An inductive proof of Fermat's little theorem.

We give a nice proof of Fermat's little theorem using induction.
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[@connorschwartz7518]

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

[@immersion6174]

There's an even simpler proof if we used Freshman's Dream as our Lemma to prove Fermat's Little Theorem ;)

[@littlefermat]

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

[@prithujsarkar2010]

lol nice one

[@Qermaq]

Aw he's just little

[@dylank6191]

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

[@egillandersson1780]

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

[@kenanwood6916]

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

[@egillandersson1780]

Ha ha !

The Little Theorem is much more useful, to be honest.

[@wisdomokoro8898]

Ha ha 😂 you shouldn't be 😜 suprised my friend

[@menohomo7716]

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.

[@GiornoYoshikage]

Never thought about such a great proof! Well done!

[@mohamedfarouk9654]

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

[@seanhunter111]

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

[@egillandersson1780]

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.

[@GiornoYoshikage]

Right

[@byronwatkins2565]

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.

[@manucitomx]

Thank you, professor.

[@blankspace1482]

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

[@ProfessorDBehrman]

Nice proof. Very easy to follow.

[@thundercraft0496]

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

[@super_matematika]

Good!

[@darreljones8645]

One noteworthy consequence of the lemma is that if p is prime, every number in the pth row of Pascal's triangle (except for the 1's at the ends) is a multiple of p.

[@binaryagenda]

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

[@wafizariar8555]

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

[@quickspace861]

Beautiful

[@FlexThoseMuscles]

great proof

[@Pika250]

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.

[@adityaekbote8498]

Nice!

[@goodplacetostop2973]

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

[@tiagomrns]

classic mp indeed

[@iwersonsch5131]

If a is in Z, then a is congruent to b mod p, with b in N.

[@eamonreidy9534]

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

[@A.Andreas]

Hey, thanks for another great video. Just wanted to confirm the final statement you made at 7:26. You said (b+1)^p ≣ b (mod p), but wrote (b+1)^p ≣ b+1 (mod p). Which is correct?

[@daniillitvinenko4348]

I think the thing written (b + 1)^p = b + 1 (mod p) is the desired result assuming that some c = b + 1. c^p = c (mod p)

[@basicallymath405]

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

[@perveilov]

I'm waiting for the big theorem later on

[@noumanegaou3227]

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)

[@CallMeIshmael999]

This is an interesting proof and I think it highlights the connection between Fermat's Little Theorem and the Freshman's Dream. But I don't think you actually needed the induction. You can write n as (n-1) + 1. Then, just like in your proof, ((n-1) + 1)^p = sum(p choose k)(n-1)^k which is congruent to (n-1) + 1 = n. It's essentially identical to your proof but it doesn't rely on induction.

[@tobybarnett5455]

Why can we say that (n-1)^k is congruent to (n-1)modP? This is something that the inductive hypothesis only assumes; we see it to be true for n=1 and so then induction allows us to prove it for consecutive natural numbers as we prove the (n+1)th case also holds, given this initial assumption. Without induction though we can’t have this assumption?

[@CallMeIshmael999]

@@tobybarnett5455 Oh right, sorry. I guess I had a brainfart and forgot the power of k on it.

[@Grizzly01]

4:25 'Fermazzletheorem' 🤣

[@FrankAnzalone]

My very first sentence would have been that's a good place to stop

[@raesavage9793]

mod is just pow upside down

[@daniillitvinenko4348]

good observation

[@KaptenKetchup]

I have a proof for this too, but it's too large to fit in a KZbin comment.

[@MrWarlls]

For the demonstration of the lemma, you have (p n) = p!/((p-n)!n!) too (easier to write) 😉

[@cinedeautor6642]

If the other day we watchched that: ----- In primitive Pythagorean triples, a or b must be a multiple of 3. Why? Since all three are squared, by Fermat's little theorem, a ^ 2 + b ^ 2 = C ^ 2 ... Is. 1 +1 = 1 Mod 3 ... Absurd. Unless a or b are multiples of 3. 1+ 0 = 1 ... So a or b has to be multiples of 3 ... No problem. --- Today say: Lemma 2 of primitive pitahogrian number...There no a¨2 + b 2 = 0 mod 7....Why.....I give a clue...See the group class of 7 Z/7....And, in this group, search the subgroup { 1, 2, 4}....