There are infinitely many prime numbers of the form 4k+3.

[@crow4277]

as someone who is just starting out with proofs and number theory, this was really helpful and easy to understand!! :D thank you!!

[@CornerstonesOfMath]

Glad it was helpful. Thank you for commenting!

[@blue_tetris]

Nice, concise explanation of this proof-by-contradiction. Great video!

[@CornerstonesOfMath]

Thank you for your kind comment!

[@d.h.y]

Awesome video 🤩🤩

[@CornerstonesOfMath]

Awesome comment! Thank you.

[@adamamrouchi8225]

Great proof.

[@CornerstonesOfMath]

Although it cannot be generalized to other forms, it still allows us to utilize some basic concepts of number theory, so yeah, I agree.

[@archangecamilien1879]

Hmm...it's also the case that of 4k, 4k+1, 4k+2, one of them, and exactly one of them has to be divisible by 3...I mean, lol, it could be any, what I said earlier was ridiculous...at any rate, the statement can be reduced to: "There is K such that for k>K, all primes are of the form 4k+1...'cause one of 4k, 4k+1, 4k+2, and 4k+3 has to be prime (all integers can be written like that for some k "after a certain", if not always, lol...I think always)...so since 4k and 4k+2 are even, the prime has to be either 4k+1 or 4k+3...so, lol, the statement can mean that all primes after a certain point are of the form 4k+1...

[@archangecamilien1879]

I think I might have solved it, etc...a bit like the Euclid (??...not sure it was Euclid, Pythagoras?...I never remember these things)...so, since "after a point", all primes will be of the form 4k+1, then let's construct a number N=4p_1*p_2*p_3* ...(4k_1 + 1)(4k_2 + 1)...(4k_n + 1) + 1 [the k_i need, of course, not be consecutive, they are the k_i that "yield" primes, lol, so to speak]...then this number isn't divisible by any of the primes, etc...oh no, lol, this doesn't work, I keep on thinking I've solved it then realizing I haven't...

[@archangecamilien1879]

Ok...N= p_1*p_2...* (4k_1 + 1)(4k_2 + 1)...(4k_n + 1) + 1 (all the primes up to and beyond the "point" where all primes are of the form 4k+1)...then this is not of the form 4j+1 because all the factors, except p_1=2, are odd...it is of the form 2k+ 1, etc, with k odd...lol, not sure that helps...

[@archangecamilien1879]

Hmm...congruences, perhaps...let's take the 3 largest primes of the form 4k+3...I mean, lol, clearly, given the list in the thumbnail, there are at least 3 such primes, so, assuming there are only finitely many of them, let's take the 3 largest ones...then (4k_big1 + 3)(4k_big2 + 3)(4k_biglast + 3)...Jesus, this isn't leading anywhere, lol...was trying to use modular arithmetic...

[@archangecamilien1879]

Ok, not sure about this, lol...write N= 2*3*5 ...*(4k_1 + 1)(4k_2 + 1)... + 1...all the primes up to 4k_2 + 1 a prime such that all the primes after it will be of the form 4k_i + 1...then N is congruent to 3 mod 4...I mean, it's either congruent to 1 mod 4 or congruent to 3 mod 4, being odd...since it is not of the form 4b + 1, lol, 2 being the only even prime, it must be of the form 4k+3 (I feel that's obvious, should be easy to prove, lol)...now, so what, lol, one might wonder...N, like any other integer (positive, I guess, lol) is the product of primes...the primes are of the form 4k+1 or 4k+3, etc, since none of them are going to be 2 (the only even prime), since N is odd...remember all the primes that can be written 4k+3 are among the primes in the first term of the sum making up N, etc, so none of those can divide n...but no matter how you multiply a prime of the form 4k+1, you will get something congruent to 1 mod 4...so that means there must be a factor, in the prime factorization of N, that is congruent to 3 mod 4...so there is a prime congruent to 3 mod 4 that wasn't originally listed...yes, that should work, if I didn't make a bad assumption or some silly mistake...therefore there are infinitely many primes of the form 4k + 3...

[@CornerstonesOfMath]

Well, I dunno if it makes you feel better, but you were at least right in terms of approaches (constructing a new number by multiplying primes, using modular arithmetic which is not directly mentioned in the video but clearly used in some way).

[@eamonnsiocain6454]

. . . without relying on the Axiom of Choice . . .

[@CornerstonesOfMath]

I was not that deep into set theory, so I'm afraid I couldn't quite get the implication of the comment. What is so problematic with AC?

[@Kiflaam]

4(9)+3?

[@CornerstonesOfMath]

What do you mean by 4(9)+3?