a prime circle

[@게르마늄-w5b]

In the part 8:41 there's a fallacy. The factorization by x+1 only holds when n+1 is odd(n is even), whereas if n is odd, there's no particular prime that divides all 2^(n+1)+1. (It is clear by the example; Fermat Primes, form of 2^2^n+1) However, we can show that n should be even. If n is odd, 2^(n+1)+1 is congruent with 2 on modulo 3. A perfect square, (p^m)^2 should be congurent with 0 or 1 on modulo 3. Hence it contradicts, so n should be even.

[@timurpryadilin8830]

i was just about to comment that !

[@kemalkayraergin5655]

thank you for explaining why n is even

[@wesleydeng71]

4:07 He should've moved p^2m to the right side which would avoid the error and lead to a quicker solution.

[@rohitg1529]

Mersenne primes are of the form 2^n - 1 not 2^n + 1 x^n - 1 = (x-1)*p(x) for some polynomial p(x), but if x=2 then (x-1) = 1 then the mersenne number is not necessarily prime What he showed in the video is accurate (but yes, only for odd n)

[@게르마늄-w5b]

@@rohitg1529 yep thanks for the point! I was meant to say Fermat Primes, by the way. But since there exists some primes, there are no common factor that divides them all

[@manucitomx]

Holy Diophantus, professor, what a way to hide a 3-4-5 triangle! Thank you!

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@SayHelllllo]

we are getting closer and closer to the Fermat's Last Theorem

[@wisdomokoro8898]

😂 The journey was Single for Wiles Andrew 💔🤫

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@markvp71]

Yeah, like having walked the first mile of a thousand miles.

[@WhattheHectogon]

At 11:25, a simpler argument would be to ask, which two powers of 2 differ by 2? Why 2^2 and 2^1, course :)

[@replicaacliper]

In fact that trick can be used as early as 7:30 and it would shorten the video by a few minutes.

[@davidblauyoutube]

@@replicaacliper I came here to make this exact comment. Well done, mathematicians of KZbin!

[@Qermaq]

I noticed right off that this is a Pythagorean triple with a and b both having one prime factor, if multiple times. As all PTs except one have at least one multi-prime composite leg, it's gotta be that triple.

[@GreenMeansGOF]

8:20 You are assuming that n is even(also the plus/minus 1 should just be +1 but whatever). I wonder if we can justify that assumption.

[@brankojangelovski3105]

You didnt show that n+1 must be odd in order to do that factorization at 8:41

[@victorclaytonbarnett2959]

Idea for a livestream: solving contest problems that you're seeing for the first time. I would love to see the process in action!

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@fracaralho]

The disdain for the idea that 0 is a natural number is so strong that he prefers to call the set of odd numbers 2Z + 1 instead of 2N + 1.

[@PubicGore]

Actually he doesn't have any disdain for the idea. In fact, he cares so little that he writes N often because he's just lazy, but has said many times that he doesn't care whether or not 0 is an element of N.

[@fracaralho]

@@PubicGore I've seen him say as much before, but I don't buy it.

[@goodplacetostop2973]

4:38 Good Place To Be 13:37 Good Place To Stop

[@Mystery_Biscuits]

Leet

[@XavierMamet]

kzbin.info/www/bejne/fZirZXSPm7SHepY is a good place to be

[@goodplacetostop2973]

@@Mystery_Biscuits You’re a man of culture as well

[@jimmykitty]

Perhaps one of the hardest geometry problems you're doing so far... 😍😍😍

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@jimmykitty]

@@quickyummy8120 Thanking you, Boss 🙏 🙌 ♥

[@TedHopp]

It's not really a geometry problem. It's a number theory problem dressed up as geometry.

[@iwonder2218]

This problem is very trivial due to the wording since it implies upfront that there is only one solution. 3-4-5 triangle comes to mind immediately where 4 is 2^2. If the problem said find all solutions then what you did was a solid method.

Depends on how you interpret the wording. I always assume that you still have to prove that there's exactly one solution.

[@Bazzzzz93]

8:10 wrong. This factorization works only for odd n+1

[@vanambalong3325]

I agree

[@emanuelvillanueva9240]

n+1 is odd it is 3

[@SlipperyTeeth]

At 7:20 you could've cut out a step by factoring the difference of squares right then and there. You'd get that p^m is trapped between two powers of 2 and thus p^m=3.

[@federicovolpe3389]

8:21 the factorization is true only if n is even (and thus n+1 is odd). You could've factored p^2m-1 into (p^m+1)(p^m-1), after that since they must be both powers of 2 and they differ by 2, they can only be 2 and 4, so p^m=3 and so p=3 m=1. Then 2^(m+1)=3^2-1, so m=2. Subbing back into the original equation you get r=5.

[@marsgal42]

I found myself thinking of the usual hint that such problems have very few solutions, often just one. We have one solution by inspection. Is it unique?

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@PunmasterSTP]

I got schooled watching the video, but even more so by reading the comments. I'm really glad that this much free educational material exists on KZbin. Thank you to everyone here!

[@thoughtfuljanitor6627]

I'm not sure I understand what's being said around 8:41 We factor 1 + 2^(n+1) into 3*N for some integer N Take the case n = 3 : then 2^(n+1) = 2^4 = 16 However 16+1 cannot be factored as a multiple of 3 So i'm not sure I understand why we can factor it that way

[@zadsar3406]

Right. It only works for odd exponents, so for even n. You could do: 2^(n + 1) = p^(2m) - 1 = (p^m - 1)(p^m + 1) Now, it is impossible that both p^m - 1 and p^m + 1 are divisible by 4. Therefore, p^m - 1 = 2 This immediately gives you p = 3 and m = 1.

[@guycomments]

Any Swans fans here just for the thumbnail?

[@quickyummy8120]

I found another amazing Olympiad problem here. kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@ZipplyZane]

Toward the end you started skipping a bit. I think you should have shown how you got x=1 and y=2, as well as actually plugging things in to get n=2. For the former, you can note that 2^(y-x) - 1 must be odd, and thus is equal to 1. So 2^x=2, and thus x=1. Plug that back in, and you get 2^(y-1) - 1 = 1, so 2^(y-1) = 2. Thus y-1=1 and y=2. Plugging in for the other you get that 3^2 - 1 = 2^(n+1) => 8 = 2^(n+1) => n+1=3 => n=2.

[@Pablo360able]

A strange corollary to this result - really, to the fact that there is one singular solution - is that no power of 2 besides 4 can be a leg of a Pythagorean triple where the other leg is a prime number.

[@caspermadlener4191]

WRONG!!! YOU MADE THE MISTAKE OF FACTORISING 2ⁿ+1!!! IT ONLY WORKS WHEN n IS ODD!!!

[@tahirimathscienceonlinetea4273]

Very nice Michael you're doing a great job.I love this kind of things which you can evaluate the radius of circle with insufficient information .keep going man 👍👍

[@_simobr]

I immediately thought about the Pythagorean triple

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@markvp71]

I started with the fact that if a^2+b^2=c^2, with a, b and c integers, then there are integers x, y, and z such that a = x^2-y^2, b = 2xy and c = x^2 + y^2. It is easy to see that z = 1 in this problem. Since a or b has to be a power of 2, it can only be b, so x = 2^s and y = 2^t. From there on it is easy.

[@AriosJentu]

And that's a good place to like

[@quickyummy8120]

I found another amazing Olympiad problem here. kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@Jared7873]

The radius is also prime. 😁

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@DouglasZwick]

13:22 "That Finnishes the solution to this problem"

[@nedaarsenijevic9479]

Молим Вас прескочите тривијалне ствари у доказима а обратите пажњу на на идеју задатка. Поздрав Милан Ребић, Србија.

[@eemelinissila1358]

Finland👍

[@doctorb9264]

The old difference of squares trick !

[@szkoclaw]

Absolutely beautiful problem, where evennness and primeness of the number lets you discard all of the bullshit and just have fun.

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@nareshmehndiratta]

YOU MUST SIMPLIFY MATHS OF ALL PHYSICS

[@quantumloc8159]

у отца пустая квартира, это значит не тждома

[@laurynastruskauskas6586]

The intro is sick

[@quickyummy8120]

I found another amazing Olympiad problem here. kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@shifta7726]

torille

[@tychophotiou6962]

I thought to myself "it must be a pythagorean triplet", so could it be 3,4,5 5,12,13 7,24,25 8,15,17.... I checked whether 3,4,5 satisfied the conditions and it did.... Solved in 15 seconds!!!

[@threstytorres4306]

Question: For positive integer n, let S(n) denote the sum of the digits of n. E.g S(24) = 2 + 4 = 6 S(92) = 9 + 2 = 11 S(200) = 2 + 0 + 0 = 2 Find the smallest positive integer satisfying, S(n) = S(n + 864) = 20 Source: Aime 2015 I Problems/Problem 8

[@rohitg1529]

The results is actually a trivial consequence of the Catalan conjecture (now a theorem) that the only solution of x^a - y^b = 1 for x,y,a,b being positive integers is 3^2 - 2^3 = 1 Proving the Catalan conjecture is another matter though :P

[@paulkohl9267]

A satisfying ending.

[@peterburbery2341]

3-4-5 triangle. This is the first Math Olympiad question I have solved myself!

[@slawaxas]

Wait this isnt a review of highly acclaimed album "Soundtracks for the Blind" by Swans

[@rogerkearns8094]

To me it was obviously the 3,4,5 triangle from the start, but perhaps that's not the point.

[@TJStellmach]

You still need to determine that that solution is unique.

[@rogerkearns8094]

@@TJStellmach Fair enough, thanks. :)

[@pierremarcotte6299]

2:38 - Why is r odd?

[@GabeKorgood]

If 3,4,5 is the only Pythagorean triple where both legs are powers of primes and the hypotenuse is odd, that should be enough to prove the problem without all the number theory

[@JoGurk]

Yeah and how do you prove that?!

[@GabeKorgood]

@@JoGurk some of the many known principles of Pythagorean triples are as follows: 1. Exactly one of a, b is divisible by 2 (is even), but never c. 2. Exactly one of a, b is divisible by 3, but never c. 3. Exactly one of a, b is divisible by 4, but never c (because c is never even). 4. Exactly one of a, b, c is divisible by 5. The 3, 4, 5 triple is the only set of numbers that satisfies these principles and the requirement that both legs be powers of primes.

[@JoGurk]

@@GabeKorgood alright, yeah. Sorry :D

[@quantumloc8159]

на тебе р

[@tomkerruish2982]

The requirement that r be odd is superfluous. If r were even, then both p^m and q^n would need to be even, which would require p=q=2, which is impossible.

[@n8cantor]

p and q could both be odd, but there are no solutions for that case either.

[@alainbarnier1995]

Extraordinaire !!

[@quickyummy8120]

I found another amazing Olympiad problem here kzbin.info/www/bejne/qJ-6e6mVit2CgLs

[@minwithoutintroduction]

واصل.قناة رائعة.

[@kevin326520]

I try to do this myself and when I reach the point at 7:20 I just make use of the Catalan conjecture since I dont know what to do even I know the answer lol

[@ΓιώργοςΚοτσάλης-σ1η]

Second