# Counting primitive solutions to a diophantine inequality

This is a refinement (perhaps a simpler version) of a question I asked here before and couldn't get an answer for.
Fix $$\alpha \in (0,1]$$ and a small constant $$c>0$$. For $$x \in [0,1]$$ and $$N\in\mathbb{N}$$, we denote by $$S(x,N)$$ the function that counts the number of pairs of integers $$(p,q)$$ such that $$|x-\frac{p}{q}|<\frac{c}{q^{\alpha+1}}\qquad(\star)$$
and $$1\leq q \leq N$$, $$gcd(p,q)=1$$.
It is known that when $$\alpha=1$$, all the solutions $$(p,q)$$ to $$(\star)$$ are in fact the convergents (i.e. truncation of the continued fraction expansion of $$x$$). The recursive formula for the denominators of those implies that they grow exponentially (asymptotically), and for all irrationals at least as fast as those for $$x$$ being the golden ratio (you can consider its fractional part in order to have it in the interval). This, in turn, implies that there exists some constant $$C>0$$ such that $$||S(x,N)||_{L^{\infty}([0,1])}\leq C \log N.$$

I'm interested in getting an upper bound on this $$L^{\infty}$$ norm for any $$\alpha\in(0,1)$$. To be more precise, I wish to show that this bound is sublinear in $$N$$.
W. Schmidt considered the more general case where the restriction $$gcd(p,q)=1$$ is dropped (I denote the corresponding counting function by $$M(x,N)$$). His asymptotic estimate, which holds for almost every $$x\in[0,1]$$ is $$M(x,N)\leq \sum_{q=1}^{N} \frac{c}{q^{\alpha}}+C(x) (\sum_{q=1}^{N} \frac{c}{q^{\alpha}})^{\frac{1}{2}}.$$ But as far as I can tell, the function $$C(x)$$ is unbounded, so there is no uniform bound.
One can also try and apply first and second moment estimates on the function $$S(x,N)$$ (using Chebyshev's inequality). This gives the desired sublinear estimate, but only up to some exceptional subset of $$[0,1]$$ which turns out to be too big for my purposes.
There seems to be a large gap between the case $$\alpha=1$$, where everything is known, and smaller values of $$\alpha$$, which I can't explain at the moment.

-- Addition --

This might be a trivial comment, but it should be emphasized that the trivial bound is $$||S(x,N)||_{L^{\infty}([0,1])}\leq C\cdot N.$$ So I'm really looking for any kind of nontrivial bound.

• I think I have an heuristic argument that for $\tau =(\sqrt{5}-1)/2$ the number of solutions scales as $S(\tau,N) \sim N^{1-\alpha}$. My argument makes use of the fact that you can expand a natural number $q$ in a Fibonacci base ($q = \sum c_k f_k$, where $f_k$ are Fibonacci numbers and $c_k \in \{0,1\}$ with $c_k c_{k+1}=0$), you then can get estimates for the difference of $\tau q$ to the next integer rather easily. Unfortunately I don’t have the time to write it up properly. And the argument seems to work only for quadratic irrationalities. $\alpha \to 1$ gives correctly $\sim \log N$. – Andreas Rüdinger Dec 21 '18 at 23:02

## 1 Answer

An elementary way of looking at the problem gives the uniform bound you want - the trick is to look at the actual fractions.

Any two non-equivalent fractions with denominator $$\le N$$ have a difference in absolute value at least $$\frac{1}{N^2}$$. So the number of primitive pairs $$(p,q)$$ where $$N \le q \le 2N$$ satisfying your inequality is:

$$\ll \frac{\frac{1}{N^{1+\alpha}}}{ \frac{1}{N^2}} = N^{1-\alpha}.$$

Putting these dyadic interval bounds together gives

$$S(x, N) \ll N^{1-\alpha}$$

uniformly on $$x$$ (for $$\alpha = 1$$ this argument also recovers your $$\log N$$ bound).