a unique and amazing iterative proof of Fermat's little theorem.

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[aweebthatlovesmath]

it's really cool to use a continuous concept to prove discrete problems.

[noahtaul]

12:45 I don’t think that logic tracks. z^a-1=0 and z^b-1=0 and a<b doesn’t mean that a divides b; it just means that z^(gcd(a, b))-1=0.

A better way to prove this claim is that if f_m(z)=f_n(z), then using one-to-oneness of f, we know f_{m-n}(z)=z. Then since k:=m-n<p, there is some u, v with u*k-v*p=1 (and we can assume u, v>0). Then applying f_k exactly u times to z, we get z = f_{u*k}(z)=f_{1+p*v}(z)=f(z) (since z is in A, meaning that f_p(z)=z). This is a contradiction, since A excludes all elements of f(z)=z.

[MCLooyverse]

I could've done with an explanation of why |A| = a^p - a. As stated, it wasn't immediately obvious to me that all the roots of the right were also roots of the left. Of course, on further consideration, I see that any fixed point of `f` is also a fixed point of `f_p`, but I couldn't see that through the polynomial view.

[MarcoMate]

At 8:45, to justify the formula |A| = a^p - a is necessary to observe that if z is such that f(z) = z, it is also such that f_p(z) = z for every natural p. This is indeed straightforward, but still it might be useful to be underlined.

[Happy_Abe]

@17:50 why does A include all parts of the loop in it. z is in A, but why is f_n(z) in A for every n less than p?

[noahtaul]

As much as I like this proof, it’s almost exactly the same as the “necklace” proof that goes as follows:

Consider all a^p necklaces of length p that are made of a different types of beads. There are a of them that are just p of the same bead over and over; the others can be divided into groups of p based on being the same under rotation. How many unique such necklaces are there? Exactly (a^p-a)/p.

It’s the same proof, because the underlying idea in group theory is the same. You have the group Z/pZ acting on a set of size a^p with a fixed points; the orbit of any element divides p, so it’s either p or 1 (by the orbit-stabilizer theorem, for example), so every other element has orbit of size p. In this proof, the set is the elements satisfying x^{a^p}=x, and the group action is raising to the a’th power. In the necklace proof, it’s the necklaces and the action is rotating them. If you can come up with another set of size a^p with a group action by Z/pZ with only a fixed points, you can have your own proof!

I do think, though, it is a nice novel way of constructing such a set. I certainly haven’t seen it before, and it brought out the idea of orbit sizes in the necklace proof

[wreynolds]

Very neat adaptation of Frobenius arguments from characteristic p. The only fact about complex numbers that was used is that the field is algebraically closed; the same argument works if we define f to be a function on any given algebraically closed field (or indeed you could define the appropriate splitting field in advance of defining f). The reason \CC is used is just that it's the most well-known, thereby keeping the prerequisites lower.

[SpeedcoreDancecore]

Why do we have z^a - 1 = 0 and z^b - 1 = 0 and a < b => a | b ? It doesn't seem true in general to me

Edit: I see why it's true, excluding 1: if for every root z of X^a - 1, we have z^b = 1 and a < b, we can say that z^(b/a) is a root of X^a-1.

But all the roots of X^a-1 are exactly {1, ξ, ξ^2, ..., ξ^(a-1)}, so both z and z^(b/a) has to be one of these.

Call z = ξ^i, then z^(b/a) = ξ^(i b/a).

So ib/a has to be an integer, so a|ib, and if we can choose i such that gcd(i, a) = 1 we can conclude a|b.

[GreenMeansGOF]

8:50 Do we know that all the roots of the second polynomial are also roots of the first polynomial?

[gerdweissenborn]

My favourite proof is in one line: Consider the order v of a in the group (Z/pZ)*. Then {1, a, ..., a^(v-1)} is a subgroup, thus v divides p-1 by Lagrange.

[JosBergervoet]

Nice, but just using the fact that we have the field F(p) and that its multiplicative group has order p-1 and the order of each element has to divide that number is easier to remember!
(Of course the claim that we have a field requires proving Bézout's identity and the claim about the orders requires some group theory of cyclic subgroups and how their cosets must fill up the group, but that's crucial knowledge anyway so it's not ballast, I mean!)

[Alan-zftt]

My comment here is influenced by some comments I read below. The video is in my opinion:
* intended for students and not designed for learned or more learned colleagues
* takes learners through step by step construction of proof leaving omissions where learners can, if they so desire, fill in the blanks
* introduces features of complex numbers to solve integer number things
* shows how different tools from different ways of doing math can be combined (breaks linearity constraints in teaching of math in separate "boxes")
I could go on but I think that is enough for now. and why not use e to the i-theta ?

[GlenMacDonald]

Probably the easiest proof is from Group Theory: Each of the non-identity elements of the group U(p) is a generator of the entire group, otherwise each would generate a non-trivial proper subgroup of U(p), violating Lagrange's Theorem (since p is prime). So each element of {2, 3, ...p-1} has order (p-1). That is, ∀a∈{2, 3, ...p-1}, a^(p-1) ≡ 1 (mod p). That is, a^p ≡ a (mod p). But trivially, 0^p ≡ 0 (mod p) and 1^p ≡ 1 (mod p), and so ∀a∈{0, 1, 2, ...p-1}, a^p ≡ a (mod p). But since we're working modulo p, ∀a∈ℤ, a^p ≡ a (mod p).

[wojciechwisniewski]

I didn't check that, but A seems to form a group (along with some operator, maybe just composition?) and that "loops" are just subgroups (even normal?) of order p, so we just have divisibility of cardinalities of these sets and it's done. Am I correct?

[marc-andredesrosiers]

Interesting application of complex analysis.

[puneetbajaj]

12:52. Z^3 - 1=0 and Z^2 -1 =0 both have 1 as a root but 2 does not divide 3. So sir, i think your claim is false

[xizarrg]

I think this is the first video on the channel to have a proof by contrapositive.

[showri.a]

Another proof I like best is this...
N^p = = N + Mp now if p is prime and N prime to p, then N^(p-1) -1 is a multiple of p. Therefore proved

[christiannavarro]

do you have any series for basic topics like pre-calculus/calculus? thanks!

[xaxuser]

what you did in the video proves that either fermat's little theorem is true, or A is empty :p