An inductive proof of Fermat's little theorem.

We give a nice proof of Fermat's little theorem using induction.

If you are going to use an ad-blocker, considering using brave and tipping me BAT!

Books I like:

Abstract Algebra:

Differential Forms:

Number Theory:

Analysis:

Calculus:

My Filming Equipment:

[connorschwartz]

Simplest proof of Fermat's Little Theorem I've ever seen, very nice!

[littlefermat]

Hi Michael,
I appreciate the fact you have shown many nice proofs of my theorem!❤️

[kenanwood]

When I first read the title, I thought it said "An inductive proof of Fermat's Last Theorem." And then I realized that was not the case. I'm disappointed now.



Jk great video as always! Keep up the good work

[dylank]

Good one, this is exactly how I proved it back in my Linear Algebra II class :)

[egillandersson]

I never read this proof, which is quite elegant and the easiest to retaining and to teach I know? ... now !
Thank you !

[seanhunter]

That's a really nice proof and well explained. THanks very much.

[GiornoYoshikage]

Never thought about such a great proof! Well done!

[menohomo]

this is one of my favorite proof, when i was around 18yo i saw the following problem featured in the anime Madoka Magica: (n+1)^p - n^p - 1 = 0 (mod p). I saw the binomial theorem right away but how do you use the fact that p is prime? I wrote it off as "probably some higher level of math required". It hit me 2 weeks later while doing something completely unrelated: you can factor out a p in the "p choose k" part and write it p * (p-1)! / k!(p-k)!

That proves that p divide into every number on the p'th row of Pascal Triangle. Turns out it's actually a characterization of the primes, because the other way around is also true: if n divide every number on the n'th row of Pascal Triangle, then n is prime!

I thought I was done with the problem but much later I realized... the (n+1)^p was congruent to n^p+1 (mod p) which itself is congruent to (n-1)^p +2 and so on... so you can go all the way down to (n-n)^p+n and (n+1)^p is congruent to n+1 (mod p).

According to wikipedia the proof is from Gauss book, and Gauss himself credit Euler.

[byronwatkins]

You did skip over something subtle in discussing your sum(mod p). p choose n is zero (mod p) which means that multiplying by integers necessarily yields divisibility by p; however, if p choose n is nonzero (mod p), then the product [(p choose n) b^n] (mod p) would cycle through the integers in [0, p=1]. Having a congruent to b(mod p) ordinarily does NOT imply that ax is congruent to b(mod p); this is necessary ONLY when b=0.

[egillandersson]

Just a little precision : it is not enough to say that n! does not divide p, but that none of the factors of n! divides p. The proof is the same.

[mohamedfarouk]

I am not the only one who thought for a second that it's Fermat LAST theorem

[blankspace]

Hey Michael, I'm a former student of yours from way back... could you make a video on the Mandlebrot set? I find it fascinating

[goodplacetostop]

4:06 Classic mp
7:33 Good Place To Stop

[thundercraft]

I was able to come up with this proof when I was trying to prove that n^p - n is always divisible by a prime p

[binaryagenda]

Can you please link to your other videos proving Fermat's little theorem?

[basicallymath]

There’s also a good group theoretic proof using LaGrange’s Theorem. Nice proof!

[Pika]

change in sign: (-m)^p = ((-1)^p)(m^p) which reduces to (-1)(m) = -m mod p. (-1)^2 = 1 which is the same as -1 mod 2, and if p is odd we obtain (-1)^p = -1.

[wafizariar]

What about some Galois theory videos? Even an introduction would be awesome!

[eamonreidy]

I know it isn't what your channel does but some advanced undergrad or postgrad module series would be amazing

[noumanegaou]

Let A negative integer and p prime
We suppose p = 2
We have A^2 and A are both even or both odd
Then A^2 = A mod(2)
We suppose p <> 2 Then p odd
A^p = ((-1)|A|)^p
= (-1)^p |A|^p
= - |A|^p = - |A| (modp)=A(modp)