How do I estimate the error term when computing the number of integers which have the fractional part of their square roots in a given interval? I'm trying to find the number of integers $n \leq N$ such that fractional part of $\sqrt n\in (\alpha,\beta]$ where $(\alpha,\beta]\subseteq(0,1]$. The approximate number is of course $(\beta-\alpha)N$, but my goal is to estimate the error term, i.e., to estimate
${\displaystyle\sum_{n\leq N,(\sqrt n)\in (\alpha,\beta]}1}-(\beta-\alpha)N$.
I've recently learnt the trick of solving these problems by reducing to exponential sums (like the divisor or the circle problem) but I can't express the error term here as exponential sums. Can someone please tell me how to do it? If I see how to express this as an exponential sum, I think I can estimate the sum itself by what I know. Thanks in advance.

Weyl's Criterion states that $a_1,a_2,\dots,$ will be equidistributed modulo $1$ if an only if $$\lim_{N\rightarrow \infty}\frac{1}{N}\sum_{n\leq N} e^{2\pi i a_n m}=0$$ for every $m$. While this is not a quantitative result, it suggests that we should examine the Fourier coefficients $$\sum_{n\leq N}e^{2\pi i m \sqrt{n} }.$$ The Erdos-Turan inequality provides what you are looking for, that is quantitative bounds on the discrepancy. Specifically, if we let  $$D(m)=\sup_{0\leq a\leq b\leq 1} \left|\frac{1\#\left\{1\leq j \leq n: a \leq \{\sqrt{n}\}\leq b\right\}}{n}-(b-a) \right|$$ be the discrepancy, then the Erdos-Turan inequality states that there exists a constant $C>0$ such that for any $k$ $$D(n)\leq C\left(\frac{1}{k}+\sum_{j=1}^k \frac{1}{j} \left|\frac{1}{n}\sum_{l=1}^n e^{2\pi i j \sqrt{l}} \right|\right).$$
