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 $12$...

If $a$ is odd, then $a = 2b+1$ for some integer $b$.
Then $a^2 + 4a + 7 = 4b^2 + 4b + 1 + 8b + 4 + 7 = 4(b^2 + 3b  + 3)$, which is evenly divisible by $4$.
Combine this with the divisibility by $3$ that you already have, and you're done.
