8. Proof: Let $k^{2} \leqslant n2 k$, therefore
$$2 k \leqslant[\sqrt{4 n+2}] \leqslant 2 k+1$$

If $\sqrt{4 n+2} \geqslant 2 k+1$, then
$$\sqrt{4 k^{2}+4 l+2} \geqslant \sqrt{4 k^{2}+4 k+1}$$

i.e. $\square$
$$4 l+2 \geqslant 4 k+1, \quad 4 l \geqslant 4 k-1$$

Since $l$ is an integer, therefore $l \geqslant k$.
So
$$[\sqrt{4 n+2}]=\left\{\begin{array}{cl} 2 k+1, & l \geqslant k \\ 2 k, & l2 k^{2}+2 k+1+2\left(k^{2}+k\right) \\ =(2 k+1)^{2} \text {. } \end{array}\right.$$