# Solution

Show that if $a$ is odd, then 12 divides $a^2 + (a+2)^2 + (a+4)^2 + 1$.

We argue directly. If $a$ is odd, then we can find an integer $k$ such that $a = 2k+1$. Consequently,

$$\begin{array}{rcl} a^2 + (a+2)^2 + (a+4)^2 + 1 &=& (2k+1)^2 + (2k+3)^2 + (2k+5)^2 + 1\\\\ &=& (4k^2 + 4k + 1) + (4k^2 + 12k + 9) + (4k^2 + 20k + 25) + 1\\\\ &=& 12k^2 + 36k + 36\\\\ &=& 12(k^2 + 3k + 3) \end{array}$$

Observing that $k^2 + 3k + 3$ must be an integer by closure, we have $12 \mid a^2 + (a+2)^2 + (a+4)^2 + 1$.

QED.