Number Theory | Euler's Totient Function and Powers of Primes

We give a formula for the value of Euler's totient function on powers of primes.

[girishab]

Hi, I am thoroughly enjoying your videos on Number Theory. Thanks for uploading these. I do have a question on your proof in this video where you build the array of p^r numbers. The entries in last but one row do not seem to be correct. E.g. The last but one entry in last column should be (p^(r-1) - 1)*p and not p^(r-1). Likewise, the next entry (first one in last row) should be p*(p^(r-1)-1)+1. While these entries may not have bearing on the final proof, it can lead to confusion while counting the exact number of coprimes. Can you check please? Thank you.

[mackenziekelly]

It's just the elements of 1 to p^k minus elements which are multiples of p. By definition, there are p^(k-1) multiples of p from 1 to p^k because p^k/p=p^(k-1). That means p^k elements minus p^(k-1) elements. Done.

[DragonKidPlaysMC]

Thank you for helping me understand this concept in such a clear proof!

[georgesadler]

Professor Penn thank you for a solid clarification of Euler Totient Function and it Powerful Powers of Primes. These lectures are fun.

[mmoose]

Although I'm certain the proof you laid out came first, this makes a lot of since from the point of view of group theory. The U-group is isomorphic to the integers mod p^n - p^(n-1) so of course because the U-group has order Euler-phi it will be equal to p^n - p^(n-1)

[lienzo]

Thanks!
there's a mistake at the lasts rows of the array. the last element of the second-last row should be (p^{r-1} - 1) * p, and that changes the other elements in the rows but it doesn't change the reasoning of the proof.

[abyzzz]

thank you so much!!! it makes my life more easier

[palakratta]

If n is even number than how to prove that summation over d/n (d*phi(d))=0

[PunmasterSTP]

Totient? More like woe, it's another wonderful video on number theory from Michael Penn! Thanks again so much for making and posting these.