Question

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

Answer 1

$[x+\frac{1}{2}][x-\frac{1}{2}]=prime$

Hence possible solutions can be $2\times 1=2$

and $-2\times -1=2$

First case-

$[x+\frac{1}{2}][x-\frac{1}{2}]=2$

where $[x+\frac{1}{2}]=2$ and $[x-\frac{1}{2}]=1$

$⟹x+\frac{1}{2}\in [2,3)$ and $x-\frac{1}{2}\in [1,2)$

$⟹x\in [\frac{3}{2},\frac{5}{2})$

Second case-

$[x+\frac{1}{2}][x-\frac{1}{2}]=2$

where $[x+\frac{1}{2}]=-1$ and $[x-\frac{1}{2}]=-2$

$⟹x+\frac{1}{2}\in [-1,0)$ and $x-\frac{1}{2}\in [-2,-1)$

$⟹x\in [\frac{-3}{2},\frac{-1}{2})$

Hence solution set is $x\in [\frac{3}{2},\frac{5}{2})\cup [\frac{-3}{2},\frac{-1}{2})$

$⟹x_1^2+x_2^2+x_3^2+x_4^2=\frac{3^2+5^2+3^2+1^2}{4}$

$=\frac{44}{4}=11$