SAUDI ARABIAN MATHEMATICAL COMPETITIONS

1) For all $n\in {\mathbb{Z}}^{+}$, we have $n\sqrt{17}\notin \mathbb{Z}$ then $[n\sqrt{17}]<n\sqrt{17}$ or $$[n\sqrt{17}{]}^2<{\left(n\sqrt{17}\right)}^2\forall n.$$ This implies that $$\begin{array}{rl}& 17n^2-[n\sqrt{17}{]}^2\ge 1\\ & \iff 17n^2-(n\sqrt{17}-\{n\sqrt{17}\}{)}^2\ge 1\\ & \iff 17n^2-\left(17n^2-2n\sqrt{17}\{n\sqrt{17}\}+\{n\sqrt{17}{\}}^2\right)\ge 1.\\ & \iff \{n\sqrt{17}\}\ge \frac{1+\{n\sqrt{17}{\}}^2}{2n\sqrt{17}}>\frac{1}{2n\sqrt{17}}\end{array}$$

2) Consider the Pell equation $m^2-17n^2=-1$, since $17$ is the prime of form $4k+1$ then this equation has infinitely many positive integer solutions. Thus $n\sqrt{17}=\sqrt{m^2+1}$ then $[n\sqrt{17}]=m$ for all $m,n\in {\mathbb{Z}}^{+}$. Then we have $\{n\sqrt{17}\}=n\sqrt{17}-m$ which means $$n\sqrt{17}-m>\frac{c}{n}\iff \frac{17n^2-m^2}{n\sqrt{17}+m}>\frac{c}{n}\iff c<\frac{1}{\sqrt{17}+\frac{m}{n}}.$$ Note that $\frac{m}{n}=\sqrt{17-\frac{1}{n^2}}$ then when $m,n\to +\infty$, we have $\frac{m}{n}\to \lim \sqrt{17-\frac{1}{n^2}}=\sqrt{17}$. This implies that $$c\le \frac{1}{\sqrt{17}+{\lim }_{m,n\to +\infty }\frac{m}{n}}=\frac{1}{2\sqrt{17}}$$ Hence, $c=\frac{1}{2\sqrt{17}}$ is the maximum constant that satisfies the given condition.