Number Theory | Lagrange's Theorem of Polynomials

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We prove Lagrange's Theorem of Polynomials which is related to the number of solutions to polynomial congruences modulo a prime.

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did you forget to cut the first 15 seconds...

changjeffreysinto

There is a small mistake in this proof. The factorization of x^i -a^i should be

(x -a)(x^i-1 + ax^i-2 + ... +a^i-1) (1)

The video has a^i.x^i-2 as the second term of the second factor of (1) which is wrong.

paul

Thank you so much for making these videos. Your method of teaching is absolutely amazing. I AM REALLY, REALLY GRATEFUL . Again thanks a million.

lydiajoshua

I dont understand the part where for a degree 1 polynomial of the form ax+b, why was there only 1 or 0 solns. I understand if gcd(a, p)=1, then there's 1 soln, and if gcd(a, p)=p, and p doesnt divide b then there is 0 soln, but what if b is a multiple of p ? Shouldnt there be p solns? That's what you said in your linear congruence video.

weisanpang

Cannot we use that (pZ/Z)[X] is an euclidian ring, so we can use f(X) = q(X)(X-a) + r. And so : f(a) = 0 => r = 0.
????

lionelguez

Confusing stuff, slowly getting the hang of it though, thank you for these videos!

prathikkannan

Lagrange? More like Lagains, because we're gaining all of this knowledge from you! Thanks again for making and posting all of these videos.

PunmasterSTP

Professor Penn, thank you for an excellent lecture, however this one is very theoretical.

georgesadler

The Fundamental Theorem of Algebra == Langrange’s Theorem (mod p)

lgooch

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