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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?

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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)
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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.

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  • $\begingroup$ I understand most of this proof, but how do you know that m+2 is 7 mod 49, for example? Doesn't the Chinese remainder theorem only guarantee unique existence? $\endgroup$ – Sultan of Quizikhstan Oct 24 '18 at 3:00
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    $\begingroup$ @SultanofQuizikhstan force $m \equiv 2 \pmod 9,$ then $m \equiv 5 \pmod {49},$ then $m \equiv 8 \pmod{121}$ $\endgroup$ – Will Jagy Oct 24 '18 at 3:08
  • $\begingroup$ To the proposer: The Chinese Remainder Theorem implies that there exists $m$ that satisfies all $n$ of the congruences. $\endgroup$ – DanielWainfleet Oct 24 '18 at 14:08

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