Prove there are finitely many pairs of integer x, y such that $|x-\sqrt{d}y|<\frac{1}{y^2}$

Question

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?

Answer 1

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! :)

Answer 2

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} $.

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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 $f'(y) > 0$.