## 2009年11月12日 星期四

### A beautiful solution

Suppose you want to prove that the sum of 1/(m^2 + n^2) diverges as m,n ranges over all positive integers. What do you do?

Here's a very beautiful solution, from one of my students in an undergraduate complex analysis class:

Every prime of the form 4k+1 is expressible as the sum of two squares. Hence the previous sum is bounded below by the sum of 1/p, where p ranges over all primes that are congruent to 1 mod 4. The latter sum diverges. Q.E.D.

#### 5 則留言:

Alex 提到...

haha my first thought was using integral test...though the solution of your student's sounds elegant, the fact that the sum of the reciprocals of primes of type 4k+1 is a non-trivial one.

Alex 提到...

correction: 'the fact that the sum of the reciprocals of primes of type 4k+1 diverges is a non-trivial one'

Polam 提到...

haha, yes I know the lower bound that the student gave was very weak, and the proof of the divergence of that lower bound is non-trivial... but I'm quite sure that the student knew the integral test, and he was just trying to "be different". It's a pleasure though to see such solutions when you're grading a pile of complex analysis homework. At least that refreshed my mind while I was grading that large pile of homework, and made the task more "meaningful".

tobywhcheng 提到...

where I can find this non-trival prove?

Polam 提到...

Toby: You can find the proof of the sum of squares theorem in Stein and Shakarchi, Complex Analysis, Chapter 10. You can find the proof of the divergence of the sum of 1/p, where p ranges over all primes that are 1 mod 4, in Stein and Shakarchi, Fourier analysis: an introduction, Chapter 8.