# 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?

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)

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.

• 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? – Sultan of Quizikhstan Oct 24 '18 at 3:00
• @SultanofQuizikhstan force $m \equiv 2 \pmod 9,$ then $m \equiv 5 \pmod {49},$ then $m \equiv 8 \pmod{121}$ – Will Jagy Oct 24 '18 at 3:08
• To the proposer: The Chinese Remainder Theorem implies that there exists $m$ that satisfies all $n$ of the congruences. – DanielWainfleet Oct 24 '18 at 14:08