Upper bound for maximal gap between consecutive numbers consisting only $4k+1$ primes Denote $A$ $-$ set of positive numbers with only prime factors of the form $4k+1$ and
$B$ $-$ set of positive numbers that can be represented as sum of two squares. $A$ is a subset of $B$ and there is upper bound $2 \sqrt{2} n^{\frac{1}{4}}$ for maximal gap between consecutive elements of $B$.
Question: Is any upper bound for maximal gap between consecutive elements of $A$ known?
 A: One can prove a bound comparable to the $O(n^{1/4})$ bound for $B$. To derive it, notice that any odd number of the form $a^2+b^2$ with $(a,b)=1$ lies in $A$. Now, for a given large number $N$, choose the largest even $a$ such that $a^2<N$. Then, of course $N-a^2=O(\sqrt{N})$. We now want to choose $b$ coprime to $a$ such that $N-a^2-b^2$ is small. Let $j(a)$ be the Jacobsthal function of $a$, i.e. the smallest $j$ such that any set of $j$ consecutive integers contains a number coprime to $a$. Consider the numbers $[\sqrt{N-a^2}], [\sqrt{N-a^2}]-1,\ldots, [\sqrt{N-a^2}]-j+1$. One of them is coprime to $a$, denote it by $b$. Then we get
$$
N-a^2-b^2\ll j(a)\sqrt{N-a^2}+j(a)^2.
$$
Next, by the result of Iwaniec (1978), $j(a)\ll \ln^2 a$, so we proved that the size of gaps in $A\cap [1,n]$ is at most $O(n^{1/4}\ln^2 n)$.
A: Short answer: yes.
In „Zur Verallgemeinerung des Bertrandschen Postulates, daß zwischen $x$ und $2x$ stets Primzahlen liegen" (Mathematische Zeitschrift, 34 (1932), pp. 505-526), R. Breusch proved that, for any  $x \geq 7$, the interval $[x,2x)$ contains at least one prime number of each of the following arithmetic progressions $\{3n+1\}_{n \in \mathbb{N}}$, $\{3n+2\}_{n \in \mathbb{N}}$, $\{4n+1\}_{n \in \mathbb{N}}$, and $\{4n+3\}_{n \in \mathbb{N}}$.
