China Western Mathematical Olympiad

Suppose that $c^2<a<(c+1{)}^2$, where $c$ is an integer greater than or equal to $1$. Then, $\lfloor \sqrt{a}\rfloor =c$, $1\le a-c^2\le 2c$, and $\{\sqrt{a}\}=\sqrt{a}-c$.

Write $\{\sqrt{a}{\}}^k=(\sqrt{a}-c{)}^k=x_k+y_k\sqrt{a}$, where $k\in \mathbb{N}$ and $x_k,y_k\in \mathbb{Z}$. We have $$S_n=(x_1+x_2+\cdots +x_n)+(y_1+y_2+\cdots +y_n)\sqrt{a}.(1)$$ We now prove that $T_n={\sum }_{k=1}^n{y}_k\ne 0$ for all positive integers $n$.

Since $$\begin{array}{rl}{x}_{k+1}+y_{k+1}\sqrt{a}& =(\sqrt{a}-c{)}^{k+1}\\ & =(\sqrt{a}-c)(x_k+y_k\sqrt{a})\\ & =(a{y}_k-c{x}_k)+(x_k-c{y}_k)\sqrt{a},\end{array}$$ we have $$x_{k+1}=a{y}_k-c{x}_k,y_{k+1}=x_k-c{y}_k.$$ Since $x_1=-c$ and $y_1=1$, we have $y_2=-2c$.

By the above equality, we have $$y_{k+2}=-2c{y}_{k+1}+(a-c^2)y_k,(2)$$ where $y_1=1$, $y_2=-2c$.

By mathematical induction, we have $$y_{2k-1}>0,y_{2k}<0.(3)$$ Combining (2) and (3), we have $$y_{2k+2}-y_{2k+1}=-(2c+1)y_{2k+1}+(a-c^2)y_{2k}<0,$$ $$y_{2k+2}+y_{2k+1}=-(2c-1)y_{2k+1}+(a-c^2)y_{2k}<0.$$ Taking the product of the above inequalities, we get $y_{2k+2}^2-y_{2k+1}^2>0$. Since $y_2^2-y_1^2>0$, we have $$|y_{2k-1}|<|y_{2k}|.(4)$$ On the other hand, we have $$y_{2k+1}-y_{2k}=-(2c+1)y_{2k}+(a-c^2)y_{2k-1}>0,$$ $$y_{2k+1}+y_{2k}=-(2c-1)y_{2k}+(a-c^2)y_{2k-1}>0.$$ Multiplying the above inequalities, we have $y_{2k+1}^2-y_{2k}^2>0$, that is, $|y_{2k}|<|y_{2k+1}|$.

Therefore, $|y_n|<|y_{n+1}|$ for all positive integers $n$. Combining (3) and (4), we have $y_{2k-1}+y_{2k}<0$, $y_{2k+1}+y_{2k}>0$ for all positive integers $n$.

Therefore, $$T_{2n-1}=y_1+(y_2+y_3)+\cdots +(y_{2n-2}+y_{2n-1})>0,$$ $$T_{2n}=(y_1+y_2)+(y_3+y_4)+\cdots +(y_{2n-1}+y_{2n})<0.$$ Hence, $T_n\ne 0$ for all positive integer $n$. This completes the proof.