there was proven a fact about b_n (mod 16) (you can find timing with b_n value table near the end, there was a circular pattern with length 4), and since 1 and -1 should stay as is after every mod we can say that b_n = -1 iff n=4k+1

@mmoncure11 11 ай бұрын

Please post video of you and spouse debating household budgeting process

@charleyhoward4594 11 ай бұрын

Lol !😄

@HagenvonEitzen 11 ай бұрын

13:08 There's not that much to check: The norm omega * omegabar is a rational prime.

@byronwatkins2565 11 ай бұрын

At 27:53, I think your sum is missing a factor of 2 due to omega omega bar... ditto for 28:39.

@garyknight8966 11 ай бұрын

However, I still can't figure out where the 2 pre-factor on the sum arose at 30:17 (not to mention the exponents of 2 rather than k)

@garyknight8966 11 ай бұрын

Not that the pre-factor matters when that whole sum is 0 !

@ankurantil6137 11 ай бұрын

He solved it, but on a slate with chalk, as he didn't have much paper and wrote the result only in his notes

@anarchostalinprimitividiag1030 11 ай бұрын

I love your content! Where in norway was the confernece you attended? And what was it all about? Im going to study math in Norway soon and would love to have you hold a seminar for me sometime :)

@21nck93 11 ай бұрын

I really have some issues with the solution that he demonstrates. Like, should there be a 2 before the sigma in 27:48 because omega*inverse-omega=2. Also, why does the exponent change from k in 30:01 to 2 in 30:21? Tbh, the ending equation just leaves me with confusion. Can anyone explains plz? 😢

@Hipeter1987 11 ай бұрын

33:10 how does the second equation imply the third equation? b4=-3 and b5=-1, so how do we get cancelation down to 0?

@Alex_Deam 11 ай бұрын

If I've understood correctly it's because the LHS is -1 and the RHS is (-1)^k +(mult of 3) (note that the exponent of 2 is a typo), and since k is odd that reduces to -1=-1+(mult of 3), which can only be true if that multiple of 3 is 0.

@dodokgp 11 ай бұрын

Video title: "The equation Ramanujan couldn't solve", Me noticing that video is 37 minutes long. Me: "Makes sense".

@thebeautifulbananabutter4662 11 ай бұрын

this is crazy

@johns.8246 11 ай бұрын

And here I thought the proof to Bertrand's conjecture was horrible.

@adamwho9801 11 ай бұрын

It is interesting that this formula breaks numerical investigation. You cannot just plug it in and test all integers.

@HagenvonEitzen 11 ай бұрын

Isn't that always the case with general statements?

@adamwho9801 11 ай бұрын

@@HagenvonEitzen No, many times it is the process that is more important than the solution. This is a competition math channel, so computers would not be used.

@gerryiles3925 11 ай бұрын

@@adamwho9801I think his point was that you can never "test all integers"...

@adamwho9801 11 ай бұрын

@@gerryiles3925 In this case it is true, but it isn't always the case. Sometimes numerical solutions can serve as a proof.

@theEx0du5 11 ай бұрын

They never do the UFD part..

@TheEternalVortex42 11 ай бұрын

Very cool video

@danilonascimentorj 8 ай бұрын

Can anyone use this and prove that for n>3, there is no solution for 5^n-121 = x^2?

@davidbrisbane7206 7 ай бұрын

Think x² + 11² and Pythagorean triples. Is it possible that the hypotenuse in this case is divisible by 5?

@dimitrosskrippka2154 11 ай бұрын

That’s funny that equations like this often have some solutions in small numbers and then no solutions whatsoever. I wonder what is the simplest Diophantus equation with a couple solutions that are all over googol

@xizar0rg 11 ай бұрын

Why mod 16? Also, why root -7 (though this I assume because +-5 didn't work)?

@pawetwardowski839 11 ай бұрын

Two reasons for sqrt(-7): first, it allows to nicely factor 2 (as the roots of x^2-x+2), second, unique factorization fails in the ring of integers of Q[sqrt(-5)].

@clearnightsky 11 ай бұрын

Tha's because the equation is x^2+7.

@at7388 6 ай бұрын

But Michael Penn could finally do it.

@VKHSD 11 ай бұрын

this video got me messed up

@giorgostarnaras5658 9 ай бұрын

can you solve a problem like this using binary nums?

@sharpnova2 11 ай бұрын

i could figure this out in seconds.

@mz1rek 11 ай бұрын

For the first time I'm commenting that the place stopped wasn't a good one; the contradiction was not very clear.

@MathsMadeSimple101 11 ай бұрын

In B4 8 year old kids in the comment section claim they can solve it

@sshenron3331 10 ай бұрын

There's ton of solution bro. First x = 1 and n = 3. And so on

@yuseifudo6075 22 күн бұрын

He is not "bro", have some respect

@NH-zh8mp 11 ай бұрын

@franzlyonheart4362 11 ай бұрын

You want us to do your homework for you ? 😂

@NH-zh8mp 11 ай бұрын

@@franzlyonheart4362 nah, I just tried every way I know and find on sites, but I’m still struggle, so I really need help

@franzlyonheart4362 11 ай бұрын

@@NH-zh8mp Can your tutors not help you? (I suppose it's a homework question from uni.) It appears quite random, because it is outside of number theory, the topic of this video here. But Penn has been known to pick up questions from many mathematical areas, maybe you'll be lucky! Thinking back to my own undergrad days, we studied Lebesgue integrals, I cannot remember Darboux, but it has been a LONG time.

@anestismoutafidis4575 4 ай бұрын

1^2+7=2^3 x=1; n=3

@pokemonanimebattle3615 11 ай бұрын

not looking at comments, x is 3 n is 4

@josephmartos 11 ай бұрын

Well, this is out of My reach :(

@gp-ht7ug 11 ай бұрын

Out of mine too

@Alan-zf2tt 11 ай бұрын

You are in very good company - with Ramanujan at least

@Sandsteine 11 ай бұрын

Which part is confusing

@gijbuis 11 ай бұрын

I'm no mathematician, but looking at that equation I see 2 variables and 1 equation - so that must be insolvable?

@DQFozz 5 ай бұрын

An unfortunate result secondary education I believe. I was taught this too. It's mostly true for linear algebra but this equation has other constraints which limit the solutions.

@ghlscitel6714 11 ай бұрын

x=3, n=4 is a solution

@GoranVedriskheops 11 ай бұрын

if x = 1, 3, and 5, then n = 3, 4, and 5

@pandabearguy1 11 ай бұрын

Were there any equations he could solve?

@brianstevens3858 11 ай бұрын

My quick solution is x=the square root of {2^n-7}. going for the quickest if n=2 then the square root of -3 or x=1.73205081 i. Is this wrong and if so how?

@조민구-h9r 11 ай бұрын

x is natural number, which means x cannot be an imaginary number.

@brianstevens3858 11 ай бұрын

@@조민구-h9r Ah. I didn't know they weren't allowed since complex numbers, are used in real-life applications, such as electricity, as well as quadratic equations.