Proof: Mersenne primes

[beyza]

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

[JunkieMonkey]

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!

[chessandmathguy]

Wow that was so elegant, thanks for posting.

[Xetaas]

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

[rgoodwinau]

Mesmerising - both in logic and use of technology.

[legendsplayground]

Elegant proof, love your content, Jesus bless.

[zanti]

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.

[matthewzarate]

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!

[m.matilda_]

Wow thanks so much for your content, Mr Woo

[nickholden]

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

[namduonghoang]

The MERSENNE PRIMES formula is part of the following formula
1. Let a, b, n be natural numbers with a>b
- is always a natural number. If is prime then a-b is prime. If a-b is composite, then 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, is always a natural number. If is prime then a+b is prime. If a+b is composite then 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.

[forthrightgambitia]

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

[markstavros]

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

[ngthanhhieu]

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

[dr.mohamedaitnouh]

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

[a]

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

[tadejsivic]

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

[carlosdanielquispec]

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!

[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.