Recall the definition of the floor function: if [tex]n[/tex] is an integer and [tex]n\le x<n+1[/tex], then [tex]\lfloor x\rfloor=n[/tex]. So [tex]\lfloor4x\rfloor=n[/tex] iff [tex]\dfrac n4\le x<\dfrac{n+1}4[/tex].
So let's consider four case.
First, suppose [tex]n\le x<n+\dfrac14[/tex] where [tex]n\in\mathbb Z[/tex]. It follows that
[tex]\lfloor x\rfloor=\left\lfloor x+\dfrac14\right\rfloor=\left\lfloor x+\dfrac12\right\rfloor=\left\lfloor x+\dfrac34\right\rfloor=n[/tex]
because at most, we have
[tex]n\le x<n+\dfrac14\implies n+\dfrac34\le x+\dfrac34<n+1\implies\left\lfloor x+\dfrac34\right\rfloor=n[/tex]
Meanwhile, [tex]\lfloor4x\rfloor=4n[/tex], which follows immediately from
[tex]n\le x<n+\dfrac14\implies 4n\le4x<4n+1[/tex]
and so
[tex]\lfloor4x\rfloor=4n=n+n+n+n=\lfloor x\rfloor+\left\lfloor x+\dfrac14\right\rfloor+\left\lfloor x+\dfrac12\right\rfloor+\left\lfloor x+\dfrac34\right\rfloor[/tex]
Then for the second case, you can consider what happens when you have [tex]n+\dfrac14\le x<n+\dfrac12[/tex]; for the third, [tex]n+\dfrac12\le x<n+\dfrac34[/tex]; and for the fourth, [tex]n+\dfrac34\le x<n+1[/tex].
