A question about the sum of two squares So I want to show a few things about the form $x^2+ y^2$, namely that for any number, $n$, we can find at least $n$ consecutive integers that are not the sum of two squares. So far, I've only shown that a number that is a sum of two squares has to be such that every prime occurring in the prime factorization that is $3 \text { mod } 4$ has an even exponent, but I'm not sure how to prove the last bit about the consecutive numbers. How should I continue?
 A: You have indicated that, given a prime $q \equiv 3 \pmod 4,$ if a positive number $n$ is divisible by $q$ but not by $q^2$ then it cannot be the sum of two squares.
Let $m+1 \equiv  3 \pmod 9.$ Then let $m+2 \equiv 7 \pmod {49}.$ Next $m+3 \equiv 11 \pmod {121}.$ Keep going, up to $m+n \equiv q_n \pmod {q_n^2}$
This is possible by repeated Chinese Remainder Theorem. 
A: The proof below is based on a paper${}^{\color{blue}{[1]}}$ by Ian Richards, I just fill in some gaps.
Given positive integer $n$,


*

*Let $[n]$ be a short hand for $\{ 1, 2, \ldots, n \}$.

*Let $\mathcal{P}$ be the subset of $[4n]$ consists of prime $\equiv -1 \pmod 4$.

*For $p \in \mathcal{P}$, let $\beta(p)$ be the integer such that $p^{\beta(p)} \le 4n < p^{\beta(p)+1}$.

*Let $P = \prod_{p \in \mathcal{P}} p^{\beta(p)+1}$ and pick a $y \in [P]$ such that $4y \equiv -1 \pmod P$.


For any $\ell \in [n]$, we claim $y + \ell$ isn't a sum of squares. 
Assume the contrary, let's say $y + \ell = u^2+v^2$ for some integers $u,v$. We have
$$4(u^2+v^2) = 4(y+\ell) \equiv 4\ell - 1\pmod P$$
Since $4\ell-1 \equiv -1\pmod 4$, $4 \ell - 1$ contains prime factors from $\mathcal{P}$ whose exponent is odd. Let $p$ be one of these prime with  exponent $2a+1$, i.e. $p \in \mathcal{P}$ which satisfies
$$p^{2a+1} | 4\ell - 1 \quad\text{ but }\quad p^{2a+2} \not| 4\ell - 1$$
By Chinese remainder theorem,
$$4(u^2+v^2) \equiv 4\ell - 1 \pmod P
\quad\implies\quad 4(u^2 + v^2) \equiv 4\ell - 1 \pmod {p^{\beta(p)+1}}\tag{*1}
$$
Since $2a+1 \le \beta(p)+1$, this leads to $p^{2a+1} | 4(u^2+v^2)$.
Since prime of the form $\equiv -1 \pmod 4$ is prime in the ring of Gaussian integers,
$$p^{2a+1} | 4(u^2+v^2)\quad\implies\quad p^{2a+2} | 4(u^2+v^2)$$
Since $2a+2 \le \beta(p)+1$, RHS of $(*1)$ tell us $p^{2a+2} | 4\ell - 1$. This contradicts with the role of $p$ and $a$.
As a result, none of the $n$ consecutive integers $y+1, y+2, \ldots, y + n$ is a sum of two squares. This means the gap between the sum of two squares can be as large as one wish.
References


*

*$\color{blue}{[1]}$ I Richards, On the gaps between numbers which are sums of two squares  - Advances in Mathematics, 1982 - Academic Press
(an online copy can be found here)

