Absolutely fascinating! This video on Fermat's little quotient blew my mind. I never realized the intricate connections between this concept and Fermat's little theorem
I tossed the generalized version into Mathematica for a few smaller integers. Interestingly, there are other solutions where a>2. (a,p) = (3,2) (3,5) (7,3) (9,2) (19,2) (26,3) (33,2). The ones where p=2 form a regular sequence of 3,9,19,33,51,73... where the difference between terms is 4n+2 so x[1]=3, x[n+1]=x[n]+4n+2. The first ones where p=3 are 2, 7, 26, 97, 362, 1351, 5042... Rather interestingly, p=5 only has a single solution for a
Okay, I gave this problem a try, and I found the proof for the formula of the solutions when p=3, it can be proved by induction and there is a recursive formula for the various families of solutions: [a+1]= 2a²-1
This integral does not converge. When you multiply what's inside the integral you get int of e^(-4x^2/(4+x^2)). As this function limits to e^(-4) in +- inf the integral cannot converge. It would have to limit to 0 (this is necessary but not sufficient condition).
This should have been Fermat's little quotient, since it arises from Fermat's little theorem.
Another possible extension would be to Euler's little quotient: What numbers n make (2^phi(n)-1)/n a perfect square?
13:17 *insert joke about Fermat’s margin note*
Brilliant! Love the use of decomposition and deductions
8:28 if n=1 => b^2=1 => b=1 => n=1? why the calculation? if n=1 then p=3. no need to find b.
Absolutely fascinating! This video on Fermat's little quotient blew my mind. I never realized the intricate connections between this concept and Fermat's little theorem
Michael is a genius❤
I tossed the generalized version into Mathematica for a few smaller integers. Interestingly, there are other solutions where a>2. (a,p) = (3,2) (3,5) (7,3) (9,2) (19,2) (26,3) (33,2). The ones where p=2 form a regular sequence of 3,9,19,33,51,73... where the difference between terms is 4n+2 so x[1]=3, x[n+1]=x[n]+4n+2. The first ones where p=3 are 2, 7, 26, 97, 362, 1351, 5042... Rather interestingly, p=5 only has a single solution for a
For p=2 you can substitute into (a^(p-1)-1)/p = x^2 and after some simple algebra you get a = 2x^2+1
Okay, I gave this problem a try, and I found the proof for the formula of the solutions when p=3, it can be proved by induction and there is a recursive formula for the various families of solutions: [a+1]= 2a²-1
A lot can come out from a little quotient, but no gnarliness.
Thank you, professor.
What if you use other numbers than 2 as a?
Found 3 and 7 from the problem's writing yay lol
What values of a makes the quotient a perfect cube and so on?
Next: fermat's little quotient
this guy is too smart for me ...
how is 1 mod p not 1?????
cool
Man, if you're gonna throw in a shout out to cryptography, should have gotten NordVPN to sponsor the video!
Can someone help me solve ? ∫ e^(- x^2/(2+ix)) * e^(- x^2/(2-ix)) dx from -inf to inf
This integral does not converge. When you multiply what's inside the integral you get int of e^(-4x^2/(4+x^2)). As this function limits to e^(-4) in +- inf the integral cannot converge. It would have to limit to 0 (this is necessary but not sufficient condition).
@@Kycilak thank you brother, notation error make this one divergent, sorry
Can anyone help me? I am a math olympian and I dont have money to buy laptop? I will give you 10x in the future?
tbh i like your videos but i'm not subscribed because you upload so many videos, and i only watch a few of those