An interval of length exceeding $\sqrt n$ contains an integer coprime to $n$ True or false? -
Let $n$ be a positive integer.  If $a$ and $b$ are integers with $b-a>\sqrt n\thinspace$, then there exists an integer $x$ coprime to $n$ in the interval $a\le x\le b$.
Comments.

*

*We may assume that $kn\le a<b<(k+1)n$ for some $k$, for if not then there is some $x=mn\pm1$ in the required range.

*And then the problem only depends on residues modulo $n$, so we may assume $0\le a<b\le n$.

*The conclusion need not be true if $a,b$ are not integers.  Trivial example: $n=2$, $a=1.1$, $b=2.9$.  Less trivial example: $n=30$, $a=1.1$, $b=6.9$.

*(Not 100% sure about this.)  If the statement is true for $n$ and $p\mid n$, then the statement is true for $pn$.  So if there is a counterexample, there must be a counterexample with squarefree $n$.

The problem was inspired by an attempted approach to this question.
 A: Note: This is an incomplete answer.
First, since you say you're not 100% sure, let's note that your last statement is true. Suppose the claim is true for $n$, and $p$ is a prime dividing $n$.  Then, consider an interval of length $> \sqrt{pn}.$ Certainly this interval has length $> \sqrt{n}$, hence contains an integer $m$ coprime to $n$ by assumption. Since $p|n,$ $m$ cannot be a multiple of $p$, hence $m$ is coprime to $pn$ as well. So, as you remark, we can restrict ourselves to considering only squarefree $n$.
Let's also look at the second bulletpoint. We can only assume $0 \leq a < b  \leq n$ (or in fact $0 \leq a < b < n$) if an interval of length $> \sqrt{n}$ can fit in $[0, n-1].$ In other words, we should have $n -1 > \sqrt{n}$, and thus $n \geq 3$. We can, however, just check that the claim is true for $n =2$ (any interval of length greater than $\sqrt{2}$ must have length at least $2,$ hence contains an odd number). Therefore, we can go ahead with this assumption (keeping in mind that $n \geq 3$ for the rest of the problem).
Now, we follow the argument given in the first answer here: Count Integers Not Greater Than $a$ Coprime To $b$
Given an interval $[a, b]$ of length $> \sqrt{n}$, the number of integers in this interval coprime to $n$ is given by
$$\sum_{m=a}^{b} \sum_{d|(m, n)} \mu(d) = \sum_{d|n}\mu(d)\sum_{a \leq m \leq b \\ d|m} 1.$$
Note that $\sum_{a \leq m \leq b \\ d|m} 1$ counts the number of integers divisible by $d$ in $[a, b]$. This is $\lfloor \frac{b-a}{d} \rfloor$ if $d \nmid a,$ and $\lfloor \frac{b-a}{d} \rfloor +1$ if $d \nmid a.$ The $+1$ terms contribute a total of $\sum_{d|n}\mu(d) = 0$, so we can ignore them.
Therefore, we are interested in estimating
$$\sum_{d|n}\mu(d)\lfloor \frac{b-a}{d} \rfloor = (b-a)\sum_{d|n}\frac{\mu(d)}{d} -\sum_{d|n} \mu(d)\left\{\frac{b-a}{d} \right\}.$$ Let $R = \sum_{d|n} \mu(d)\left\{\frac{b-a}{d} \right\}.$
We have
\begin{align}
(b-a)\sum_{d|n}\frac{\mu(d)}{d} -\sum_{d|n} \mu(d)\left\{\frac{b-a}{d} \right\} &= (b-a)\frac{\phi(n)}{n} - R\\
&> \sqrt{n} \cdot \frac{\phi(n)}{n} - R\\
&= \frac{\phi(n)}{\sqrt{n}} - R.
\end{align}
If we can show that $\frac{\phi(n)}{\sqrt{n}} - R \geq 1,$ we are done. In other words, we need to show that
$$\sum_{d|n} \mu(d)\left\{\frac{b-a}{d} \right\} \leq \frac{\phi(n) - \sqrt{n}}{\sqrt{n}}.$$ At this point, I don't know how to continue.
