Proof: Mersenne primes

[@JunkieMonkey321]

im done with math in school, but i still watch your videos for entertainment. your enthusiasm is so infectious. thanks for making the world a better place, especially for those lucky enough to have you teach them in person!

[@beyza2004]

This is so genius but logical and easy at the same time... wow that's just why I love maths

[@ashwinparthiban4561]

One of your best works

[@rgoodwinau]

Mesmerising - both in logic and use of technology.

[@chessandmathguy]

Wow that was so elegant, thanks for posting.

[@legendsplayground7017]

Elegant proof, love your content, Jesus bless.

[@nickholden585]

That was awesome, I was never good at math in school but this was a pretty straight forward logical deduction.

[@photographymaniac2529]

You just nailed it buddy 🤩

[@Xetaas]

Thanks for the video! I find these super interesting to watch

[@zanti4132]

I think the easiest way to prove this is to think in base 2: 2ⁿ - 1 = 111...111 - there will be exactly n 1s in the base 2 representation of the number. So if n is composite, i.e. n = xy for integers x and y greater than 1, then 2ⁿ - 1 must have 2ˣ - 1, i.e. the number with x 1s in its base 2 representation, as a factor.

[@ARKGAMING]

Ik it might be trivial but if you let n=ab with a,b≠1 you exclude the case for n=1 which is also part of your contra positive, which I can only assume means you'd need to address it separately later.

[@adrianpilot159]

1 isnt a prime number therefore you can ignore the case where n=1

[@ARKGAMING]

@@adrianpilot159 but we're using the contra positive where n is *not* prime

[@adrianpilot159]

@@ARKGAMING my bad, u would need to check the case n=1 later

[@chessandmathguy]

True, 1 is non-prime so has to be verified separately.

[@m.matilda_3415]

Wow thanks so much for your content, Mr Woo

[@matthewzarate8851]

This question came up on my final exam last night (I wasn't able to solve it)--now this video came up on my feed. Funny how that works!

[@softvibes1602]

Thank You.

[@markstavros7505]

Awesome video! Could you do a video on why the Lucas-Lehmer test is correct also?

[@namduonghoang6301]

The MERSENNE PRIMES formula is part of the following formula 1. Let a, b, n be natural numbers with a>b - {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is always a natural number. If {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is prime then a-b is prime. If a-b is composite, then {[a^(a-b)]-[b^(a-b)]}/[(a-b)^2] is composite. - If [(a^n)-(b^n)]/(a-b) is prime, then n is prime. If n is composite then [(a^n)-(b^n)]/(a-b) is composite. 2. Let a, b, n be natural numbers where n is odd - With a+b being odd, {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is always a natural number. If {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is prime then a+b is prime. If a+b is composite then {[a^(a+b)]+[b^(a+b)]}/[(a+b)^2] is composite. - If [(a^n)+(b^n)]/(a+b) is prime, then n is prime. If n is composite, then [(a^n)+(b^n)]/(a+b) is composite.

[@dr.mohamedaitnouh4501]

How do you make this video please with tablet and you explaining. Very nice! can you prove that there no perfect odd number??

[@a357]

"This is a really cute sort of question" in other words this is too easy you should challenge yourself

[@ngthanhhieu88]

Thank you so much, but can you share me what kind of camera do you use?

[@64faapts]

Oh Boy.... geography is hard! 😂

[@tadejsivic534]

Forgot at the end? QED! There we go, now it's done.

[@forthrightgambitia1032]

Amazing. Things like this make you remember the cube identity for ever.

[@ilirllukaci5345]

Thanks.

[@mukundabharadwaj852]

Awesome!!!!!!

[@carlosdanielquispec3759]

Si n es entero positivo y no es primo entonces puede ser la unidad o compuesto. Faltaría analizar el caso n=1. ¡Muy buen video!

[@pablogonzalez3705]

Ese caso es trivial.

[@lipeshff]

Wow! Amazing proof. ❤ from 🌏 😊

[@MathZoneKH]

love your content prof

[@gggg-cu8gh]

This was in 2022 HSC!

[@TusharGupta1991]

Which app is being used?

[@patrickkaunda1406]

Please can you solve this for me logy base 3=x. Whereby xy= 30

[@Qermaq]

13:11 you said a, b belong to Z not N. Thus there could be negative powers. If there are negative odd powers, then we are adding a bunch of stuff but some of that stuff is negative. So it's not really proven here that the right expression cannot be 1.

[@avivwachman3207]

They still aren’t negative. Example: 5^(-2)= 1/25

[@forpublicstuff728]

Cool! Thank you=) I was searching for the explanation of the exercise 25 in chapter 5 in the Book Of Proof (3rd edition) by Richard Hammack - just in case someone is doing the same =)

[@philippenachtergal6077]

Hum. So n not prime -> 2^n -1 not prime If n is an even number bigger than 2 we have n = 2p -> 2^(2p) - 1 = (2^p-1)*(2^p+1) , since n > 2 --> p > 1 and 2^p - 1 > 1. So we have a decomposition into prime factors and 2^n-1 is not prime If we try for n multiple of 3 bigger than 3 itself, we have n = 3p -> 2^(3p) - 1 = (2^p - 1 ) * ( 2^(2p) + 2^p + 1) which means as above that 2^n-1 with n a multiple of 3 > 3 is not prime Now, I "just" need to proof that 2^(kp) - 1^(kp) can be factored as (2^p - 1) ( some positive integer) for any number k or any prime number k. 2^(kp)-1^(kp) = ( 2^p - 1 ) ( 2^((k-1)p) + 2^((k-2)p) + ... + 2^p + 1 ) with both side > 1 as long as p > 1 If you distribute the product above, you'll see that you obtain 2^(kp) - 1^(kp) This means that as long as n is not prime, it can be expressed as n=k*p with k>1 and p > 1, and 2^n-1 is not prime Eheh, seems quite similar to his proof :)

[@تيتانالصغيره-ك4ب]

My teacher, why did the comments in Arabic distinguish his position? I see you from the Arab world, Egypt

[@qwertyTRiG]

Nice argument. I don't think number theory was covered much when I was in school.

[@TheAcer4666]

Both the statement of the question and the proof given are a little sloppy. First up, it doesn't state what set n lives in (e.g. n = log_2(12) satisfies 2^n - 1 prime, but n is not prime). It should say that n is in N (or Z). Then in the proof, n=ab is written with a,b in Z, so I guess the assumption is that n is in Z. In that case, we should really say a and b are not 1 OR -1. [7 = (-7)*(-1), for example] But with a, b possibly negative, you can't then use the factorisation rule given below. So a better thing to do would be to assume n is positive, pick a, b from N, then the proof is alright. Deal with the case of n negative separately. C-

[@Mathskylive]

This problem is nice about an arithmetic property: If 2^n-1 is prime then n is prime.

[@chanchalkumar0001]

Sir love from india

[@navodalwickramawardhana9040]

👍👍

[@تيتانالصغيره-ك4ب]

I do not understand you because I am an Arab

[@beyza2004]

Luckily, math is an universal language :)

[@aashsyed1277]

@@beyza2004 yes

[@Grizzly01]

@arqam station But manages to post a comment in perfectly cromulent English?

[@andrewboldi47]

@@Grizzly01 Probably using Google translate