Prove there are finitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y^2}$ Prove there are finitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y^2}$ where $d$ is a non-square natural number. I know there're infinitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y}$ according to Dirichlet's approximation theorem. Any hints?
 A: Original Answer
Roth's theorem states that if $a$ is an irrational algebraic number, then for every $\epsilon > 0$, the inequality
$$\left\lvert\alpha - \frac{p}{q}\right\rvert < \frac{1}{q^{2 + \epsilon}}$$
can have only finitely many solutions in co-prime integers $p$ and $q$.
Divide both sides of the inequality in your question to get
$$\left\lvert\sqrt{d} - \frac{x}{y}\right\rvert < \frac{1}{y^3}$$
If we take $\alpha = \sqrt{d}$, then $\alpha$ is an irrational algebraic number because $d$ is not a perfect square and $\sqrt{d}$ is the root of the polynomial $P(x) = x^2 - d$ of degree $2$.
Then, we take $p = x$, $q = y$ and $\epsilon = 1$ in Roth's theorem to get the desired result.
Update
Also, since $\alpha$ is algebraic of degree $2$ (the polynomial $P(x)$ has degree $2$), then by definition of irrationality measure, the irrationality measure of $\alpha$ is $\mu(x) = 2$. Hence, we can go by this definition as well to show that the inequality has at most finite solutions $\frac{p}{q}$ for integers $p$ and $q$. This is explored as suggested by @AlexeyBurdin in his comment, and also to address @EmmaJohnson's follow-up comment to provide a proof without invoking Roth's theorem.
Update
As @Jyrki Lahtonen has mentioned, the requirement that $x$ and $y$ must be co-prime does not necessarily imply that there are infinitely many pairs $(mx, my)$ that satisfy the inequality. This is because the scaling factor $m$ also affects the bound $\frac{1}{y^2}$. Hence, there are indeed finite pairs $(x, y)$ that satisfy the inequality. Thanks @Jyrki Lahtonen! :)
A: I will show that
there are only a finite number of solutions
if
$|x-y\sqrt{d}|
\lt \dfrac1{y\,f(y)}
$
where
$f(y) \to \infty$,
$f(1) > 0$,
and $f'(y) > 0
$.
If
$f(y) = y$
(this question)
then
$y
\lt \sqrt{d}+\sqrt{d+1}
$.
If
$f(y) = y^c$
with $c > 0$
then
$y
\le (4\sqrt{d})^{1/c}
$.
If
$f(y) = \ln(y)$
then
$y
\le e^{4\sqrt{d}}
$.
If $d$ is not a square then
$|x^2-dy^2|
\ge 1$
so
$1
\le |x^2-dy^2|
=|(x-y\sqrt{d})(x+y\sqrt{d})|
$
so
$|x-y\sqrt{d}|
\ge|\dfrac1{x+y\sqrt{d}}|
$.
If
$|x-y\sqrt{d}|
\lt \dfrac1{y^2}
$
then
$-\dfrac1{y^2}
\lt x-y\sqrt{d}
\lt \dfrac1{y^2}
$
and
$\dfrac1{y^2}
\ge \dfrac1{x+y\sqrt{d}}
$
or
$y^2
\lt x+y\sqrt{d}
\lt (y\sqrt{d}+\dfrac1{y^2})+y\sqrt{d}
\lt 2y\sqrt{d}+1
$
so
$y^2-2y\sqrt{d}
\lt 1
$
so that
$(y-\sqrt{d})^2
=y^2-2y\sqrt{d}+d
\lt d+1
$
so
$y
\lt \sqrt{d}+\sqrt{d+1}
$.
More generally,
if
$|x-y\sqrt{d}|
\lt \dfrac1{y^{1+c}}
$
where $c > 0$
then
$-\dfrac1{y^{1+c}}
\lt x-y\sqrt{d}
\lt \dfrac1{y^{1+c}}
$
and
$\dfrac1{y^{1+c}}
\ge \dfrac1{x+y\sqrt{d}}
$
or
$y^{1+c}
\lt x+y\sqrt{d}
\lt (y\sqrt{d}+\dfrac1{y^{1+c}})+y\sqrt{d}
\lt 2y\sqrt{d}+1
$
so
$1
\gt y^{1+c}-2y\sqrt{d}
= y^{1+c}(1-\dfrac{2\sqrt{d}}{y^c})
$
so that,
if
$y^c
\gt 4\sqrt{d}
$
then
$1
\gt \dfrac{y^{1+c}}{2}
$
which is false.
Therefore
$y \le (4\sqrt{d})^{1/c}
$.

Even more,
if
$|x-y\sqrt{d}|
\lt \dfrac1{y\ln(y)}
$
then
$-\dfrac1{y\ln(y)}
\lt x-y\sqrt{d}
\lt \dfrac1{y\ln(y)}
$
and
$\dfrac1{y\ln(y)}
\ge \dfrac1{x+y\sqrt{d}}
$
or
$y\ln(y)
\lt x+y\sqrt{d}
\lt (y\sqrt{d}+\dfrac1{y\ln(y)})+y\sqrt{d}
\lt 2y\sqrt{d}+1
$
so
$1
\gt y\ln(y)-2y\sqrt{d}
= y\ln(y)(1-\dfrac{2\sqrt{d}}{\ln(y)})
$
so that,
if
$\ln(y)
\gt 4\sqrt{d}
$
or
$y
\gt e^{4\sqrt{d}}
$
then
$1
\gt \dfrac{y\ln(y)}{2}
$
which is false.
Therefore
$y \le  e^{4\sqrt{d}}
$.
This works for
$|x-y\sqrt{d}|
\lt \dfrac1{y\,f(y)}
$
where
$f(y) \to \infty$,
$f^{(-1)}(y) \to \infty$,
$f(1) > 0$,
and $f'(y) > 0$.
