Prove that if m is an odd integer, then (m4 + 4m2 + 11) / 16 is an integer.

Let $m$ be an odd integer. We can write $m = 2k+1$ for some integer $k$.

First, find $m^2$:

$$m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$

Since $k(k+1)$ is always an even number (product of two consecutive integers), let $k(k+1) = 2j$ for some integer $j$.
So, $m^2 = 4(2j) + 1 = 8j + 1$.

Next, find $m^4$:

$$m^4 = (m^2)^2 = (8j+1)^2 = 64j^2 + 16j + 1 = 16(4j^2 + j) + 1$$

Let $P = 4j^2 + j$, which is an integer. So, $m^4 = 16P + 1$.

Now, substitute $m^4 = 16P+1$ and $m^2 = 8j+1$ into the given expression:

$$\frac{m^4 + 4m^2 + 11}{16} = \frac{(16P+1) + 4(8j+1) + 11}{16}$$ $$= \frac{16P + 1 + 32j + 4 + 11}{16}$$ $$= \frac{16P + 32j + 16}{16}$$

Factor out 16 from the numerator:

$$= \frac{16(P + 2j + 1)}{16}$$ $$= P + 2j + 1$$

Since $P$ and $j$ are integers, their sum $P+2j+1$ is also an integer.
Therefore, the number $\frac{m^4 + 4m^2 + 11}{16}$ is an integer.

$\boxed{\text{The expression simplifies to an integer } P+2j+1.}$