# Does the interval $[x, x + 10 x^{1/4})$ contain a sum of two squares?

Is it true that there exist an integer $N \in \mathbb N$ , such that $\forall x > N$ , the interval $[x, x+10x^{1/4} )$ always contains a sum of two squares ?

I know that $n \in \mathbb N$ is a sum of two squares iff it is of the form $n=q_1^{2a_1}...q_m^{2a_m}p_1^{k_1} ... p_r^{k_r}$ where $q_1,...,q_m,p_1,...,p_r$ are primes such that $q_i \equiv 3( \mod 4)$ and $a_1,...,a_m, k_1,...,k_r$ are non-negative integers ; where $m , r\ge 0$ . So it is enough to prove that for large enough $x$ , the interval $[x , x+10x^{1/4})$ always contains such a positive integer . But I don't know how to prove this.

• Reading about the Gauss circle problem would be a good starting point. – Daniel Fischer Jul 30 '17 at 19:01
• For me to start: where did you get the problem/conjecture? Is it stated as something definitely true or as an open problem? It is reasonable, by the way, as the count of numbers up to large $z$ that are sums of two squares is about $$\frac{Cz}{\sqrt {\log z}}$$ with (known) positive constant $C.$ – Will Jagy Jul 30 '17 at 19:19
• @WillJagy : For the source of the problem ; it was asked as an open ended question in the class by our professor . Where do you get the asymptotic formula about the count ? Could you please give any reference ? Thanks – user Jul 31 '17 at 15:54
• store.doverpublications.com/0486425398.html – Will Jagy Jul 31 '17 at 18:20

## 1 Answer

You might try to proceed greedily. Suppose we want to find a number that is "near" (but less than) $x$ that is a sum of two squares. Maybe one of these squares will be the largest square less or equal to $x$, which I'll denote by $a$. (So $a = \lfloor \sqrt x \rfloor^2$), where $\lfloor \cdot \rfloor$ is the "floor" function, or the "greatest integer below function"). In this case, the second square, which I'll call $b$, should be chosen to be the greatest square less than or equal to $x - a$. (So $b = \lfloor \sqrt{x - a} \rfloor^2$).

This gives us a candidate sum of squares $a + b$, which is relatively close to $x$. Now we ask, how close does this get us? How large is $x - a - b$?

This really asks, how close is $x$ to the largest square up to $x$? That is, how close is $x$ to $a = \lfloor \sqrt x \rfloor^2$?

Note that $$x - \lfloor \sqrt x \rfloor^2 = (\sqrt x - \lfloor \sqrt x \rfloor)(\sqrt x + \lfloor \sqrt x \rfloor) \leq (1) \cdot (2 \sqrt x) \leq 2 \sqrt x.$$ So the largest square up to $x$ is no more than $2 \sqrt x$ away from $x$.

Thus $x - a \leq 2 \sqrt x$. Iterating, we have that $$(x - a) - b \leq 2 \sqrt{x - a} \leq 2 (\sqrt {2 \sqrt x}) = 2^{3/2} x^{1/4}.$$

We have shown that given an $x$, the nearest "greedy" sum of squares is at most $2^{3/2} x^{1/4}$ away. So we've shown that there is a sum of squares in intervals of the shape $(x - 2^{3/2} x^{1/4}, x]$, which clearly implies the original question.

Additional Notes

In comments, Will Jagy noted that the density of numbers which are sums of two squares is of the shape $Cx/\sqrt{\log x}$, where $C$ is the Landau-Ramanujan constant. It is conjectured that the gaps between numbers which are sums of squares shouldn't deviate too far from what is expected from the density, and in particular is bounded by $x^\epsilon$ for any $\epsilon > 0$. In particular it is conjectured that there is a number which is a sum of two squares in the interval $[x, x + x^{\epsilon})$ for any $\epsilon > 0$, for $x$ sufficiently large.

This answer shows that the gap between numbers which are sums of squares is at most $2^{3/2}x^{1/4}$, which is a far cry from $x^{\epsilon}$.

In general, there are not good uniform improvements over what's presented above. But there are remarkable on-average bounds due to Hooley, showing for instance that almost all (in the density sense) gaps between sums of squares is no more than $(\log x) (\log \log x)$.