8. Let $m, n$ be integers, $53<m, n<100$, and let $x=(m+\sqrt{m}+1)^{n}, y=(m-\sqrt{m}+1)^{n}$.
If the integer part of $y$ is 2013, and the remainder when 2013 is divided by $m$ is 53, then the remainder when the integer part of $x$ is divided by $m$ is $\qquad$
8. $m-52$.

Let $x=(m+1+\sqrt{m})^{n}$
$$
=A+B \sqrt{m}\left(A, B \in \mathbf{N}_{+}\right) \text {. }
$$

Then $y=(m+1-\sqrt{m})^{n}=A-B \sqrt{m}$.
If $m$ is a perfect square, let $m=a^{2}$, we get
$$
\begin{array}{l}
y=(m+1-\sqrt{m})^{n}=\left(a^{2}+1-a\right)^{n} \\
=2013=3 \times 11 \times 61 .
\end{array}
$$

Thus, $n=1$, which is a contradiction.
Therefore, $m$ is not a perfect square.
Hence, $y$ is not an integer, and $2013<y<2014$.
(1) + (2) gives
$$
\begin{array}{l}
x+y=2 A \\
\Rightarrow x=2 A-y=(2 A-2014)+(2014-y) .
\end{array}
$$

Then $[x]=2 A-2014=x+y-2014$
$$
\begin{array}{l}
=(m+1+\sqrt{m})^{n}+(m+1-\sqrt{m})^{n}-2014 \\
=2 \sum_{k=0}^{\left[\frac{n}{2}\right]} \mathrm{C}_{n}^{2 k}(m+1)^{n-2 k}(\sqrt{m})^{2 k}-2014 \\
\equiv 2 \mathrm{C}_{n}^{0}(m+1)^{n}-2014 \equiv 2-53-1 \\
\equiv-52 \equiv m-52(\bmod m)
\end{array}
$$