Divisibility Proof (2 of 2: Using algebraic expansion)

[@andrearredondo]

Recently, Eddie visited our school. On behalf of our school we would love to say thank you for your visit and time.

[@criped3251]

Yoo its andre

[@TheWasian]

Okay Andre A

[@conamore7887]

I've learned a lot from watching your free KZbin videos. I wonder why people have been brain washed into believing that valuable things are always not for free. I'm among the lucky ones in Uganda who have access to this quintessential mathematics teacher . Te doy las gracias por todos, Eddie woo .

[@davidplanet3919]

The last expression is classic binary addition. For n = 4 for example, 2^4 = (2^3 + 2^2 + 2^1 + 1) + 1. Or in binary 10000 = 1111 + 1.

[@lutfaralif5880]

My favourite..an absolute best at explaining and making things fun💜

[@omerayhan7751]

thank you for enlightening us professor

[@arimermelstein9167]

I have a small nitpick. I assume that you mean that if a^(n-1) is prime for all n, then a=2. Because, for example, if a=8, a^1-1 = 7 which is prime. This is because the second part is 1.

[@kentonjones5394]

In (ii), can n = 1 be another solution

[@SeeTv.]

Really cool coincidence how Flammable Maths and you made a video about it over the span of a few days.

[@aashsyed1277]

what video?????

[@SeeTv.]

@@aashsyed1277 the newest one I think

[@sundarasubramanian9058]

Sir could u say what r the equipments u r using for maths online class sir...

[@BillyShears230]

He is using an iPad Pro with an Apple Pencil, and the app he is writing on is notability,not sure if that answers your question

[@nova4452]

Love the signature on my arm thank you

[@criped3251]

Hello Josh

[@nova4452]

@@criped3251 Sup

[@zilvarro5766]

You should probably mention that n>1 in (ii)

[@jamiewalker329]

(x^n -1)/(x-1) is the sum to n terms of a geometric series first term 1 and common ratio x. Done.

[@santiagopabstmeschini5033]

But 3^1 -1 is not equal to 2^(some integer) - 1. So It doesn't have to be a = 2

[@Cyrathil]

My immediate thought was a = 8 (8^1 - 1 = 7 is prime, but a != 2) but similar. Maybe I'm missing something, but I don't really understand the hand-waiving of the right product never being equal to 1. If n = 1, the right-side is always equal to 1, so the left product can be basically anything.

[@erik_forsmoo]

For n=1 the equation becomes a^1 - 1 = (a - 1) * 1, or a - 1 = a - 1. So at that point we're just stating that if a-1 is prime, then a-1 is prime, which is redundant. It would probably be best to just exclude n=1 from the equation.

[@jamiewalker329]

Eddie doesn't do special cases.

[@jesusalycastrobernal7386]

It has to be a=2 for all primes that are of the form a^n-1, the case for a=3 is the exception but only works for n=1, despite the case for a=3, there is no other integer a distinct of 2 that satisfies a^n-1 is prime.

[@jesusalycastrobernal7386]

@@Cyrathil in that case, 8=2^3 so a=2 and n=3, is like saying that 32^1-1=31 is prime, but 32=2^5

[@aashsyed1277]

8:18 not really true what if the other part is not prime..............

[@lumist9671]

Really like your teaching and your class!! Hope I can get some insights from you in terms of my channel if possible, haha (:

[@TeamNolex]

Interesting...

[@Ashu-bv3ej]

What is your IQ sir

[@omtripathi5]

Hello sir I am 12grade student and jee aspirant

[@None-vb4vr]

When are you coming to our school

[@Serendipity-nh2uk]

I- I’m early........😌💅

[@orni2throrinque105]

First, but understood nothing

[@rekhabansal2182]

You are teaching wrong tricks to my brother he is changed and I don't like that 😡