Elementary Proof of Landau's count on number representable as sum of two squares

Question

In Analytic Number Theory by Iwaniec and Kowalski, there is an elementary proof of Landau's result of $\#\{n \le x: \exists a, b,\ s.t.\ n = a^2 + b^2\} \sim Cx/\sqrt{\log x}$ with an explicit constant $C$. However, it is left as exercise 4 in Chapter 1, and the hint is to use Thm 1.1. However, there is a restirction that $\kappa > -1/2$ in Thm 1.1, and if my calculations are correct, no matter whether I select primes congruent to 1 or 3 mod 4, I will have to take $\kappa = -1/2$ in Thm 1.1, which makes the error term as large as the main term. Of course the error term can be made smaller if we use PNT (basically the $O$ becomes $o$), but I feel that it is not the intended approach and is not quite "elementary". Does anyone have any hint?

Answer 1

This is a good question that has been unanswered for a long time! As Will Jagy mentions, there is a proof in LeVeque but it is not elementary.

Let $f(n)$ be the characteristic function of positive integers that are representable as the sum of two squares. Then $f(n)$ is multiplicative, but we can't apply Theorem 1.1 directly to $f(n)$. However, we can apply the theorem to $f(n)/n$, and we get the following asymptotic: $$\sum_{n \leq x} \frac{f(n)}{n} \sim 2C (\log x)^{1/2},$$ where $C$ is the Landau-Ramanujan constant.

It remains for us to somehow get an asymptotic for $\sum_{n \leq x} f(n)$ using the asymptotic for $\sum_{n \leq x} f(n)/n$. This seems hard -- summation by parts will not work.

Surprisingly, we can do this if there are some bounds on the growth of $f$ at the prime powers. There is a very nice lemma from Wirsing's 1961 paper (Hilfssatz 2 on page 93).

Proposition Let $f(n)$ be a multiplicative function such that:
(i) $f(n) \geq 0$,
(ii) $f(p^k) \leq ab^k$ for some $b < 2$ and for all primes $p$ and all $k \geq 2$,
(iii) The sum over the primes has an asymptotic $$\sum_{p \leq x} f(p) \sim \tau \frac{x}{\log x}.$$ Then $$\sum_{n \leq x} f(n) \sim \tau \frac{x}{\log x} \sum_{n \leq x} \frac{f(n)}{n}.$$

The proof of this proposition is entirely elementary, but a bit too long to reproduce here.

Applying this proposition, we can conclude with Landau's formula: $$\sum_{n \leq x} f(n) \sim C \frac{x}{(\log x)^{1/2}}.$$