3:55 LOL moment

@Happy_Abe

4 ай бұрын

😂😂😂

When factoring y^2+11 as (y+sqrt(-11))(y-sqrt(-11)), those factors actually won't be relatively prime if y is odd, since then they're each divisible by 2. But then if y is odd, x is even, and then y^2=7(mod 8), which is impossible.

@jovanradenkovic2852

4 ай бұрын

y^2+11==4(mod 8) is never a cube. Similarly, 11 can't divide y.

@MathFromAlphaToOmega

4 ай бұрын

@@jovanradenkovic2852 Thanks for reminding me - I forgot about sqrt(-11) being a common divisor.

Nice video as always. Note to editor, please increase video audio volume. Adds on YT play with way higher volume than the video, and when an add plays while you are following along, it is not great to say the least. Makes watching the video a bad experience.

This video teaches me how much I have still to learn!

At 5:35 there shouldn’t be a cube on the 4a+3… or did I miss something?

@Happy_Abe

4 ай бұрын

Yeah that looks like a mistake

@krisbrandenberger544

4 ай бұрын

It is a mistake.

Mordell curves? More like "Magnificent lecture, that improves"...our knowledge! 👍

😡

@scebsy6524

4 ай бұрын

caught you

@Maths_3.1415

4 ай бұрын

22:37 There was a time when he used to say "and that's a...."

@mikeschieffer2644

4 ай бұрын

I guess there just isn't a good place to stop on Mordell curves!

It's pretty trivial to construct Mordell curves with integer solutions. You pick any value of x, and calculate values of n in which x^3 + n produces a square. There are infinitely many Mordell curves that have integer solutions as there are infinitely many squares.

@vinesthemonkey

4 ай бұрын

yes, but it's more interesting there are finitely many solutions (or none) for each n

I'm a little late to the chat but doesn't the n=0 case have infinitely many solutions? If x^3 = y^2, then you get (0,0) and solutions of the form (m^2, ±m^3) such as (1,±1), (4, ±8), etc. which there are of course infinitely many of. Edit: nevermind, according the Wikipedia n is a non-zero integer so this doesn't happen.

Now, as a follow up: is there some pattern or recipe for which values of n give a Mordell curve with (or without) integer points, or is it just happenstance?

@MathFromAlphaToOmega

4 ай бұрын

I'm pretty sure the values of n for which there are integer points are distributed essentially randomly. You could come up with lots of values of n that work, like y^2=x^3+a^2, which has (0,a) and (0,-a) on it, but determining whether there are integer points for a general n is not a simple matter.

@Alan-zf2tt

4 ай бұрын

∞∀∞ google has a good bit on info on this. A search on Mordell equation might help you on your quest. Trying out a new signature ∞∃∞∀∞

@kennethforeman6164

4 ай бұрын

n can be generated by considering all integers m>=0. It should be clear that x^3+/-m^3 has at least one solution when y=0 and x=-/+m. But we can add 1 [x^3+/-m^3+1] and then y=+/-1 becomes a solution for the same x=-/+m value. In general, we can add k^2 to the +/-m^3 and have x as -/+m depending on the sign chosen in the original equation and y=+/-k. Generically, n={m^3+k^2 | m,k integers}It should be clear that no other n will give integer solutions.

a^3+b^3 = (a+b)(a^2-ab+b^2)

a²-a=2(mod 8) have 2 solutions 2,7 but you don't check 7, afterwords that isn't got us a solution for a problem.

@leif1075

4 ай бұрын

What do you mean sorry can you clarify? It depends on the value of a.

Why is the extension adjoining 1/2(1+sqrt(11)) and not just sqrt(11) also why is it a UFD? Also i know that equations like that have at most finitely many solutions (16 as far as i know) but why? That trick won't work on large numbers where the quadratic extension isn't a UFD

Is this anything to do with Birch and Swinnerton-Dyer?

@tetramur8969

4 ай бұрын

No, BSD considers rational points, not integral ones

Number theory my beloved

@Bemajster

4 ай бұрын

Have you tried going on a therapy?

@lgooch

4 ай бұрын

Geometry better

@zhabiboss

4 ай бұрын

@@Bemajster no, i tried to derivate to find the slop of my life’s curve to see how fucking fast it’s going down

@Bemajster

4 ай бұрын

@@zhabibossgood

Why is the audio so low?

Why is a^2-a+1 congruent to 1,3 mod 8 if it has a prime factor like that?

@aadfg0

4 ай бұрын

Michael's argument applies for every p which divides the term. Since every p is 1 or 3 mod 8, the product of all factors is 1 or 3 mod 8.

@spkersten

4 ай бұрын

@@aadfg0thanks!

@leif1075

4 ай бұрын

What he says at 10:00 is fucking unclear and impenetrable what the heck does he mean by negstive 2 or negstive 1 is not a perfect square mod p..of course negstive numbers are NEVER perfect squares ..so what does that even mean..I don't see anyone understanding this.

@leif1075

4 ай бұрын

​@aadfg0 also a bunch of terms that equals 3 mod8 multiplied together the remainder would then be 3^n mod 8 where n is the number of terms not just 3 mod 8 any longer.. 1^n is obviously just 1 so that product would retain the remainder of 1 mod 8 but nkt any other since they get multiplied by each other..unless you meant something else.

@aadfg0

4 ай бұрын

Try 3^n mod 8, you get a cycle of 1,3,1,3,...

By the way, there are at most 12 integer solutions to an elliptic curve equation

@kennethforeman6164

4 ай бұрын

Hmm. I think this may be true only when n is non-zero in this particular class of curves. Of course, that's because we might not consider the curve elliptic if n were 0. Do you know off-hand if we can restrict the maximum number of solutions possible for this specific elliptic form to less than 12 for non-zero n or which specific examples give all 12?

@fartoxedm5638

4 ай бұрын

@@kennethforeman6164 the torsion subgroup of an elliptic curve over Q is finite and is isomorphic to several.groups of orders 2 to 10 and 11 to 12(if I remember correctly). And also it consists of integer points and nothing more. I remember that you can actually find them by a homomorphism to rational squares. Also there is an n-descend method but I was too weak in algebra to fully understand it during reading about it

@kennethforeman6164

4 ай бұрын

@@fartoxedm5638 Thanks. I'll have to go back and look that up too.

Sir we are so pleased to beyond your imagination.... Thanks a lot from Ramakrishna mission students....from India Also due respect ❤

Why use 4a+1 and 4a+3 and not 2a+1?

@Zebinify

4 ай бұрын

because eventually, we are trying to use the fact that a^2 not landing on 3 mod4 or 2 mod4; if using base 2, then it's a free for all, a^2 can land anywhere mod2.

@bjornfeuerbacher5514

4 ай бұрын

Because these two sets of numbers behave differently in this problem, as Michael shows.

@leif1075

4 ай бұрын

​@@ZebinifyI don't understand what you mean at all. What donyou mean yb land anywhere..isn't the actual reason because we have squared and cited terms so any even number or odd will have a w squared hence you wrote as 4a instead of 2a..that is the reason so not sure what you meant if you could please clarify..

I guess it wasn't a good place to stop.

I apologize for being so immature, but… that curve perfectly captures my teenage years’ most embarrassing moments!

@letswait30days

4 ай бұрын

Apologies do not fix the inappropriateness of this comment.

Pretty tough to catch up. Requires a lot of background work.

"All odd numbers are of the form 4a + 1 or 4a + 3"... don't you mean 2a+1? And where does 4a+3 come in? Don't 2a+1 and 4a+3 just generate the same number for a different a?

@bjornfeuerbacher5514

4 ай бұрын

All odd numbers can be written as 2a+1, yes. But often (like here) it's convenient to divide the odd numbers into two classes - the ones which can be written as 4a+1 and the ones which can be written as 4a+3. And no, 2a+1 and 4a+3 do _not_ generate the same numbers. 2a+1 generates _all_ odd numbers (1, 3, 5, 7, 9, 11, 13, 15, ...), whereas 4a+3 generates only every second odd number (3, 7, 11, 15, ..., but not 1, 5, 9, 13, ...)

@leif1075

4 ай бұрын

Still doesn't explain why anyone would think of 4a plus 3well except for the fact we have x squared so 2a becomes 4a when squared..I believe that's the only reason

@vinesthemonkey

4 ай бұрын

because it's the next power of 2 from just considering even/odd and often gives a shortcut for quadratic residue type congruences. See also lifting-the-exponent lemma

@bjornfeuerbacher5514

4 ай бұрын

@@leif1075 The reason why people would think of that is because it's well known in number theory that numbers of the form 4a+1 behave totally different than numbers of the form 4a+3, both when one considers square and when one considers prime factorizations.

@bjornfeuerbacher5514

4 ай бұрын

@@vinesthemonkey Thanks for mentioning that lemma, I didn't know that. :)

I.dont see why amyone would extend to imsgonsety numbers suddelnyl..whybwould they..why not complete the swuare with 11? Aurely thats what ppl wpuld.think of..come on now

@forcelifeforce

3 ай бұрын

* suddenly

We don't want mordell curve,we want mordell theorem 😎

@charleyhoward4594

4 ай бұрын

We do ????????

@omargaber3122

4 ай бұрын

Yes , not just we , you too😎​@@charleyhoward4594

Why use mod 4 instead of mod 2..think it would be more obvious tonuse 2

@ConManAU

4 ай бұрын

Presumably if you try mod 2 you don’t actually get a useful result. Notice how different the 4a+1 and 4a-1 cases are.

@williamwarren5234

4 ай бұрын

The squares mod 2 can be anything, 0 or 1. mod 4 they can only take certain values, as Michael shows

Hello Professor Michael, I know that you are a person who only does what your mind tells you and does not listen to the requests of others, but I hope you will accept this request of mine, as I believe that you are the only one capable of that. We need an explanation of the limited gaps between prime numbers by the Chinese Zhang and the French Maynard. We also need an explanation of the scientific paper (Bertrand's Postulate for Carmichael Numbers Daniel Larsen) Please do it 😢

@bjornfeuerbacher5514

4 ай бұрын

Who is "we", and why do you need it?

@winnablebtw459

4 ай бұрын

James Maynard is English and you could always read the papers yourself and follow the citations as needed.

@omargaber3122

4 ай бұрын

​@@bjornfeuerbacher5514 I and me

@omargaber3122

4 ай бұрын

​@@winnablebtw459he look like French 😎

I never like these type of problems

@mr.bacteria7148

4 ай бұрын

Why not?

@charleyhoward4594

4 ай бұрын

@@mr.bacteria7148 because it requires far, far too much reflection to make sense of what's being talked about

These lectures always spend too much time writing out basic algebra, and not enough time on the actually useful tricks, like quadratic residues mod 4, mod 8, and other mods

@vinesthemonkey

4 ай бұрын

the quadratic residues and the quadratic fields are much richer and more interesting than individual cases

@adamnevraumont4027

4 ай бұрын

Sure - he's doing math accessible to 1st year undergraduates? Hence working out a bunch of details he could elide for a more advanced audience. The practice is good.

@vinesthemonkey

4 ай бұрын

@@adamnevraumont4027 I'm pretty sure undergrads can understand and verify the casework mod 4 without it needing to be written out line by line for 15 minutes. Put another way, anyone studying elliptic curves is already well-versed enough in modular arithmetic to do these simple computations.

@adamnevraumont4027

4 ай бұрын

@@vinesthemonkey I said first year - so fresh out of high school with a variable level of preparation, but some talent. This kind of "show the steps" gets the kids to be able to skip the steps themselves. And you can watch it at 2x speed if you prefer (many youtube videos are better that way). This material always feels aimed at a bright and mathematically naive first year undergraduate. It goes over the details with a grinding stubbornness I recall from my courses back then, and I personally really enjoyed seeing math actually worked out from something approximating first principles finally. When he asks you to take something on faith he does so explicitly and cites it, and only does so when the material required would be too bulky to solve the cute problem he's working on. Watching this guy is like asmr to me. :)

asnwer=2 isit