The solution set of the equation (x + 2)^2 + [x-2]^2 = (x-2)^2 + [x+2…

We have,

$[x+n]=[x]+n$ where $n∈ Z$ and $x ∈ R$.

$∴(x+2)^2+[x-2]^2=(x-2)^2+(x+2]^2$

$⇒(x + 2)^2 - (x-2)^2 = [x+2]^2 -[x-2]^2$

$⇒(x + 2)^2 - \{(x+2)-4\}^2 =[x + 2]^2 -[(x+2) - 4]^2$

$⇒ (x+2)^2 - \{(x+2)-4\}^2=[x+2]^2-\{[x+2]-4\}^2$

$⇒8(x+2)=8 [x + 2]$

$⇒(x+2)-[x + 2]=0$

$⇒\{x + 2\} = 0$, where {x} denotes the fractional part of x

$⇒x+2∈Z ⇒ x∈Z$

Hence, the solution set of the given equation is the set of integers.