Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$? Given an odd integer $a$ , establish that $a^2+(a+2)^2+(a+4)^2+1$ is divisible by $12$?
So far I have:
$a^2+(a+2)^2+(a+4)^2+1$ 
$=a^2+a^2+4a+4+a^2+8a+16+1 $
$=3a^2+12a+21$ 
$=3(a^2+4a+7) $
where do I go from here.. the solution I have is divisible by $3$ not $12$...

Hint: 
If $a$ is odd then by the Division algorithm $a = 2k + 1$ for some integer $k$. 
