I really like the teaching style here of motivating the proofs by analyzing small values and finding patterns! This makes the proof feel less magical like they appear out of thin air, and more natural like they can actually be found by a human.
Another proof of the result at 22:36 is to note that we can write the residue classes mod n as the product ab uniquely, where b is coprime to n. Hence for a fixed a there are exactly phi(n / a) choices of b, so \sigma_{a | n} phi(n / a) = \sigma_{a | n} phi(a) = n.
I really like the teaching style here of motivating the proofs by analyzing small values and finding patterns! This makes the proof feel less magical like they appear out of thin air, and more natural like they can actually be found by a human.
Thanks for sharing your knowledgment with us. Hugs from Brazil!
Another proof of the result at 22:36 is to note that we can write the residue classes mod n as the product ab uniquely, where b is coprime to n. Hence for a fixed a there are exactly phi(n / a) choices of b, so \sigma_{a | n} phi(n / a) = \sigma_{a | n} phi(a) = n.
good lord, why are textbooks and other videos so opaque about this material? thank you for uploading this!
I THANK GOD FOR THE EXISTENCE OF PROFF RICHARD E BORCHERDS
Thanks!
yeeeeeeeeeeeeeeeeeeee
You again, mysterious yeee man
@@ethanjahan780 yeeeee