I also have a video where I prove there are infinitely many primes of the form 4n+1: th-cam.com/video/qhi0w2_SewM/w-d-xo.html. This uses a different argument using the notion of quadratic residues.
Professor Penn, I am learning from watching and do many problems in different Number Theory books. These lectures are very helpful for a deep understanding of Number Theory.
My conjecture based off this: Given m>n integers, where m,n are coprime, there are infinitely many primes of the form mk+n for some k integer. Has this been proven true or false already or is it still a conjecture?
3 is clearly of the form 4n+3 so we know it is on the list p0,p1,....,pk. Since the argument requires us to take a product that does not include 3, namely p1p2..pk it is convenient to name the elements of the list so that p0=3. We could really put it anywhere, but this makes everything work cleanly.
@@MichaelPennMath Does p0 really have a meaning? I mean, if processed by claiming n=0 in 4n+3 would give you 4.0+3=3 but n is defined to be a natural number. True that it goes into the list p0,p1,p2,...,pn. Any odd prime would be of the form 4n+1 or 4n+3 but how does 3 get included in this set if n is defined to be a natural number?
@@chandeepadissanayake6975 it is necessary for the proof. The proof relies on the fact that if there is finitely many primes of form 4n + 3, then N must have at least one prime of the form 4n + 3. And clearly 3 is one of the possibilities. Thats why you need to handle it.
There are problems with what you did here. In the proof, N is the product of each prime of the form 4n + 3. If 4n + 3 is not prime, we don't need to consider it as a factor of N.
I also have a video where I prove there are infinitely many primes of the form 4n+1: th-cam.com/video/qhi0w2_SewM/w-d-xo.html. This uses a different argument using the notion of quadratic residues.
Your chalk board is super not level
That flip was so cool
Professor Penn, I am learning from watching and do many problems in different Number Theory books. These lectures are very helpful for a deep understanding of Number Theory.
I feel very 4tunate to have come across your number theory videos. I am learning so much, and thank you so much for making all of them!
n element of N so first p0 would be 7 not 3 right?
Yes!
My eyes was going towards biceps after every 5 seconds
Great explanation Sir, huge respect from India
The Kurzgesagt t shirt is looking good
Yeah. Its always good to see the best give recognition to those that try their best.
My conjecture based off this:
Given m>n integers, where m,n are coprime, there are infinitely many primes of the form mk+n for some k integer.
Has this been proven true or false already or is it still a conjecture?
Clever proof!
Thank you! Love your videos as always!
Thank you, very clear proof.
does the same argument work to show that there are infinitely many primes of the form 4n+1?
try it
its the same but n cant be an prime
Good video!!
How to assume that p0 is 3?
3 is clearly of the form 4n+3 so we know it is on the list p0,p1,....,pk. Since the argument requires us to take a product that does not include 3, namely p1p2..pk it is convenient to name the elements of the list so that p0=3. We could really put it anywhere, but this makes everything work cleanly.
@@MichaelPennMath Does p0 really have a meaning? I mean, if processed by claiming n=0 in 4n+3 would give you 4.0+3=3 but n is defined to be a natural number. True that it goes into the list p0,p1,p2,...,pn. Any odd prime would be of the form 4n+1 or 4n+3 but how does 3 get included in this set if n is defined to be a natural number?
@@chandeepadissanayake6975 it is necessary for the proof. The proof relies on the fact that if there is finitely many primes of form 4n + 3, then N must have at least one prime of the form 4n + 3. And clearly 3 is one of the possibilities. Thats why you need to handle it.
We define the set to have 3. It does not really matter if you don't consider it as a 4k+3 prime.
thanks man
Good
Today I got this question in my exam
For n = 10000 : 4×10000+3 = 40003 = 109×367 .•. Not prime :(
the converse is not always true !
There are problems with what you did here. In the proof, N is the product of each prime of the form 4n + 3. If 4n + 3 is not prime, we don't need to consider it as a factor of N.
We don't state that every 4n+3 is prime. We state that out of all numbers of the form 4n+3, there are infinitely many that are prime