How many squares??

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Опубликовано:

1 дек 2021

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Комментарии : 51

@goodplacetostop2973 2 года назад

0:03 I think you already uploaded this video 16:33 Almost 200K subs!

@hayimshamir8426 2 года назад

I also remember this

@gniedu 2 года назад

In the end, k-2m and k+2m can't be 1 for one and -1 for the other, since the product is 1. They have to be both 1 or both -1. It doesn't affect your solution since this case won't give a solution anyway

@p.abhijitsuhas26 2 года назад

Didnt michael already upload this vid?

@NeilGirdhar 2 года назад

You don't need to solve for the closed form at all if you reduce everything mod 8 first. Then you can easily solve that a_n is congruent to 3 or 5 mod 8 for all n >= 2, which as shown in the video can't be perfect squares. This is because whatever a is, a_1 must be 1 or 5 mod 8, a_2 must be 3 mod 8, a_3 must be 5 mod 8, etc.

@NeilGirdhar 2 года назад

(That said, the use generating functions to solve for the closed form was illuminating.)

@franksaved3893 2 года назад

I remember the same problem, or a similar one, in the past videos.

@noahtaul 2 года назад

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-lTmYayfPYBU.html

@alejandrobuendia8477 2 года назад

15-oct

@alexbush9250 2 года назад

It’s because this concepts fits perfectly with what he has been teaching in his number theory course. It’s a great example pulling together multiple topics from the course.

@user-jc2lz6jb2e 2 года назад

@@noahtaul it's not just the same problem, but the exact same video 😂

@christianaustin782 2 года назад

Dude as the video went on i kept checking the upload date of this one to make sure it wasn't an old one because it seemed really familiar. Glad I checked the comments 😂😂

@Prinrin 2 года назад

You could also note that for even n>1, a_n is 3 mod 4 (not a square), and for odd n>1, a_n is 13 mod 16 (not a square since it's 5 mod 8).

@SanketAlekar 2 года назад

One suggestion - this could be solved in fewer steps if we use the fact that odd squares can only be 1 mod 4 and 1 mod 8. 4a_n + (-1)^n is always odd, so after a_0 every a_n is guaranteed to be odd. As a result every term is 4*odd +/- 1. If it's 4*odd + 1, 8 cannot divide 4*odd, so it can't be of the form 8k +1. If it's 4*odd - 1, then it's of the form 4k - 1 which is impossible for an odd square. So the only way for there to be squares in the sequence is a_0 being an even square and a_1 being an odd square. But if a_0 is a square, that means 4a_0 is also a square, which makes 4a_0 + 1 impossible to be a square unless 4a_0 is 0 (0,1 are the only two consecutive numbers that are squares). So a_0 = 0, a_1 = 1.

@colbyforfun8028 2 года назад

Easy way to prove the corollary at 9:45, non-rigorously. The last digit of powers of 4 is 4, 6, 4, 6 etc so by adding in alternating +1 and -1 for each power of 4, the last digit becomes 5, 5, 5, etc which is obviously divisible by 5.

@user-cr4fc3nj3i 2 года назад

Even easier way is to take mod 5 4ⁿ-(-1)ⁿ≡(-1)ⁿ-(-1)ⁿ≡0 (mod 5) Done, it's just that easy

@bobbyhanson346 2 года назад

Nice. Here's how to make it rigorous: note that 4^2 = 6 (mod 10) and (6)(4) = 4 (mod 10).

@zero-starting373 2 года назад

We don't need the solution by general function in solving the such an easy recurrence. We could find a(n) in 30 seconds. Just rewrite the recurrence as a(n+1)/4^{n+1} = a(n)/4^n + (1/4)(- 1/4)^n. I would solve the recurrence in this way,just in 10 seconds, a (n+1) = 4 a(n) + (-1)^n a (n+1) + (1/5)(-1){^n+1} = 4 { a(n) + (1/5)(-1)^n}, yielding a(n) + (1/5)(-1)^n = 4^{n}[a(0) + (1/5)(-1)^{0}] ......

@rialtho_the_magnificent 2 года назад

how many squares? Not many!

@user-pn8vw8rr3m 2 года назад

see Michael - push like

@adandap 2 года назад

This is another good example where a summing factor - the discrete equivalent of the integrating factor in a first order DE - can be used to turn the equation into an exact difference. In this case the summing factor is 1/[product_(j=0 to n) 4] = 4^(-(n+1)). Then sum both sides from n=0 to N and the exact difference telescopes and you get a_(n+1) = 4^(n+1) ( a - sum(0 to n) (-1/4)^n) etc.

@studyforyou6794 2 года назад

Very well explained 🥰🥰

@bernieg5874 2 года назад

Michael -- please please please do some videos about how people INVENT these problems

@mowskii5791 2 года назад

Reupload?

@seroujghazarian6343 2 года назад

You can prove that 4^n-(-1)^n is a multiple of 5 because 4^n-(-1)^n=5(\sum_{k=0}^{n-1}{(4^{n-1-k})((-1)^k)})

@oida10000 2 года назад

Actually 1*(-1)=-1

@charleyhoward4594 2 года назад

Today's date of December 2 is indeed an anniversary of sorts; this is after all the 79th anniversary of the creation of the world's first nuclear chain reaction in Chicago under the leadership of Enrico Fermi. A tremendous milestone for scientific and technological progress.

@orenfivel6247 2 года назад

Can u PLS solve a Linear difference Eq. when the external input (non homogeneous part) is f_n=A*cos(θ*n)+B*sin(θ*n) [instead of (-1)^n] , using generating func.?

@orenfivel6247 2 года назад

and also consider a two step recursion with complex roots in homogeneous part, using generating func., like a_(n+2)+4*a_n=0 a_(n+2)+a_(n+1)+a_n=0 a_(n+2)-a_(n+1)+1.5*a_n=0

@user-qn4fs2og7k 10 месяцев назад

c(n) = a(n)/(-1)^n

@josephmartos 2 года назад

The geometric series sum formula dsnt work only if the absolute value of the ratio is less than 1? What i am missing pls help

@user-cr4fc3nj3i 2 года назад

Usually we ignore about this restriction when we are dealing with generating functions. You can also work some details out and you will see that whether x is between -1 and 1 or not doesn't really matter at all.

@bobbyhanson346 2 года назад

We can just restrict the domain of A(x) such that the series converges. We don't actually care about the series itself anyway, only its coefficients.

@josephmartos 2 года назад

@@bobbyhanson346 ..... So we just conviniently set a continous Interval for X where the series A(x) converges right?

@bobbyhanson346 2 года назад

@@josephmartos Yes, exactly.

@jfcrow1 2 года назад

what if a=2 get one perfect square which is max.

@franksaved3893 2 года назад

You have to maximize the numbers of terms to be squares, not to get a square with the maximum value.

@threstytorres4306 2 года назад

12MINS LATE, Because the Subscription Page had an Error, and the only thing that i can reach on this channel by searching un the search box

@nHans 2 года назад

Yup, annoying. Even I got the same error-the subscription page kept coming up blank. Tried F5; close-and-reopen browser; clear cache and cookies; using a different browser, and other common troubleshooting steps. On the RU-vid home page, there was a new notification saying something about changes to terms & conditions. Clicked it and scrolled till the end. Don't know if that had anything to do with it, but the problem seems resolved now.

@lynescopete5911 2 года назад

@physicsmaths6781 2 года назад

Hello sir i am your big fan

@udic01 2 года назад

12:27 the 4^n/5 disappears only because 1/5=5^(-1)=5(mod8) That explanation is missing at that time.

@bobbyhanson346 2 года назад

It disappears because 4^n = 0 (mod 8) when n >= 2.

@udic01 2 года назад

@@bobbyhanson346 only because there exists a reciprocal

@accountdefunct4193 2 года назад

anyone else get lost in the mods?

@martinschulz6832 2 года назад

Maybe, I missed something, but did he leave out the case, when a1 is a perfect square?? If yes I'd like to add that this only happens if a = k^2+k. So the maximal number of perfect squares is 1. Just in case somebody wonders...

@martinschulz6832 2 года назад

Ah I see. Thanks. Didn't watch the whole video...

@physicsmaths6781 2 года назад

I am from india