Thanks so much. Happy to have you on the channel. Check out the proof using combinatorics that's linked at the end of the video; that's a really cool one.
This was extremely useful! Could you upload a video where you prove the same using Euler's totient theorem? (Any help from others in the comment section would be appreciated too!)
Not only intuitional but it is also shown why a*2a*3a*4a*5a...(p-1)a has the same remainder with 1*2*3*4*5*6*...(p-1) when divided by p. Excellent explanation, sir
I am a late career person trying to master cryptography, so I need to firm up my number theory. Thank you so much, Professor, for a lucid layout of this proof!!
Nice explanation. You used the primality near the end to say that a^(p-1) - a was divisible by p necessarily because the other factor, (p-1)!, can't be divisible by p, because p has no factors less than it except 1.
I guess this explains the intuition behind Euler's generalization, as we would only take those numbers less than n, for some n, which are coprime with n. In that case, again the product of all of these can't be divisible by n because they have no common factors other than 1, so a^(phi(n)) - 1 has to be divisible by n... Where did I miss that n and a have to be coprime?
Your intuition is right on track for Euler's Theorem; similar proof idea. The primality (and in Euler the coprimeness) comes up when we claim a,2a,3a,...,(p-1)a all have no common factors with p. In Euler's Theorem, you instead consider the product of a with numbers from 1 through n-1 that are coprime with n.
Nice proof ! And this gave me a question : Let P is a natural number If for any integer a we have a^p -a congruent to 0 mod p Then p is a prime ? In other words Does Ferma's little theorem give us double implication ? Thanks.
Thank you ! I have been looking for an answer for a while now As an 18 years old student in highschool my professors can't answer questions like these so really thank you Unfortunately , it doesn't work the other way around ! Or it maybe could have helped Keep up the good work !
Thanks Sri! Do you mean Euler's Theorem, that a^n and a are congruent mod n if a is relatively prime to n? I will be presenting that in a video that's coming in a few weeks :)
That's great. You have a preview now! Yes p-adic numbers are actually very interesting and lead to a variety of interesting avenues in abstract algebra.
Awesome explanation! I’d forgotten why the theorem works, and this was the perfect refresher for me. Keep up the good work!
Thanks so much. Happy to have you on the channel. Check out the proof using combinatorics that's linked at the end of the video; that's a really cool one.
still confused but less confused, thanks for the good explanation
one of the best channeles on youtube for explaining discreate math topics
This was extremely useful! Could you upload a video where you prove the same using Euler's totient theorem? (Any help from others in the comment section would be appreciated too!)
Excellent video, consider making a Patreon so we can support you more.
Hi Billy. Thanks! I'll have to look into this suggestion. Several people have made the same suggestion on here in the past.
Thanks again for the suggestion. I made one! www.patreon.com/ProfOmarMath
Not only intuitional but it is also shown why a*2a*3a*4a*5a...(p-1)a has the same remainder with 1*2*3*4*5*6*...(p-1) when divided by p.
Excellent explanation, sir
Thanks Yaren!
When my friend showed me a similar proof for euler's theorem that had to do with rearrangements of the units, my mind was BLOWN. Such a good time.
It all coalesces in the group theory proof
Beautiful, absolutely beautiful. Thank you so much!
No prob!
I am a late career person trying to master cryptography, so I need to firm up my number theory. Thank you so much, Professor, for a lucid layout of this proof!!
Wish I could like it more than once!
Thanks Juliano!
Nice explanation. You used the primality near the end to say that a^(p-1) - a was divisible by p necessarily because the other factor, (p-1)!, can't be divisible by p, because p has no factors less than it except 1.
I guess this explains the intuition behind Euler's generalization, as we would only take those numbers less than n, for some n, which are coprime with n. In that case, again the product of all of these can't be divisible by n because they have no common factors other than 1, so a^(phi(n)) - 1 has to be divisible by n... Where did I miss that n and a have to be coprime?
Your intuition is right on track for Euler's Theorem; similar proof idea. The primality (and in Euler the coprimeness) comes up when we claim a,2a,3a,...,(p-1)a all have no common factors with p. In Euler's Theorem, you instead consider the product of a with numbers from 1 through n-1 that are coprime with n.
Also professor im not sure if you are aware but your channel is set up for kids?
So we can't hit the bell to get notifications
Oh WOW, thanks so much for informing me about this, I had no idea!
@@ProfOmarMath no problem
This is excellent
Thanks Kevin!
Nice proof ! And this gave me a question :
Let P is a natural number
If for any integer a
we have
a^p -a congruent to 0 mod p
Then p is a prime ?
In other words
Does Ferma's little theorem give us double implication ?
Thanks.
Check out Lehmer's Theorem 😀
If I remember from when I was in undergrad, p=561 is a counterexample if you restrict to gcd(a,p)=1
Thank you !
I have been looking for an answer for a while now
As an 18 years old student in highschool my professors can't answer questions like these so really thank you
Unfortunately , it doesn't work the other way around !
Or it maybe could have helped
Keep up the good work !
Thanks for giving an intuitive proof of Fermat's theorem. I believe if p is relatively prime to this argument still works. Is that true?
Thanks Sri! Do you mean Euler's Theorem, that a^n and a are congruent mod n if a is relatively prime to n? I will be presenting that in a video that's coming in a few weeks :)
This is awesome!
I haven't taken this class yet 🤣🤣
Trying to keep up.
Also-- have you heard of p-adic numbers?
That's great. You have a preview now!
Yes p-adic numbers are actually very interesting and lead to a variety of interesting avenues in abstract algebra.
@@ProfOmarMath my friend mentioned them to me--he has a degree in number theory. You can get very big numbers vry fast? From what I remember
Awesome. Thank you, prof
Thanks Ardilo!
Salam from Indonesia, Prof.
Best video ever!
Check out some other ones 😍
You're still the best!
Joe!!
Very well explained :)
Thanks Harshit!
very well explained
Thanks!
I have seen the proof with mathematical induction method.
Does this use the binomial theorem arguing that p choose k has p as a factor?
@@ProfOmarMath Yes, sir!
Very nice video!
Thanks!
Sir plz teach quadratic reciprocity.
Cool idea Yash!
Yes.. I understand it.
Great Nawus!
simply the best.
Thanks Joe!
big brain time
😄
Why prove it? Why not just believe it? Much more powerful
😅