4. Given a positive integer $m$ such that $m^{4}+16 m+8$ can be expressed as the product of two or more consecutive integers. Find the maximum value of $m$.
(Proposed by Indonesia)
4. If $m \equiv 0(\bmod 3)$, then
$$
m^{4}+16 m+8 \equiv 2(\bmod 3) \text {; }
$$

If $m \equiv 1(\bmod 3)$, then
$$
m^{4}+16 m+8 \equiv 1(\bmod 3) \text {; }
$$

If $m \equiv 2(\bmod 3)$, then
$$
m^{4}+16 m+8 \equiv 2(\bmod 3) \text {. }
$$

Thus, regardless of the value of $m$, $m^{4}+16 m+8$ is never a multiple of 3, and therefore, it cannot be expressed as the product of three or more consecutive integers.
Considering the factorization of the form
$$
m^{4}+16 m+8=\left(m^{2}+n\right)\left(m^{2}+n+1\right)(n \in \mathbf{N})
$$

we have
$$
(2 n+1) m^{2}-16 m+\left(n^{2}+n-8\right)=0,
$$

with the discriminant
$$
\Delta=4\left(64-(2 n+1)\left(n^{2}+n-8\right)\right) \text {. }
$$

When $n=0,1,2,3$, we have
$$
\begin{array}{l}
\Delta_{0}=4 \times 72, \Delta_{1}=4 \times 82, \\
\Delta_{2}=4 \times 74, \Delta_{3}=4 \times 36 ;
\end{array}
$$

When $n \geqslant 4$, $\Delta_{4}<0$.
Thus, only when $n=3$, $m$ takes integer values, at which point,
$$
m=\frac{16+\sqrt{4 \times 36}}{2 \times(2 \times 3+1)}=2 \text {. }
$$