# Is there an integer points sequence tending to $x+\sqrt 2y=\frac{1}{2}$?

I find an claim says that $sup\{x+\sqrt 2y|x,y\in\mathbb{Z},x+\sqrt 2 y\le\frac{1}{2}\}=\frac{1}{2}$.
I want to show for infinitely many large enough $N$, the equation $\frac{1}{2}-\sqrt 2y-x=\frac{1}{N}$ have integer solution $x,y$. But I'm confusing about how to show that.

The bit about $1/N$ is impossible, it is rational.
There are infinitely many solutions in positive integers to $$u^2 - 8 v^2 = -7.$$ Begin with $(1,1).$ We get an infinite sequence of solutions by repeating $$(u,v) \mapsto (3 u + 8 v, u + 3 v).$$ You ought to check that for yourself. Note that $u$ is always odd.
By taking $2x-1 = u, y = - v,$ we get an infinite sequence of solutions to $$(2x-1)^2 - 8 y^2 = -7,$$ with $x > 0$ but $y < 0.$ Well, $$2x-1 - y \sqrt 8$$ becomes an arbitrarily large positive real number. Then $$2x-1 + y \sqrt 8 = \frac{-7}{ 2x-1 - y \sqrt 8}$$ becomes a negative real of arbitrarily small absolute value. So $$x- \frac{1}{2} + y \sqrt 2 = \frac{-7}{ 2 \left( 2x-1 - y \sqrt 8 \right)}$$ is also a negative real of arbitrarily small absolute value. You could write $$x- \frac{1}{2} + y \sqrt 2 = -\delta$$ and $$x + y \sqrt 2 = \frac{1}{2} -\delta$$
There is a sequence of rational points $p_n/q_n$ with $p_n,q_n\in\mathbb{Z}$, $\gcd(p_n,q_n)=1$ such that $\lim_{n\to\infty}p_n/q_n=\sqrt2$. We may assume without loss of generality that $p_n$ is odd and $q_n$ is even. Then $$\frac{1-p_n}{2}+\frac{q_n}{2}\,\frac{p_n}{q_n}=\frac12,\quad \frac{1-p_n}{2},\frac{p_n}{q_n}\in\mathbb{Z}.$$
• I wonder do we still need to show $\frac{q_n}{2}(\sqrt 2-\frac{p_n}{q_n})$ tends to 0? Since $q_n$ may be very large. Commented Nov 28, 2015 at 21:51
• Yes, that inequality is needed. It follows from the fact that $\sqrt2$ is an algebraic number of degree $2$ and hence of irrationality measure $2$. This implies that $|\sqrt2-p/q|\le q^{-\mu}$ for $1<\mu<2$ and all $q$ large enough. Commented Nov 28, 2015 at 23:15