Problem on Step functions

We know that $0 \leq \{x\} < 1$

$\Rightarrow 0 \leq \sqrt{\lfloor x \rfloor} - 1 < 1$

$\Rightarrow 1 \leq \sqrt{\lfloor x \rfloor} < 2 \Rightarrow 1 \leq \lfloor x \rfloor <4 \Rightarrow \lfloor x \rfloor \in \{1,2,3\}.\;\text{Now putting the values of }\lfloor x \rfloor \text{ in the equation }\sqrt{\lfloor x \rfloor} - 1 = \{x\}\text{ we will get the corresponding values of }\{x\} \text{ as }\{x\} = \{0,\sqrt{2}-1,\sqrt{3}-1\}$ .

$\text{Now we know }x = \lfloor x \rfloor + \{x\}\text{ .By putting the corresponding values of }\lfloor x \rfloor \text{ and }\{x\} \text{ we get }$

$x_1 = 1 \hspace{2cm} x_2 = 1 + \sqrt{2} \hspace{2cm} x_3 = 2 + \sqrt{3}$

$\text{Similarly solving for }y\,,$

$0 \leq \sqrt{\lceil y \rceil} - 1 < 1$

$\Rightarrow 1 \leq \sqrt{\lceil y \rceil} < 2 \Rightarrow 1 \leq \lceil y \rceil <4 \Rightarrow \lceil y \rceil \in \{1,2,3\}.\;\text{Now putting the values of }\lceil y \rceil \text{ in the equation }\sqrt{\lceil y \rceil} - 1 = \{y\}\text{ we will get the corresponding values of }\{y\} \text{ as }\{y\} = \{0,\sqrt{2}-1,\sqrt{3}-1\}$ .

$\text{We know }\lfloor y \rfloor = \lceil y \rceil - 1\,,\, \text{when } y \notin \mathbb Z\text{ and }\lfloor y \rfloor = \lceil y \rceil = y \,,\, \text{when }y\in \mathbb Z$

Using these equations, we get the corresponding values of $\lfloor y\rfloor \text{ as }\lfloor y\rfloor = \{1,1,2\}$

$\text{Again using }y = \lfloor y \rfloor + \{y\}\text{ .By putting the corresponding values of }\lfloor y \rfloor \text{ and }\{y\} \text{ we get }$

$y_1 = 1 \hspace{2cm} y_2 = \sqrt{2} \hspace{2cm} y_3 = 1+\sqrt{3}$

Now, $\lfloor x_2 - y_2 \rfloor - \{x_1 - y_1\} + \lceil x_3 - y_3\rceil = \lfloor (1 + \sqrt{2}) - \sqrt{2} \rfloor - \{1 - 1\} + \lceil (2 + \sqrt{3}) - (1+\sqrt{3})\rceil = 1 - 0 + 1 = 2$ .

$\Rightarrow a = 2$

$\text{Finding all the possible values of }2y_{b_3} - x_{b_4} \text{ we get }$

$2y_1 - x_1 = 2(1) - 1 = 1 \hspace{4cm} 2y_1 - x_2 = 2(1) - (1+\sqrt{2}) = 1 - \sqrt{2} \hspace{4cm} 2y_1 - x_3 = 2(1) - (2 + \sqrt{3}) = -\sqrt{3}$

$2y_2 - x_1 = 2\sqrt{2} - 1 \hspace{4cm} 2y_2 - x_2 = 2\sqrt{2} - (1 + \sqrt{2}) = \sqrt{2} - 1 \hspace{4cm} 2y_2 - x_3 = 2\sqrt{2} - 2-\sqrt{3}$

$2y_3-x_1=2(1+\sqrt{3})-1=1+2\sqrt{3}\hspace{3cm}2y_3-x_2=2(1+\sqrt{3})-(1+\sqrt{2})=1+2\sqrt{3}-\sqrt{2}\hspace{3cm}2y_3-x_3=2(1+\sqrt{3})-(2+\sqrt{3})=\sqrt{3}$

$\text{Now finding all the possible values of }x_1^2x_{b_1} - y_1y_{b_2}^2 \text{ we get }$

$\because x_1 = y_1 = 1\text{ our expression becomes }x_{b_1} - y_{b_2}^2$

$x_1-y_1^2=1-1^2=0\hspace{5cm}x_1-y_2^2=1-(\sqrt{2})^2=-1\hspace{5cm}x_1-y_3^2=1-(1+\sqrt{3})^2=-3-2\sqrt{3}$

$x_2-y_1^2=(1+\sqrt{2})-1^2=-\sqrt{2}\hspace{2.2cm}x_2-y_2^2=(1+\sqrt{2})-(\sqrt{2})^2=\sqrt{2}-1\hspace{2.2cm}x_2-y_3^2=(1+\sqrt{2})-(1+\sqrt{3})^2=\sqrt{2}-3-2\sqrt{3}$

$x_3-y_1^2=(2+\sqrt{3})-1=1+\sqrt{3}\hspace{3cm}x_3-y_2^2=(2+\sqrt{3})-(\sqrt{2})^2=\sqrt{3}\hspace{3cm}x_3-y_3^2=(2+\sqrt{3})-(1+\sqrt{3})^2=-2-\sqrt{3}$

$\text{Matching all the expressions we get }2\text{ possible values of }\{b_1,b_2,b_3,b_4\}$

$\{b_1,b_2,b_3,b_4\}_1=\{2,2,2,2\} \text{ and }\{b_1,b_2,b_3,b_4\}_2=\{3,2,3,3\}\text{ but according to question all the }b_i \text{ are not equal. So first case is not considered. }$

$\text{So, }b_1=3\;,\;b_2=2\;,\;b_3=3\;,\;b_4=3$

$\text{Hence our answer is }2+3+2+3+3 = \boxed{13}$