The solution set of the equation `(x+1)^(2)+[x-1]^(2)=(x-1)^(2)+[x+1]^(2),` where [x] and (x) are the greatest integer and y are prime numbers and `x^(2)-2y^(2)=1,` is

Options

A. N
B. Z
C. Q
D. R

Answer

B

Explanation

We have,
`[x+n]=[x]+n` where `n in Z` and `x in R`.
`:. (x+2)^(2)+[x-2]^(2)=(x-2)^(2)+[x+2]^(2)`
`implies (x+2)^(2)-(x-2)^(2)=[x+2]^(2)-[x-2]^(2)`
`implies(x+2)^(2)-{(x+2)-4}^(2)=[x+2]^(2)-[(x+2)-4]^(2)`
`implies (x+2)^(2)-{(x+2)-4}^(2)=[x+2]^(2)-{[x+2]-4}^(2)`
`implies 8(x+2)=8[x+2]`
`implies(x+2)-[x+2]=0`
`implies {x+2}=0` , where{x} denotes the fractional part of x
`implies x+2 in Z implies x in Z`
Hence, the solution set of the given equation is the set ofintegers.