A simple way to bound the density of sums of two odd squares

Question

Define $$S(x) ~=~ \# \left\{ n^2+m^2\leq x : n,m\in\mathbb{N}\right\}$$ Landau (1908) proved that with $$ B(x) ~=~ K\,\frac{x}{ \sqrt{\log x}} ~~\text{ one has}~~~ \lim \limits_{x\to \infty} \frac{S(x)}{B(x)}~=~1,$$ where $K$ is the Landau-Ramanujan constant. Because
$$ a= n^2 +m^2 \Leftrightarrow 4\,a= (2n)^2 +(2m)^2,$$ one can immediately derive for the number of sums of two even squares $$S_0(x) ~=~ \# \left\{ (2n)^2 +(2m)^2\leq x : n,m\in\mathbb{N}\right\},$$ that $S_0(4 x)=S(x)$, and consequently for $\epsilon>0$ and sufficiently large $x$ $$S_0(x) \geq \frac{K}{4+\epsilon}\,\frac{x}{ \sqrt{\log x}}. $$ Question: Is there a similarly simple argument to bound $$S_1(x) ~=~ \# \left\{ (2n-1)^2 +(2m-1)^2\leq x : n,m\in\mathbb{N}\right\}?$$

Answer 1

A number $n$ is a sum of two squares if and only if $2n$ is (this goes back to Fermat and Euler). This implies $$S_{\text{even}}(x):=|\{n^2+m^2: n,m \ge 0, \, n^2+m^2 \le x,\, n^2+m^2 \text{ even}\}| =S(x/2).$$ If $n^2+m^2$ is even then $n$ and $m$ have the same parity. The number of $n^2+m^2$'s where $n$ and $m$ are even is, as you observed, $$S_{0}(x):=|\{n^2+m^2: n,m \ge 0, \, n^2+m^2 \le x,\, n\text{ and }m \text{ even}\}| =S(x/4)$$ since we can divide $n$ and $m$ by $2$. This means that $$S_1(x) \ge S_{\text{even}}(x) - S_{0}(x) = S(x/2)-S(x/4) \sim S_{0}(x).$$

In fact $S_1(x) = S_{\text{even}}(x) - S_{0}(x)$, because $n^2+m^2$ is congruent to $0$ modulo $4$ when $n,m$ are both even and is congruent to $2 $ modulo $4$ when $n,m$ are both odd, so the sets counted by $S_{0}$ and $S_1$ do not intersect.