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

Answer: 11

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 $\left[x+\frac{1}{2}\right]=2$
$\therefore 1\le x-\frac{1}{2}<2\text{ and }2\le x+\frac{1}{2}<3$
$\frac{3}{2}\le x<\frac{5}{2}\text{ and }\frac{3}{2}\le x<\frac{5}{2}$
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 $\left[x-\frac{1}{2}\right]=-2$
$\therefore -1\le x+\frac{1}{2}<0$ and $-2\le x-\frac{1}{2}<-1$
$-\frac{3}{2}\le x<-\frac{1}{2}$ and $-\frac{3}{2}\le x<-\frac{1}{2}$
Hence $x\in \left[-\frac{3}{2},-\frac{1}{2}\right)$
from (1) and (2) $x\in \left[-\frac{3}{2},-\frac{1}{2}\right)\cup \left[\frac{3}{2},\frac{5}{2}\right)$
$\therefore x_1^2+x_2^2+x_3^2+x_4^2=\frac{9}{4}+\frac{1}{4}+\frac{9}{4}+\frac{25}{4}=\frac{44}{4}=11$