Question

Let $[ x ]=$ the greatest integer less than or equal to $x$. If all the values of $x$ such that the product $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ is prime, belongs to the set $\left[x_1, x_2\right) \cup\left[x_3, x_4\right)$, find the value of $x_1^2+x_2^2+x_3^2+x_4^2$.

Answer 1

Product of two integers is prime if one of them is 1.
now $\left[x-\frac{1}{2}\right]\left[x+\frac{1}{2}\right]$ is to be prime
Case-I: Let $\left[x-\frac{1}{2}\right]=1$ and $\left[x+\frac{1}{2}\right]=2$
$\therefore 1 \leq x-\frac{1}{2}<2 \text { and } 2 \leq x+\frac{1}{2}<3 $
$\frac{3}{2} \leq x< \frac{5}{2} \text { and } \frac{3}{2} \leq x<\frac{5}{2} $
Hence $x \in\left[\frac{3}{2}, \frac{5}{2}\right)$....(1)
Case-II: Let $\left[x-\frac{1}{2}\right]=-1$ and $\left[x+\frac{1}{2}\right]=-2 $ (we will find no solution)
Case-III: Let $\left[x+\frac{1}{2}\right]=1$ and $\left[x-\frac{1}{2}\right]=2 $ (we will find no solution)
Case-IV: $$ Let $\left[x+\frac{1}{2}\right]=-1$ and $\left[x-\frac{1}{2}\right]=-2$
$\therefore -1 \leq x+\frac{1}{2}<0 $ and $-2 \leq x-\frac{1}{2}<-1$
$-\frac{3}{2} \leq x<-\frac{1}{2} $ and $-\frac{3}{2} \leq x<-\frac{1}{2}$
Hence $x \in\left[-\frac{3}{2},-\frac{1}{2}\right)$
from (1) and (2) $ x \in\left[-\frac{3}{2},-\frac{1}{2}\right) \cup\left[\frac{3}{2}, \frac{5}{2}\right)$
$\therefore x_1^2+x_2^2+x_3^2+x_4^2=\frac{9}{4}+\frac{1}{4}+\frac{9}{4}+\frac{25}{4}=\frac{44}{4}=11$