an example right from my number theory class.

[@DavidSavinainen]

I love the graphical notes that appear through the video! Very well done!

[@goodplacetostop2973]

0:09 Is 1 prime now? 🤔 (5:10 Oh!) 0:42 New effect on board? 12:13 Don’t forget to stay hydrated and to take care of yourself, don’t work too hard! Have a good day ✌️

[@Frank9412co]

1 is prime if we're working with integers. In fact, that is the only prime in Z (the usual primes have 3 divisors, themselves, their additive opposite and 1)

[@Frank9412co]

4*, forgot to put -1

[@f5673-t1h]

@@Frank9412co that is completely wrong: en.wikipedia.org/wiki/Prime_element By definition, primality excludes 0 and invertible elements (in the case of Z, that's 1 and -1). Primaliy does NOT mean having only 2 divisors; that is not what primality is about, and we don't care about that at all. The definition of prime is "an element (number) that is not 0 and not invertible, such that p dividing xy means p divides x or p divides y". The primes in Z are the usual primes (2, 3, 5, 7, 11, etc.) and their negatives (-2, -3, -5, -7, -11, etc.).

[@charlottedarroch]

@@f5673-t1h Another way of defining primality is through the use of an equivalence relation using the units. We let x ~ y iff x = uy for some unit u. Then a non-zero, non-unit p is prime if it is only divisible by units and things in its unit equivalence class. In the integers the units are {1,-1}, so the equivalence class of any non-zero n is just {n,-n}, so if p is prime, then so is -p. In the Gaussian integers, the units are {1,-1,i,-i}, so for non-zero a+bi with a,b non-zero the equivalence class is {a+bi,-a-bi,b-ai,-b+ai}, so for primes we have 4 different unit equivalent copies of that prime. When it comes to prime factorisation, one can obtain unique factorisation (when it's possible, which isn't always) by picking one representative of each equivalence class to be 'the' prime. In the integers, this amounts to always choosing the positive member of the equivalence class. In the Gaussian integers there are a few useful choices for the canonical representatives. This definition of prime is equivalent to the one given above, though the one given above is often much more useful to apply.

[@WillCummingsvideos]

5:10 Mike is the boss hahahah

[@littlefermat]

I was going to comment about Dirichlet's theorem when the graphical note appeared! (Michael: Everything is going according to my predictions😂)

[@jplikesmaths]

0:42 new effect is appealing

[@drliorsilberman]

That's a terrible proof: you didn't reveal how you came up with the polynomial x^4-x^2+1. Here's a much better version of the argument: To start with, if x was an element of order 12 mod p, then by Lagrange's Theorem the order p-1 of the multiplicative group mod p would be divisible by 12, so p would be 1 mod 12. So the goal is to force x to have order 12 mod p. It turns that p | x^4-x^2+1 exactly encodes this fact, which is obscured by your presentation of the argument. To discover this we begin with the initial attempt of making p a prime divisor of x^12-1. Would that work? Not -- that would only make the order of x mod p *divisible* by 12, not equal to 12 -- and we need a way to avoid the divisors of 12 other than 12. For that let's consider the expression (x^12-1)/(x^6-1) = x^6+1. If p divides this number then x still has order dividing 12 mod p (because p | x^6+1 | x^12-1) but x does not have order dividing 6 because p cannot not also divide x^6-1: we can have (x^6-1,x^6+1) = (x^6-1,2) = 1 as long as we ensure x is even (by making x = 2p_1\cdots p_n instead of as in the video). But even then it would still be possible that x has order 4. To find the polynomial encoding this fact, note that it's equivalent to x having order dividing 4 but not dividing 2, i.e. the polynomial (x^4-1)/(x^2-1) = x^2+1. Thus for our proof we want to use the polynomial (x^6+1)/(x^2+1) = x^4-x^2+1. To conclude: let x = 6p_1\cdots p_n. If p divides x^4-x^2+1 then p isn't one of the primes p_1,...,p_n. First, p divides x^12-1 so x has order mod p dividing 12; second p does not divide x^6-1 so the order of x doesn't divide 6, and finally p does not divide x^2+1 either (because (x^4-x^2+1,x^2+1) = (3,x^2+1)=0 by the choice of x) so the order mod p of x isn't 4. So x is indeed an element of order 12 and p is a new prime congruence to 1 mod 12. This avoids the need to invoke quadratic reciprocity, and also shows how to make the proof work for any congruence condition p \equiv 1 (n) (by just constructing the nth cyclotomic polynomial, which is what we did above).

[@glennberry4829]

I got stuck a bit on the statement we could be sure we had a p>3 dividing N. N is clearly odd so 2 is not a prime factor, much less the only. But 3? Each p_sub_i from the finite list is congruent to 1 mod 3, so x is congruent to 1 mod 3, so N is also congruent to 1 mod 3, so 3 is not a prime factor of N.

[@angelowentzler9961]

There is always a p | N. At worst N itself is prime but then p = N will work. Because N = (multiple of the product of the list of primes) + 1, N is relative prime to all primes in the list and so p cannot be on the list and must be new. The +1 is key.

[@RexxSchneider]

More generally, all squares are congruent to 0 or 1 mod 3; and so the fourth power of the same integer is congruent to the same value, 0 or 1. That means that X^4 - X^2 is always congruent to 0 mod 3, whatever the value of X. So (X^4 - X^2 + 1) will always be congruent to 1 mod 3, and hence 3 cannot divide N. It's worth noting that X^4 - X^2 ≡ 0 or 2 mod 5, so 5 can't divide N either. Similarly for 7 and 11, but 13 can divide it for X ≡ {2, 6, 7, 11}. (X^4 - X^2 + 1) is a useful polynomial for eliminating divisibility by small primes (which are sometimes the "awkward" ones).

[@wesleydeng71]

x^4-x^2=x^2(x+1)(x-1)==0(mod 3) (product of 3 consecutive integers is a multiple of 3). So, 3 can't divide N.

[@superhightemperature2941]

You are the same as me. Originally, I also got stuck at 4:53 which says p>3 and at 10:22 which says p=p_sub_i. After that, I figure it out. We have to carefully read again the assumption. The assumption said that there are only finite numbers of prime of the form 12k+1 and we suppose that there are only n such prime number and so we set them as p1, p2,…, pn. Then we set x=p1p2p3…pn. When we multiply all of these, the product will also be the form of 12k+1. For example, p1p2=(12a+1)(12b+1)=144ab+12a+12b+1=12k+1. And so, when we multiply all the n brackets, the product will also be the form of 12k+1. And so x=12k+1. Then, N=x^4-x^2+1. But, x^4 is of the form 12k+1. And x^2 is also of the form 12k+1. And so x^4-x^2 is of the form 12k. And so, N=x^4-x^2+1=12k+1. Because 12 is a multiple of 2 and 3 but N=12k+1 and so N is not a multiple of 2 or 3. And so, p>3. But after a series of proof, we find that p=12k+1 and so at 10:22 we said that p=pi for some i because from the assumption there are only n primes of the form 12k+1 and we have already collect all of them and set them as p1, p2, …, pn and so p=pi. And as a result, p divides N and x at the same time which will lead to a contradiction because N is a multiple of x and then +1.

[@noumanegaou3227]

We can use cyclotomic polynomial

[@eric3813]

Such a great Proof! Thank you so much for this awesome Video!

[@Mephisto707]

Is there any known family of primes of which there are only finitely many of them? Besides even primes, of course.

[@demenion3521]

you can come up with a lot of silly examples like "all primes smaller than N" for some N. but you could also consider the set of pandigital primes in any given base which is finite if you don't allow for repeating digits

[@Tehom1]

Fermat primes are a good non-silly example. 65537 is probably the largest Fermat prime, though it's unproven at this time.

[@felaxwindwalker206]

Found it interesting that you called 1 a prime at the beginning of this.

[@srikanthtupurani6316]

It happens sometimes. He is an amazing communicator of math. He explains things so clearly.

[@felaxwindwalker206]

@@srikanthtupurani6316 Agreed.

[@damisolamakinde7541]

5:10

[@felaxwindwalker206]

@@damisolamakinde7541 Right, that comment wasn't in there when I posted this.

[@mathflipped]

I love these kinds of problems!

[@MichaelPennMath]

Me too!

[@threstytorres4306]

I THINK 12k+1 is a composite when k is congruent to 2(mod 5) like 2, 7, 12, 17...

[@demenion3521]

that is at least one family of composite numbers of this form since 12=2 (mod 5) and hence 12*k+1=2*2+1=0 (mod 5) if k=2 (mod 5). another family would be k=4 (mod 7) since 12*k+1=5*4+1=0 (mod 7) and you can continue this game for any prime and try to find more composites

[@charlottedarroch]

Sure! You can prove this quite simply. If n = 12k+1 and k = 5j+2, then n = 12(5j+2)+1 = 60j+25 which is divisible by 5. And a similar relation exists for all primes p >= 5. In particular, if k is congruent to -12^(-1) mod p (which exists for p >= 5 since then gcd(12,p) = 1), then 12k+1 is divisible by p.

[@tinaryan4023]

I haven't tested it fully yet but multiples of 5 results in this form divided by 5 .eventually reduce to prime

[@synaestheziac]

Quick suggestion: can there be a little sound effect that plays when one of the graphics appears on the screen? I’m not always looking at my phone when watching videos and I don’t want to miss anything. Thanks!

[@saberasafa3927]

We say that a positive integer n is good if any integer that can be written as sum of squares of n integers with each of them is divisible by n can also be written as sum of squares of n integers with none of them is divisible by n. Find all good integers.plz solve it

[@xc0xupx]

There is an alternative method to prove the same argument by using Fermat's little theorem and multiplicative order. x^6+1=(x^2+1)(x^4-x^2+1), so we have x^6=-1 (mod p) and x^12=1 (mod p). Let d be the order of x modulo p: that is, d is the smallest positive integer such that x^d=1 (mod p). We have d ∤ 6 and d | 12, which implies d is either 4 or 12. (Case 1) d=4. p | x^4-1=(x^2-1)(x^2+1). We have p ∤ x^2-1 by the definition of order, so p | x^2+1 and x^2=-1 (mod p). Thus p | x^4-x^2+1 = (-1)^2-(-1)+1 = 3 → p=3. But that's not possible, because the order of any number modulo p should be at most p-1 by FLT. (Case 2) d=12. With d = 12 and FLT x^(p-1)=1 (mod p), we have 12 | p-1, which means p = 1 (mod 12).

[@haziqthebiohazard3661]

Well 1 used to be a prime :(

[@elaadt]

Yup. And Pluto used to be a planet.

[@hyperboloidofonesheet1036]

8:34 I had to rewatch this three times to figure that out.

[@luisbenites4825]

Isn't that just the definition of fact 2?

[@malawigw]

Is 1 prime?

[@dimy931]

Can someone explain why the videos are in old school 3d now ? It annoys me a little

[@ilanlevin463]

5:09 , no, but that 1 did trigger me. Wait a minute...

[@shahinjahanlu2199]

Thx dear professor

[@Rick.Fleischer]

Is the class in number theory being taught online? Archived and available?

[@MichaelPennMath]

The course is taught in-person, but most lecture material is online -- here is the playlist: kzbin.info/aero/PL22w63XsKjqwn2V9CiP7cuSGv9plj71vv During class time, I usually give an extra quick example or two and then students work through problems collaboratively. Everyone is having a great time!

[@jondancyger8126]

1 is a prime?!

[@gennarobullo89]

Great explanation, it's a bit hard to assess which of the initial assumptions was indeed false! Keep up the good work!

[@tonyhaddad1394]

good job , 👍👍👍👍

[@tomholroyd7519]

Isn't that more of an induction than a bwoc? Given n primes, find another that's not already on the list?

[@Pastroni89]

With induction you suppose that the statement is true for an arbitrary term of the formula (set n=k), and then you want to show that the following term (k+1) is just the previous (k) combined with the base case (+1). If it is, then the statement always holds for any value of the variable n. We don't see it not work in these videos, but say you're trying to find a closed form for a series and guessed wrong, you shouldn't be able to show that the k+1 term is a combination of the k term with the base case (there'll be something left/missing). I don't think this statement can be proven with induction, since not all the numbers generated by the formula are primes. We know we can generate infinitely many of these numbers, just not which ones are prime and if there is ever a "last" prime.

[@roshnikabir4782]

Thanks a lot Take love from Bangladesh💜

[@admink8662]

Ah yes, 1 is prime now.

[@angelowentzler9961]

1 is not a prime

[@angelowentzler9961]

Watching the entire vid before commenting is for patient people!

[@elvistheawesome4864]

Sorry Michael please help this homework On number theory 5m^2 - 6mn + 7n^2 = 1985 n^3 - 5n + 10= 2^k

[@محمودابوعلي-ش8س]

realy beautiful❤❤ thank you

[@candamir26]

Thank you that you stopped these animations in the later videos. They were very distracting and not so helpful. The new animations that appear over the whole screen are quite good, though.

[@jimallysonnevado3973]

dirichlet is overkill for this

[@namasteanti]

Sir your vedio was just watched by me.( don't worry about subscribers) ((----+----))

[@digxx]

Can anyone pin the video to fact#1?

[@silknx]

I would also like some hint to prove fact #1 and fact #2, please.

[@digxx]

@@silknx Fact#2 is fairly simple: Given a generator g (mod p) we know g^{p-1}=1 and p-1 is the order (since it's a generator). Hence g^{(p-1)/2} can only be -1 mod p i.e. there exist an integer x s.t. x^2=g^{(p-1)/2} mod p and taking the square root we see that the integer would be of the form x=+-g^{(p-1)/4} which implies fact#2.

[@vanandreyambalong628]

First

[@thatkindcoder7510]

Not first in the comment section B)