, 2008

Solution:
We will prove that $n^{4}+8 n+11$ is never a multiple of $3$. This is clear if $n$ is a multiple of $3$. If $n$ is not a multiple of $3$, then
$$n^{4}+8 n+11 = (n^{4}-1) + 12 + 8 n = (n-1)(n+1)(n^{2}+1) + 12 + 8 n,$$
where $8 n$ is the only term not divisible by $3$. Thus $n^{4}+8 n+11$ is never the product of three or more consecutive integers.

It remains to discuss the case when $n^{4}+8 n+11 = y(y+1)$ for some integer $y$. We write this as
$$4(n^{4}+8 n+11) = 4 y(y+1)$$
or
$$4 n^{4} + 32 n + 45 = (2 y + 1)^{2}.$$
A check shows that among $n = \pm 1$ and $n = 0$ only $n = 1$ satisfies the requirement, as $1^{4} + 8 \cdot 1 + 11 = 20 = 4 \cdot 5$.

Now let $|n| \geq 2$. The identities
$$4 n^{4} + 32 n + 45 = (2 n^{2} - 2)^{2} + 8(n+2)^{2} + 9$$
and
$$4 n^{4} + 32 n + 45 = (2 n^{2} + 8)^{2} - 32 n(n-1) - 19$$
indicate that for $|n| \geq 2$, $2 n^{2} - 2 < 2 y + 1 < 2 n^{2} + 8$. But $2 y + 1$ is odd, so it can equal $2 n^{2} \pm 1$, $2 n^{2} + 3$, $2 n^{2} + 5$, or $2 n^{2} + 7$. We investigate them one by one.

If $4 n^{4} + 32 n + 45 = (2 n^{2} - 1)^{2}$:
$$n^{2} + 8 n + 11 = 0 \implies (n+4)^{2} = 5,$$
which is impossible, as $5$ is not a perfect square.

If $4 n^{4} + 32 n + 45 = (2 n^{2} + 1)^{2}$:
$$n^{2} - 8 n - 11 = 0 \implies (n-4)^{2} = 27,$$
which also fails.

Also, $4 n^{4} + 32 n + 45 = (2 n^{2} + 3)^{2}$:
$$3 n^{2} - 8 n - 9 = 0 \implies 9 n^{2} - 24 n - 27 = 0 \implies (3 n - 4)^{2} = 43,$$
fails.

If $4 n^{4} + 32 n + 45 = (2 n^{2} + 5)^{2}$:
$$5 n^{2} - 8 n = 5 \implies 25 n^{2} - 40 n = 25 \implies (5 n - 4)^{2} = 41,$$
which also fails.

Finally, if $4 n^{4} + 32 n + 45 = (2 n^{2} + 7)^{2}$, then
$$28 n^{2} - 32 n + 4 = 0 \implies 4(n-1)(7 n-1) = 0,$$
whence $n = 1$ that we already found.

Thus the only solution is $n = 1$.