Prove that for any odd integer m, m4 + 4m2 + 11 is divisible by 16.

✅ Correct.

Your solution correctly uses modular arithmetic to show that the numerator is a multiple of 16.

Here's a breakdown of your steps:

1. You correctly established that if $m$ is an odd integer, then $m^2 \equiv 1 \pmod{8}$. This is because $m^2 = (2k+1)^2 = 4k(k+1)+1$, and $k(k+1)$ is always even, so $4k(k+1)$ is a multiple of 8.
2. You correctly deduced that $m^4 = (m^2)^2 \equiv 1^2 \equiv 1 \pmod{16}$. This follows from $m^2 \equiv 1 \pmod{8}$, which implies $m^2 = 8j+1$. Then $m^4 = (8j+1)^2 = 64j^2 + 16j + 1 = 16(4j^2+j)+1 \equiv 1 \pmod{16}$.
3. For the term $4m^2$: Since $m^2 \equiv 1 \pmod{8}$, $m^2$ can be $1, 9, 17, \dots$. If $m^2 \equiv 1 \pmod{16}$, then $4m^2 \equiv 4(1) \equiv 4 \pmod{16}$. If $m^2 \equiv 9 \pmod{16}$, then $4m^2 \equiv 4(9) = 36 \equiv 4 \pmod{16}$. In both cases, $4m^2 \equiv 4 \pmod{16}$. Your use of $4(1)$ in the calculation $1+4(1)+11$ is effectively correct for the modulo 16 result.
4. You correctly summed the terms modulo 16: $$m^4 + 4m^2 + 11 \equiv 1 + 4 + 11 \pmod{16}$$ $$m^4 + 4m^2 + 11 \equiv 16 \pmod{16}$$ $$m^4 + 4m^2 + 11 \equiv 0 \pmod{16}$$
5. Your conclusion that $\frac{m^4 + 4m^2 + 11}{16}$ is an integer is correct because the numerator is a multiple of 16.