a) Lemma: For each positive integer $n$ we have $n-1$ prime numbers that are $3(mod 4)$ and that their product is at most $2^{2^{n}}$. We will prove by induction. Base cases: $n=1$: $n=2$: {3} $n=3$: {3, 7}. Induction step: multiply all the numbers, subtract either $2$ or $4$ to make the number $3(mod 4)$ and pick a prime divisor of it $3(mod 4)$ and add to the group. Pick with CRT a number $x$ s.t. $x+i-1=p_i (mod$ $p_i^2)$. and $x+n<p_1^2*p_2^2*...*p_{n-1}^2<2^{2^{n+1}}$ Pick $O=(0, 0)$, $r=\sqrt{x-1}+a$, $R=\sqrt{x+n-1}-a$. $S=\pi(R^2-r^2)=\pi(n-2a)>3n$ for sufficiently small $a$. $R<2^{2^n}$ This works because the only integer points $(a, b)$ in the ring satisfy $x+n-1>a^2+b^2>x-1$, but no number in the range $(x, x+n-2)$ is a sum of 2 squares since they all are divisible by some $3(mod 4)$ prime and not divisible by the square of it so we are done. b) $100R<S^2=\pi^2(R^2-r^2)^2$. $R^2-r^2>10\sqrt{R}>\sqrt{r}$. Let's move $O$ to a the square between $(0, 0), (0, -1), (-1, -1), (-1, 0)$, and mark it's coordinates by $(-x', -y')$. We want to prove that exists a number $d^2$ in the range $(r, x+10\sqrt[4]{r})$ such that there is a lattice point $(x, y)$ with distance $d$ for $r>16$ (for $r<16$ the inequality given is impossible unless $R-r>1$ but then there clearly is a lattice point in the ring). Say $(n+x')^2+y'^2<r\leq (n+1+x')^2$, and mark $r=(n+x')^2+y'^2+p$ . We want to find a nonnegative integer $m$ s.t. $y'^2+p\leq (m+y')^2\leq y'^2+p+10\sqrt[4]{r}$. If $p=0$ just pick $m=0$. Otherwise, say $k$ is the largest nonnegative integer s.t $(k+y')^2<y'^2+p$. clearly $p+y'^2\leq (1+k+y')^2$, so we just want to prove $(1+k+y')^2\leq y^2+p+10\sqrt[4]{r}$. For that we can just prove $(1+k+y')^2-(k+y')^2<10\sqrt[4]{r}$. Note $k^2\leq (k+y')^2<2n+1$. So $(1+k+y')^2-(k+y')^2<2k+3\leq 2\sqrt{2n+1}+3<4\sqrt{n}+3<4\sqrt[4]{r}+3<10\sqrt[4]{r}$ and we are done. The lattice point $(n, k+1)$ has a distance $d^2$ s.t. $r^2<d^2<r^2+10\sqrt[4]{r}<R^2$.