For each real number $ x$< let $ \lfloor x \rfloor$ be the integer satisfying $ \lfloor x \rfloor \le x < \lfloor x \rfloor +1$ and let $ \{x\}=x-\lfloor x \rfloor$. Let $ c$ be a real number such that \[ \{n\sqrt{3}\}>\dfrac{c}{n\sqrt{3}}\] for all positive integers $ n$. Prove that $ c \le 1$.
Raja Oktovin wrote:
For each real number $ x$< let $ \lfloor x \rfloor$ be the integer satisfying $ \lfloor x \rfloor \le x < \lfloor x \rfloor + 1$ and let $ \{x\} = x - \lfloor x \rfloor$. Let $ c$ be a real number such that
\[ \{n\sqrt {3}\} > \dfrac{c}{n\sqrt {3}}\]
for all positive integers $ n$. Prove that $ c \le 1$.
Consider the pell equation $ 3x^2 - y^2 = 2$ which have infinitely many solutions with $ y$ as great as we want (start with $ (1,1)$ and use the transformation $ (x,y)\to(2x + y,3x + 2y)$)
Let then $ (x,y)$ a solution : $ 3x^2 - y^2 = 2$ $ \implies$ $ x\sqrt 3 = \sqrt {y^2 + 2}$ and so $ [x\sqrt 3] = y$ and $ \{x\sqrt 3\} = \sqrt {y^2 + 2} - y$
So $ \{x\sqrt 3\}x\sqrt 3 = y^2 + 2 - y\sqrt {y^2 + 2}$ $ = 1 + \frac {1}{y^2 + 2 + y\sqrt {y^2 + 2}}$
And so $ 1 + \frac {1}{y^2 + 2 + y\sqrt {y^2 + 2}} > c$
And since we have solutions in which $ y$ may be as great as we want, we get $ 1\ge c$
Q.E.D.
I was using pell equations too but the wrong ones and they weren't giving me anything. using the one you used should have been so obvious . thanks, pco